UNIT 4 BIOLOGY NOTESpage CHAPTER 7 – GREEN SHEETS1 History of

Heredity Definitions of Chromosomes

Mendel – HeredityWatson and Crick, Franklin and Chargraff – Structure of DNASomatic cells, Gamete, Genes, Locus, Alleles, Diploid,Haploid, Chromatin, Chromatids, Histones, Centromere, Homologous, Autosomes, Sex chromosomes(X,Y) Karyotype,

2 Structure in Eukaryotes

Picture of chromosome pairs normally and when dividing, with parts labelled.

* Karyotype Cut out the chromosomes (doubled) and finding their homologous pair.3 Structure in

ProkaryotesCell division Overview

Structure of chromosomes and genetic material in prokaryotes (bacteria)Asexual Versus Sexual reproductionOverview of Mitosis and meiosis.

4 Monohybrid crosses

Mendels study of pea plants and words used -Purebreeding, Parent, F1,F2 Dominant, Recessive, Genotype, Phenotype,Homozygous, Heterozygous

5 Monohybrid cross, Punnet square, Probabilities

An example of a monohybrid cross used by Mendel to show that characteristics where inherited from each parent and weren’t a blend of the parents (codominance not discovered then) but were discrete as dominant and recessive. The F1 generation carried the hidded recessive which revealed itself again in F2

6 PhenotypeTEST CROSS

CODOMINANCE

Phenotype = Genotype + EnvironmentTest Cross – To find out if an organism displaying a DOMINANT phenotype is HOMOZYGOUS or HETEROZYGOUSOr Incomplete dominance result in a THIRD phenotype for the Heterozygote.

7 MULTIPLE ALLELESLETHAL ALLELESPOLYGENES

DIHYBRID CROSS

When there are more than two alleles for a gene eg blood type has 3 alleles A,B and OThis is when a particular combination of alleles is lethal (kills) the organisms so the ratio of offspring will be different to the expected because some die.This is when there are many genes for one characteristic which shows a CONTINUOUS change eg HEIGHT in humans and skin COLOUR.This the study of two characteristics at a time. In the text book it was height and flower colour of pea plants. You need a large punnet square.

8 INHERITANCE PATTERNSPEDIGREES

If the characteristic is studied on the AUTOSOMAL chromosomes of the X –LINKED (Sex linked)This studies inheritance over generations to establish if a characteristic is Autosomal, X-linked , Dominant or Recessive.

9 PEDIGREE AUTOSOMAL

Working through two Autosomal conditions to find the genotype of the people in the various generations. Recessive can skip generations.

10 PEDIGREE X-LINKED

Working through two X-linked conditions to find the genotype of the people in the various generations. Recessive Mother-son, Dominant Father-Daughter.

11 Practice working through pedigrees12 RECIPROCAL

CROSSGenotype symbols

How to see if a characteristic is carried on the Autosomal chromosomes or the X chromosomes.

REPRODUCTION.

DNA REPLICATIONSTAGES

MITOSIS (Somatic-Normal 2 Diploid Daughter cells) MEIOSIS (Gametes-Eggs/Sperm 4 Haploid Daughter cells)Semi-conservative Replication, DNA helicase, DNA polymerase,

I.P.M.A.T Interphase, Prophase, Metaphase, Anaphase, Telophase. CytokinesisMITOSIS 1 Division- Chromosomes condense Chromotids, Chromatin, Centrioles/SpindleMEIOSIS 2 Divisions- IPMAT 1 and 2. Homologous pairs cross over at Prophase,

Homologous pairs line up along the equator at Metaphase 1 MENDELS LAWS Law of Segregation, Independent Assortment, Linked genes ,

Recombination/CrossoverLINKED GENES Dihybrid cross results if genes on separate chromosomes and if linked.

CHAPTER 8 DNA YELLOW SHEETS1 TRANSCRIPTION

TRANSLATIONText book questions. ***Must be done. Transcription(premRNA) -Splicing (introns,exons)- Translation (polypeptide)

2 PROTEIN SYNTHESIS BIOZONE

A summary of the processes and structures used in Transcription (RNA polymerase, Codons, mRNA, Splicing Exon/Intron,Translation (Ribosome, tRNA, Anticodon, Amino Acid,

3 GENE REGULATION

Lactose metabolism in bacteriaSwitches on/off, Regulatory Gene, Regulatory protein, Promoter, Operator,

4 MUTATIONS Somatic/Germline

GENE –(Point)Substitution, Inversion, Insertion/Deletion (Frameshift)CHROMOSOME –Deletions, Inversions, TRANSLOCATION, Duplications

ANEUPLOIDY

CAUSES

(NON-DISJUNCTION) -Chr 21 Downs Syndrome Klinefelters XXY, Turners XO, Polyploidy(Plants) Mutation rate, Mutagens, Xrays, U.V light,

CHAPTER 9 BIOTECHNOLOGY BLUE SHEETS1 Prokaryotes (Plasmids)/ Eukaryotes, Overview of Tools and Techniques2 RESTRICTION ENZYMES

GEL ELECTROPHORESISPCR

Cutting DNA at specific recognitions sites Separating DNA pieces based on their size and charge(Polymerase chain reaction)-making multiple copies of DNA pieces.

3 DNA ligaseRECOMBINANT DNATRANSFORMATION

Glue. Rejoining DNA) Foreign DNA inserted Eg A gene inserted into a plasmid Foreign DNA inserted into a bacteria.

4 DNA SEQUENCINGDNA PROFILING

Finding the order of the nucleotides (A,T,C,G)using STR’s Short tandem Repeats eg ATCCG ATCCG ATCCG

5 GENE PROBES – searching DNA for a complementary piece6 GENETIC TESTING Restriction Fragment Length Polymorphism (RFLP)

Restriction sites lost/gained so can use it to identify organisms.7 TRANSGENIC ORGANISM

CLONINGSTEM CELLS

Transformations with EukaryotesEmbryo splitting, Nuclear transferUndifferentiated cells which can turn into all cell types

CHAPTER 10 EVIDENCE OF CHANGE PINK SHEETS1 FOSSILS and DATING

COMPARITIVE ANATOMY

BIOGEOGRAPHY

MOLECULAR COMPARISON

Divergent evolution (Homologous features)Convergent evolution (Analogous features)Vestigial Features, EmbryologyClosely related species live close together. If separated it may have been due to ancestors on Gondwana or Pangea before they split.Nuclear DNA (Sequencing and Hybridisation)Mitochondrial DNA, Chloroplasts and Proteins.

2 FOSSILS AND DATING FOSSIL FORMATIONDATING – Comparitive techniques, Absolute Techniques- Radio-isotopes C14 , K40

Video - EVOLUTION A worksheet summary of evolution.

CHAPTER 11 CHANGING POPULATIONS PINK SHEETS1 DARWINS THEORY NATURAL SELECTION

Adaptation, Variation, Selective pressures,Evolution,Gene pool-Mutation, Natural selection, Gene flow, Genetic drift

2 ALLELE FREQUENCY Change- Species, Allopatric SpeciationFactors Preventing Reproduction –Pre and Post zygotic

Wolves to Dogs Game-Selecting for a lack of aggression causes allele change1 HUMAN EVOLUTION Homologous features Primates, monkeys apes, Chromosome 2 HOMININ PHYLOGENY Hominins, BIPEDALISM, LARGER BRAIN3 HOMININS Spread from Africa, Nomadic-settled, Cultural change HUMAN INTERVENTION IN EVOLUTIONARY PROCESS1 SELECTIVE BREEDING Issues on the use of techniques which could alter human 2 TRANSGENIC ORGANISMS Evolution. 3 4CLONING,SCREENING STEM CELLS, GENE THERAPY

CH. MATERIAL EXPERT Done 1 7 STRUCTURE OF CHROMOSOMES

Prokaryotes, Eukaryotes, Genes, Locus, alleles, Genotype/Phenotype, Phenotype= genotype + environmentHomozygous/Heterozygous, homologous, Autosomes, Sex chromosomes, KARYOTYPE, Chromatids, Polyploid, Apoptosis.

Matthew

2 7 CELL DIVISION DNA. Replication (Semi-conservative, DNA helicase, DNA polymerase, DNA ligase, MEIOSIS -I.P.M.A.T x2 (Haploid) ,MITOSIS (Diploid)Crossover,

Dino

3 7 CROSSESCodominant, Homozygous/Heterozygous, Genotype/Phenotype)Test Cross for genotype(or Back cross), Dihybrid cross, Linked genes Multiple alleles (Blood type) , Lethal alleles,Polygenes (continuous variation – Height ,colour)

Nadia

4 7 Monohybrid cross Mendels crosses.(Dominant, Recessive)X linked cross (Reciprocal) Dominant/recessive, Autosomal/Sex linked

Casey

5 8 PEDIGREES Dominant/recessive, Autosomal/Sex linked Kandace6 8 PROTEIN SYNTHESIS

TRANSCRIPTION- Nucleus,RNA polymerase,Codons, mRNA Exon/IntronTRANSLATION – Ribosomes(rRNA and Proteins),tRNA, Anticodons,Reverse Transcriptase – RNA to DNA

Kayla

7 8 DNA Structure and Discovery Watson/Crick, Rosalind Franklin, Chargraff’s bases, Chromatin, Histones,GENE REGULATION – Gene switched on/off, Promoter, Operator, regulatory genes ,

Shirelle

8 8 MUTATIONS – Chromosome (Crossover during meiosis, Deletions,Inversions, TRANSLOCATION, Duplication ANEUPLOIDY (Non-Disjuction-Trisomy, Klinfelters, Turners) PolyploidyGene- Substitution, Inversion, Insertion/deletion, Frameshift, Mutagens, Mutation rates, Somatic or Germ line

Natalia

9 9 GENETIC ENGINEERING- Tools – Restriction enzymes(sticky/blunt ends), DNA ligase, Recombinant DNA , Vectors- plasmids, viruses, Transgenic organisms, Transformations, Antibiotic resistance markers,

Jessica

10 9 P.C.R ,Gel electrophoresis, DNA sequencing, Human Genome project, DNA profiling(fingerprinting)-STR’s and VNTR’s ,

Kireeti

11 912

Genetic Testing- Restriction length polymorphisms(RFLP), Gene Probes, Artificial evolution– Selective breeding,Cloning(Embryo splitting, nuclear transfer) Stem cells, Transformation, Screening, Gene therapy

Alula

12 10 FOSSILS- sedimentary rock, trace, Dating- absolute (radioisotopes), Comparative (index fossils) Pangea-Gondwana Biogeography, Wallaces line

Rachel

13 10 EVOLUTION – Adaptation, Recent common ancestors, , Divergent- homologous features, Convergent-analogous features, Phylogenetic tree (cladogram) COMPARISONS- Structural(Embryology, homologous features) Biochemical(DNA Nuclear/ Mitochondrial, Molecular clock, proteins-cytochrome C, haemoglobin, Hybridisation) Biogeography, Extinction,

Robert

14 11 NATURAL SELECTION–Variation, Selection pressure, Gene Pool, Wild-type allele, Polymorphism, Mutation, Genetic drift, Bottleneck effect, Founder effect,Darwin, Lamarck, Wallace, - Speciation, Isolating mechanisms (pre-reproductive and post reproductive) Allopatric speciation, Heterozygote fitness (sickle cell)

Blake

15 12 Hominids to Hominins – Primates, Apes and monkeys, Fossil structure, Bipedalism, opposable thumb, Australopithecus(afarensis,africanus), Paranthropus, Homo habilis, Homo erectus, Homo neanderthal, Homo florensis, Homo sapiens, Cultural evolution, tools, fire, painting, rituals, language,

David

CHAPTER 7 STRUCTURE OF CHROMOSOMES PART 1Prokaryotes, Eukaryotes, Genes, Locus, alleles, Genotype/Phenotype, Centromere, homologous, Chromatids, Diploid, Haploid, Polyploid, Autosomes, Sex chromosomes, KARYOTYPE,

Questions 1. Describe the relationship between DNA, Genes and chromosomes.2. What is the difference between a gene and an allele?3. What is the difference between the terms genotype and phenotype? Do two organisms with the same genotype necessarily have the same phenotype. Give an example to support your answer. Do two individuals with the same phenotype necessarily have the same genotype? Give an example to support your answer.4. How many chromosomes do humans have in a) normal body cells (somatic cells) b) sex cells (gametes).5. The chromosomes sets of normal human males and females are similar but different. Explain.6. What differences in chromosome structure are used to sort chromosomes for a karyotype?7. What features visible under a microscope can be used to distinguish a human X chromosome from a Y chromosome?8. Referring to Recessive conditions what is meant by a carrier?9. What is Apoptosis and how is it different to necrosis?

APOPTOSISApoptosis is controlled cell death. It can happen during development when the body is being shaped (cells between fingers undergo apoptosis to separate fingers), when the cell is infected or not functioning correctly or as a control for rapidly dividing cells like blood or skin. It is initiated by chemicals either from in or out-side the cell and results in enzymes dismantling the cell whose parts are consumed by phagocytes. KARYOTYPEThe analysis involves comparing chromosomes for their length, the placement of centromeres (areas where the two chromatids are joined), and the location and sizes of

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How many chromosomes? What sex? Are there any chromosome mutations – non disjunction? What is this condition called?

Question 1From the photograph it is reasonable to conclude thatA. the fetus is a female.

B. there are 44 autosomes.

C. the DNA sequence is identical along each of the pair of chromosomes labelled 13.

D. the centromeres of the chromosomes in Group A are near the ends of the chromosomes.

2. In rabbits, a single autosomal gene determining coat colour has four different alleles. At this locus the maximum number of different alleles a rabbit can have is-A one B two C three D four

3. (Q122 Q7 2000)The number of different genotypes possible with respect to this gene for coat colour isA four B six C eight D ten

4. (15. From a karyotype it is A impossible to identify the sex of the organism.B normal to have 44 autosomes in humans.C possible to say the DNA sequence along a homologous pair is identical.D normal to view the centromeres in the middle of the chromosomes.

5. (Question 1)The domestic cat, Felis catus, has a diploid number of 38. a. How many chromosomes would a gamete of a cat contain?

6. (Question 22)In a group of organisms, individuals genetically identical at a particular single gene locus show a variety of phenotypes for the trait. It is reasonable to conclude that the variation in phenotypes for this trait is the result ofA. codominance.B. polygenic inheritance.C. environmental influences.D. multiple alleles at the locus.

7. ( Question 22)Homologous chromosomes contain the sameA. DNA sequences. B. number of guanine and adenine nucleotides. C. alleles. D. genes.

8. Alleles A are made of genes. B are usually expressed individuallyC code for specific proteins D are made of proteins

9. (Question 9)For humans, the term nuclear genome refers to all the genes inA. an autosome. B. an X chromosome. C. a Y chromosome. D. a set of autosomes plus the sex chromosomes.

10. (Question 1)In eukaryotic organisms genes areA. composed of DNA. B. alternative forms of an allele.C. composed of DNA and protein. D. the same length as a chromosome.

11. (Question 2)In prokaryotic organismsA. translation occurs at the ribosome. B. transcription occurs in the nucleus.C. chromosomes are usually linear. D. DNA is only found in plasmids.

12. (Question 14)This process is called A. mitosis. B. necrosis. C. apoptosis. D. binary fission.

13. (Question 15)The process shown in the diagram aboveA. is initiated only within the cell itself.B. involves the total destruction of structures Y.C. is under the control of a single enzyme.D. involves the destruction of parts by phagocytes such as X.

14. (Q1, 2001) The diploid number of the platypus is 52. The number of chromosomes present in a somatic cell of a platypus isA 13 B 26 C 52 D 104

15. (Q8, 2000)A goldfish has 94 chromosomes in its somatic cells the diploid number for the gold fish is A 47 B 94 C 188 D 376

16. The diagram showsA One chromosome B Two chromosomesC An homologous pair of chromosomes D One chromatid

The open circle in the middle is a symbol for theA Locus C ChromatidB Centromere D Histone

17. Question 2Bay scallops (Argopecten irradians) have three shell colours: orange, yellow and black. It is known that thecolour is under the control of one gene locus with three alleles.a. What is a gene locus?b. If you compared the alleles of this locus at the molecular level, what would be different?

CHAPTER 7 CELL DIVISION PART 2 DNA. Replication (Semi-conservative, DNA helicase, DNA polymerase, DNA ligase, MEIOSIS -I.P.M.A.T x2 , Daughter cells – Haploid MITOSIS X1 Daughter cells - DiploidLinked genes , Crossover,

Questions1. Describe the DNA replication process.2. When does replication occur in the life cycle of the cell?3. Compare DNA replication and PCR by listing the similarities and differences between the two processes.4. Why is meiosis significant to sexually reproducing organisms? 5. Compare Meiosis and Mitosis – number of cell divisions, chromosome number in daughter cells (n,2n) Stages when crossover occurs, Non-disjunction can happen. Note- when chromosomes are doubled the whole thing is called a chromosome and the parts are called chromatids.6. How are the terms ‘ segregation of alleles” and ‘independent assortment of chromosomes’ related?7. During meiosis, crossing over and recombination occur between homologous chromosomes. Describe the outcome of recombination.8. What are linked genes?9. Describe what may happen to linked genes during the ‘crossing over’ process. Why is crossing over important?10. Which is most likely to be affected by crossing over- closely or distantly linked genes?

MEIOSIS

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1. (Q110 (Q8,2000)A goldfish has 94 chromosomes in its somatic cells. The diploid number for gold fish isA 47 B 94 C 188 D 376

2 (Question 4)The cell could come fromA. skin. B. liver. C. testes. D. bone marrow.

3. During DNA replicationA. messenger RNA (mRNA) is produced.B. reverse transcriptase enzymes play an important role.C. bonds between phosphate and sugar molecules break.D. each of the DNA strands acts as a template strand.

4. (Question 3)In specialised cells in the ovary and testis, cells divide by the process of meiosis to produce gametes. A cell with a diploid number of 4 underwent meiosis. The following images illustrate different stages throughout the total process of meiosis in this cell

a. Using the letters under each cell (A–F), put the cells in order commencing with the earliest stage of meiosis shown.

b. During meiosis, crossing over and recombination occur between homologous chromosomes. Describe the outcome of recombination.

c. During meiosis the nucleus undergoes two divisions.

i. Which of the cells, E or F, represents anaphase 1? Explain

The list 1–4 below describes events and outcomes of the replication of DNA within a eukaryotic cell.1. Complementary nucleotides bind to each of the two strands.2. Sugar phosphate bonds form between the nucleotides.3. The newly formed DNA molecules are semi-conserved.4. Unwinding of the DNA molecule forms two single strands.5. (Question 12)The correct order of these events during DNA replication, with the earliest event first, isA. 1, 2, 3, 4B. 1, 4, 3, 2C. 4, 2, 1, 3D. 4, 1, 2, 3

6. (Question 1)In some species of locusts the female has a ‘diploid chromosome number’ of 16.a. What is meant by the diploid chromosome number?

b. In this species a male locust has a diploid number of 15.

What is the chromosome number of each of the two daughter cells produced during a mitotic division in a male?The difference in chromosome number between the sexes is the result of a difference in the number of sex chromosomes (X chromosomes in this case). Female locusts have two X chromosomes (XX) and male locusts have only one X chromosome and no other sex chromosome (XO).c. At the end of a meiotic division, how many chromosomes would you expect in the gametes of

i. a female? ii. a male?

7. (Question 5)In bees, females are diploid and males are haploid.This means that male beesA. produce gametes with half the haploid number of chromosomes.B. produce gametes by meiosis.C. produce gametes by mitosis.D. do not produce gametes.

8. During meiosisA Genetically identical daughter cells are produced.B Homologous chromosomes line up along the equatorC DNA is replicated when the chromosomes are visible.D One diploid cell makes four haploid cells.

9. During mitosisA Homologous chromosomes line up along the equatorB Haploid cells are producedC Non-disjunction produces greater variation.D Crossing over occurs.

I II III

10. Which of the following statements is correct?

A Cell 1 could be either mitosis or meiosisB Cell 11 could be either mitosis or meiosisC Cell 11 could be either mitosis or meiosis.D The order of cells in mitosis would be II,I, III.

11. Question 14DNA was incubated with radioactive nucleotides. After one cycle of replication the distribution of radioactiveand non-radioactive nucleotides in the DNA would be

12. Question 1The Zenkey is a hybrid animal produced from a cross between a species of zebra with a diploid number of 44and a donkey with a diploid number of 62.a. What is the diploid number of the Zenkey?

b. By what process are gametes formed?

c. Starting with the cell shown below with a pair of homologous chromosomes, draw what happens to thecell and its chromosomes during the process by which gametes are produced.

d. Zenkeys are unable to produce offspring. Using your knowledge of gamete formation, suggest why theZenkey is sterile.

CHAPTER 7 CROSSES PART 3 Codominant, Homozygous/Heterozygous, Genotype/Phenotype) Test Cross for genotype(or Back cross), X linked cross (Reciprocal) Dihybrid cross, Multiple alleles (Blood type) , Lethal alleles,Polygenes (continuous variation – Height ,colour)

Questions1. Distinguish between the terms complete dominance and partial dominance.2. What is the major difference between monohybrid and dihybrid cross problems?3. Does the genotype of one offspring influence the genotype of future offpring?4. Coat colour in cattle is co-dominant. Cattle can have 3 phenotypes –red, white or roan. Using ‘CR’ as the symbol for allele for red hair and ‘CW’ for white hair. Calculate the following probabilities if a roan cow “CRCW’ is crossed with a roan ‘CRCW’ bull.a) the cattle produce a heterozygous calf,b) the cattle produce a white calf.c) the cattle produce a red calf.d) the cattle produce two roan calves.e) the cattle produce three red calves.

5. Think carefully in natural populations are individuals with a dominant phenotype always more common than individuals with a recessive phenotype? (remember Natural selection!)

6. Consider two genes with two alternative alleles. Both are on different chromosomes. A gene with the alleles B (abnormally high cholesterol and b (normal cholesterol) is on chromosome 19. Another gene controls Rhesus blood type (D- Rhesus positive and d- Rhesus negative ) Its locus is on chromosome 1.A couple are heterozygous for the two genes. What is the expected phenotypic ratio amongst their offspring given they produce a huge number of children.In another couple the father is heterozygous for both characteristic and the mother has low cholesterol and is rhesus negative. Calculate the expected phenotypic ratios.

7. Genetic variation can be monogenic or polygenic. The resulting variation can be discontinuous or continuous. Explain the terms using examples.

1. (Question 6)In the Australian human population, when collecting data about the frequency of different phenotypes at the ABO blood group locus, it is possible to group all members of the population into four phenotypic classes.This is an example ofA. hybridisation.B. continuous variation.C. polygenic inheritance.D. discontinuous variation.

2. (Question 7)A fisherman was surprised to catch a fish which had no scales (nude). To investigate the origin of this phenotypethe nude fish was mated several times to fish with scales and the result of each cross was recorded. In the crossesof nude with scaled, a third phenotype appeared, which was later called linear. The linear phenotype has onlya single line of scales down one side of the body. The outcomes of these crosses are shown in the table.

From the data it can be concluded thatA. there is incomplete dominance between the nude and scaled phenotype.B. the environment is the reason for the loss of scales in the nude Þ sh.C. all of the linear fish are homozygous.D. the nude fish are heterozygous.

3. (Question 8)In guinea pigs two genes have the following alleles.Gene 1: hair length Gene 2: hair typeS : long hair W : straight hairs : short hair w : wavy hairA breeder carried out the cross SSWW × ssww and obtained a number of SsWw offspring.The breeder then carried out many test crosses involving these offspring to find out if the two genes were on the same chromosome. If the genes were on the same chromosome, close to each other, you could reasonably expect the offspring from the test crosses to includeA. more with long, wavy hair than short, wavy hair.B. more with short, straight hair than short, wavy hair.C. approximately equal numbers of long, straight hair and long, wavy hair.D. approximately equal numbers of long, straight hair and short, wavy hair.

One of the human blood groups is the MN group. There are two alleles, LM and LN, at this gene locus which determine the presence of an antigen, M or N, on the surface of the red blood cells. The heterozygote LMLN has a different phenotype from each of the homozygotes.4. (Question 13)If an individual of blood type M and one of blood type MN have children, the number of different phenotypes possible in their offspring isA. 1 B. 2 C. 3 D. 4

5. (Q 119 Q2 ,2000)A cross that could confirm that a prize-winning dog was a carrier for albinism is A AA x AA B aa x aa C Aa x Aa D Aa x aa

6. (Q 127 Q6,2001)A trait such as egg size in chickens is polygenic. This means that A one gene locus with many alleles is involved.B the population can be sorted into one or two phenotypic classesC the distribution of phenotypes is continuous.D one gene locus with incomplete dominance is involved

7. (Q142 (Q2, 2003) Purebreeding guinea pigs with rough textured black coats were crossed to purebreeding guinea pigs with smooth textured white coats. The F1 were all rough textured with black coats. The F1 were allowed to interbreed to produce an F2. The numbers of each phenotype produced in the F2 are shown in the tableF2 offspring NumberRough, black 95Rough, white 32Smooth, black 27Smooth,white 11Total 165

a There are two genes involved in these crosses. What information provided in these results confirms that there are two genes each with two alleles? b Use allelic symbols R and r for the texture locus, and B and b for the colour locus. i) What are the genotypes of the purebreeding parental guinea pigs? Indicate in your answer which genotype matches which phenotype. ii What is the genotype of the F1 guinea pigs? iii Give one genotype for an F2 smooth, black guinea pig

8. (Question 7 2005)Variation in some traits is due to the action of many genes.These traits are said to beA. polymorphic. B. polypeptic. C. polygenic. D. polyploid.

The ABO blood group system is under the control of a single gene with the allelesIA : presence of protein A on red blood cells IB : presence of protein B on red blood cellsi : neither protein A or B on red blood cells9. (Question 9)With respect to this geneA. an individual could have one of eight different genotypes.B. an individual could have one of six different phenotypes.C. a child could have protein A even if both parents lacked protein A on their red blood cells.D. a child with neither protein A nor B could have a mother with protein A and a father with protein B.

10. A particular species of plant has the following genes and alleles.Leaf shape L : normal shape l : wrinkled shapeNumber of seed cases P : one seed case p : three seed cases

Question 10The genotypes of the two parents would be

A. ppll x PPLLB. Ppll x PPLLC. PpLl x PpLlD. PPLl x PpLl

The hair colour of Australian shepherd dogs is under genetic control. The colour of the hair found inside the ears, on the legs and under the tails, is under the control of a gene that has the following alleles.AW whiteaS sable colourac copper colourWhite is dominant to both sable and copper colour. Sable colour is dominant to copper colour.11. (Question 2)If two dogs with the genotypes AW ac and aS ac are mated, the resulting offspring could haveA. 4 genotypes and 4 phenotypes.B. 4 genotypes and 3 phenotypes.C. 3 genotypes and 4 phenotypes.D. 3 genotypes and 3 phenotypes.

12. (Question 12)In the Australian sheep blowfly, the length of the wing is a polygenic trait.This means thatA. there is a small number of clearly defined phenotypes.B. wing length is controlled by one gene with many alleles.C. the wing length phenotype shows continuous variation.D. the phenotype is controlled by many genes on the same chromosome.

13. (Question 4)With respect to the ABO blood group locus it is possible to produce children of four different phenotypes if the parents areA. type B x type B. B. type A x type B. C. type O x type AB. D. type AB x type AB.

Two genes in watermelons with their alternate alleles areGene 1 S : spots s : solid colour Gene 2 B : bitter fruit b : sweet fruitThe two genes assort independently. Two plants, both heterozygous at each gene locus, were crossed and 1600 seeds were collected.14. (Question 3)When plants were grown from these seeds, it would be reasonable to expect that aboutA. 100 of the plants produced spotted-coloured, bitter fruit.B. 300 of the plants produced solid-coloured, bitter fruit.C. 300 of the plants produced solid-coloured, sweet fruit.D. 1600 of the plants produced spotted-coloured, sweet fruit.

Milk production in cows is under genetic control. The daily volume of milk produced by each cow in a large herd of cows was recorded and the results graphed. The results were as follows.

15. (Question 8)A trait showing this kind of distribution isA. not found in humans.B. called a discontinuous trait.C. under the control of many genes.D. not influenced by environmental factors.

In lentils the seed coat pattern is determined by a gene with three alleles. The phenotypes are marbled, spotted and clear. Four crosses were repeated many times. The crosses and the outcomes of these crosses are shown in the table below.

16. Question 3The number of different genotypes possible at the locus for seed coat pattern isA. 3 B. 4 C. 5 D. 6

17. Question 4From the data it is possible to conclude thatA. spotted is recessive to clear.B. all of the clear offspring are heterozygous.C. two thirds of the marbled offspring in cross 3 are heterozygous.D. the marbled parents in cross 1 have the same genotype as the marbled parents in cross 3.

18.Coat colour in mice is under the control of a single gene with two alleles. Many crosses between yellow coated mice and mice with grey coats gave the following results. The mice with grey coats were known to be homozygous.Parental cross yellow x grey First generation 50% yellow : 50% greyMany crosses were carried out between the first generation yellow mice.d. What genotypic ratio and phenotypic ratio would we expect to see in the offspring of the cross betweenthe first generation yellow mice? Make sure you indicate the allelic symbols you are using for this genelocus.Scientists performed this cross many times and the result they observed was always a ratio of 2 yellow to1 grey mouse.e. How can this result be explained?

Question 3In a particular insect species, sex is determined by a single gene. The male insect has the genotype Mm and thefemale insect is mm. This gene is linked to another gene that determines the body colour of the insect.a. What is meant by linked genes?The linked gene determining body colour in these insects has two phenotypes, black body colour and bronzebody colour. Black body colour is the dominant phenotype.When a heterozygous black male was crossed with a bronze female the following offspring were produced.

b. What are the expected ratio’s for this cross? Which traits are linked?CHAPTER 7 MONOHYBRID/SEXLINKED PART 4 Monohybrid cross Mendels crosses.(Dominant, Recessive) X linked cross (Reciprocal)Dominant/recessive, Autosomal/Sex linked

Questions 1. A gene’s locus is on the X chromosome. Does this affect its inheritance patterns?2. A man and woman are heterozygous for a gene. The gene has two possible alleles and there are two phenotypes. Use ‘D’ to represent the dominant allele and ‘d’ for the recessive allele. Calculate the following probabilitiesa) the two individuals produce a heterozygous dominant offspring.b) the two individuals produce a homozygous dominant offspring.c) the two individuals produce a homozygous recessive offspring.d) the two individuals produces two homozygous recessive offspring.e) the probability the child is heterozygous given the two individuals produce a child with the dominant phenotype.f) the expected ratio for dominant to recessive phenotypes.

3. Show a cross involving a homozygous recessive man and a woman heterozygous for the gene. What are the possible genotypes and phenotypes of their offspring and in what proportions.?4. Explain why sex-linked recessive conditions, like colour blindness, are more often expressed in men than woman5. A normal woman with a colour-blind father has children with a man of normal vision. a) What is the probability that a child is colour blind?b) What is the probability that a son is colour blind?c) What is the probability that a daughter is colour blind?6. Geneticists, when experimenting with fruit flies, pea plants, mice etc like large numbers of offspring before making conclusions. Relate this to differences between expected results, observed results, sample size and probability.7. Describe the drawing of pedigrees. What symbols are used and how are generations numbered?8. What patterns do you look for in a pedigree to distinguish between autosomal recessive and autosomal dominant inheritance?9. What patterns do you look for in a pedigree to distinguish between x-linked recessive and X-linked dominant inheritance?

PEDIGREES - Is the trait Dominant D, Recessive r ex-linked dominant XD, sex-linked recessive Xr ?

RULES – Does it skip generations?

If YES – RECESSIVE – Autosomal or Sex – linked ? You can only prove that it is autosomal, you just suspect it is sex-linked. If a mother has the recessive trait then her sons must get it to be sex- linked – otherwise its autosomal. Xr Xr x XY

Xr YIf NO – It may be DOMINANT (although can still be RECESSIVE) DOMINANT - If the offspring have a trait that is dominant at least one parent must also have it. - If two parents have it but not all their offspring then the trait can’t be recessive it must be dominant.

SEX-LINKED? – If the father has it his daughter must have it for a sex-linked dominant trait.XD Y X XX

↓ XDX

1. (Q5, 2002)Colour blindness is inherited as an X-linked recessive condition. It is reasonable to claim that A colour - blind male must have a colour blind mother.B colour - blind male must have a colour-blind grandfather.C colour - blind female must have a colour-blind father.B colour – blind female must have a colour – blind grandmother. 2. (Q117 Q7 2001) Numbers of offspring of each colour

Cross brown GreenCross A Brown x green 12 11Cross B Brown x brown 7 1Cross C Brown x brown 18 0Cross D Green x green 0 21It is possible to conclude from these results that the brown phenotype is dominant because-A approximately equal numbers of brown and green offspring are produced in cross AB one green offspring is produced in cross BC only brown offspring are produced in Cross CD only green offspring are produced in Cross D.

3. (SA Q4, 2001)In tomatoes the shape of the fruit is an inherited trait. In a cross between plants which are pure breeding for spherical fruit, and other which are pure breeding for oval fruit, all offspring have spherical fruit.From this information you can conclude that-A a cross between tomatoes, each heterozygous, for spherical fruit, could produce offspring with two phenotypes.B oval-shaped fruit is the dominant trait.C any cross between tomatoes with spherical and oval fruit will produce offspring with one phenotype.D all tomatoes with spherical fruit are homozygous. (1 mark)

4. (Q76, Q 1,2004). An autosomal gene controls the shapes of tabby stripes in a cat’s fur. It has the alternative allelesT : vertical strips of colour (called mackerel tabby) t : swirly stripes of colour (called blotched tabby) Several crosses were carried out between mackerel and blotched cats. All of the mackerel tabby cats were heterozygous. In total, the cats produces 100 kittens. Among these kittens, it would be reasonable to expect-A approximately 50 mackerel tabby femalesB approximately 25 blotched tabby femalesC more mackerel tabby females than mackerel tabby malesD more blotched tabby males than blotched tabby females.

(1 mark)

5. (Question 1)In humans the presence of a dimple in the chin is dominant.The first child of a couple, each with a dimple, does not have a dimple but their second child does have a dimple.a. Use alleles D and d to show the genotype corresponding to the phenotype of the parents and the child without a dimple. Genotype of parents genotype of child without a dimple

6. A plant has two phenotypes, herbicide resistant and herbicide sensitive. A farmer wanted to establish the patternof inheritance for this trait and performed the following crosses. Assume this trait is from one gene with two alleles.

b. i. Which one of the crosses, on its own, allows you to conclude which is the dominant phenotype?

ii. Explain your choice in b.i.

c. Use appropriate allelic symbols to show the genotypes of the parents and offspring of cross 3.

7. (2005) The gene responsible for the autosomal recessive condition PKU controls the production of an enzyme thatconverts the amino acid phenylalanine to tyrosine. The gene has the allelesP : produced enzyme p : no enzyme producedIn a person lacking the enzyme, phenylalanine accumulates in toxic levels throughout the body and the diseasedevelops. Development of the disease is prevented by placing a baby, who lacks the enzyme, on a special diet a few days after birth. This diet is low in phenylalanine and is maintained for as long as possible, preferably for life.Question 3With respect to PKU,A. the special diet for babies with the disease would also contain reduced levels of tyrosine.B. each person in a population would have one of three possible genotypes at the PKU gene locus.C. a PKU individual treated from soon after birth would not be able to pass allele p onto their offspring.D. because the disease can be treated, one would expect the number of babies born with PKU to decline over time.

8. (Q141 Q1, 2003)The back of the leopard frog can be either patterned or non- patterned. Several patterned frogs were allowed to breed and they produced 75 patterned offspring and 25 non-patterned offspring.a i Which of the phenotypes, patterned or non-patterned, is dominant? ii Explain your answer to i. (BE CAREFUL!) b Using your own allelic notation, show the genotypes with their respective phenotype for the parents and offspring of the cross between the patterned frogs described above. Draw a punnet square using these alleles and give the expected genotype and phenotype ratios.

Crosses between different patterned and non-patterned frogs were performed. Not all crosses produced the same outcome. The results are shown in the table below. For both cross A and cross B there were large numbers of offspring produced.

Parents OffspringCross A Patterned x non-patterned All patternedCross B Patterned x non-patterned ½ patterned ; ½ non patternedc The parents in crosses A and B have the same phenotypes. Explain why the outcome of the crosses A and B are different .

CHAPTER 7 PEDIGREES PART 5 1. (Question 2)Ptosis is a weakness in the muscles of the eyelid.a. Using the information below, construct a pedigree to fit the description of the inheritance of ptosis in the family.Bill had ptosis. Bill married Jill who did not have ptosis and they had two children, a boy Ben who was unaffected and a girl Daisy who had ptosis. Daisy married Bob who did not have ptosis and they had three daughters, two girls with ptosis and one girl without ptosis.b. The pedigree below shows another family in which ptosis was found.

i. Based on the information in this pedigree, what is the most likely mode of inheritance of ptosis?

ii. Provide two pieces of evidence from the pedigree that support this conclusion.

2 (Q157 Q3, 1999) Human pedigreesA survey of male and female Biology students found that some individuals have one whorl of hair at the crown of the head whereas others have two whorls. This was investigated further in two families, Family A and Family B. Data collected from these families were recorded in two pedigrees shown in Figure 2.40.

PEDIGREE A PEDIGREE B - one whorl - two whorls

a i Which of the pedigrees, A or B will fit more than one pattern of inheritance for one whorl? ii Explain how the pedigree chosen in a i fits more than one pattern of inheritance.b i Consider the pedigree not chosen in part a i. Explain what feature of this pedigree allows you to decide that the inheritance of a single whorl is dominant ii Describe one feature of the pedigree considered in part b i which indicates that the inheritance of one whorl is autosomal dominant and not X-linked dominant.

I

II

III

12

3 4 52 6

1 2

2 3 4 5

3. (Question 1)With respect to the PKU gene locus, you could reasonably conclude that individualA. I – 1 is homozygous. B. I – 2 is homozygous. C. II – 5 is heterozygous. D. III – 1 is heterozygous.4. (Q124 Q3, 2000)The pedigree consistent with X-linked recessive inheritance of the trait represented by the shaded symbols is

A B

C D

I

II

III

12

32 4

1 2 3 4

I

II

III

12

32 4

1 2 3 4

I

II

III

12

32 4

1 2 3 4

I

II

III

12

32 4

1 2 3 4

5. (Question 2)Coat colour in cocker spaniel dogs varies. Four of these colours are black, liver, red and lemon. These fourcolours result from the interaction of two particular autosomal genes.The pedigree below (Figure 9) shows the inheritance of coat colour in a group of cocker spaniels.

a. I–1 and I–2 are heterozygous at both the R and B locus. What evidence from the pedigree supports this conclusion?

b. i. What is the specific genotype of II–4?ii. What is the specific genotype of III–4?

c. Explain how many different phenotypes could be expected in the offspring of a mating between individualsII–4 and III–4. Show all working.

6. (Question 3)Haemophilia is an X-linked recessive condition. The following pedigree shows a portion of a family in whichsome members have haemophilia. Those on the pedigree with haemophilia are shaded.

a. Use appropriate allele symbols from XH, Xh and Y to indicate the genotype of each of the followingindividuals. I1 and I2 b. The couple II1 and II2 have a son. What is the probability that the child has haemophilia?c. The couple II5 and II6 have a son. What is the probability that the child has haemophilia?The following pedigree is of a family in which one member (shaded) has an autosomal recessive condition. The alleles of the gene locus involved are G and g.

d. Give a possible genotype for each of the four members of the family.I1I2II1II2

CHAPTER 8 PROTEIN SYNTHESIS PART 6 TRANSCRIPTION- Nucleus, RNA polymerase, Codons, mRNA (exon/Intron)TRANSLATION – Ribosomes(rRNA and Proteins),tRNA, Anticodons,

Questions 1. How do prokaryote and Eukaryote genes differ?2. What are triplets, codons and anticodons?3. In terms of nucleotides and triplets, what is a gene?4. What are the major differences between RNA and DNA?5. How many different codons are possible in the genetic code?6. Are all codons needed? How many amino acids are there?7. Do all codons code for amino acids?8. List the major steps in protein synthesis including the names of each stage, where each stage occurs , the products formed and any enzymes required.9. Introns and exons are parts of the eukaryotic gene sequences. Which parts seem to have the most importance?10 What does Reverse transcriptase do?

TRANSCRIPTION / TRANSLATIONUse the photocopied Biozone sheet for the overviewTable of mRNA codons pg 253What is the difference between RNA and DNA codes and structures?Convert the following DNA code to an Amino Acid chainD.N.A - TAC TTT TGC CTA ATC_____ -A.A -

CODONS - need 3 nucleotides bases to code for 1 amino acid – Why 3?1 base codes for 4 Amino Acids because there are four different bases A,C,T,G2 bases code for 4 x 4 = 16 different combinations so 16 A.A’s3 bases code for ________________________________________

Figure 3 shows portion of a cell engaged in protein synthesis. The various parts of the cell are not drawn to scale.

1. (Question 10)It is reasonable to conclude thatA. process N is translation.B. structure P is made of t-RNA.C. compound Q is messenger RNA.D. structure R is the site of protein synthesis

Haemoglobin consists of four polypeptide chains. In normal haemoglobin, two of these chains are beta chains, each comprising 146 amino acids. Variations exist in the amino acid composition of these chains and this results in different kinds of haemoglobins. One of these variants is called haemoglobin S.The first seven amino acids in the beta chains of these two haemoglobins, and the amino acid at position 143 are given below. The amino acids at each of the remaining positions are the same for each haemoglobin.

2. (Question 13)It is reasonable to conclude that during transcription of normal haemoglobin, the mRNA codon sequence could be

A. GTT for amino acid 1.

B. ACT for amino acid 4.

C. CTT for amino acid 3.

D. CAC for amino acid 2.

3. (Question 2)The following diagram outlines events associated with the production of a polypeptide chain in a eukaryotic cell.

a. What is the name of the process at step 1?

b. i. Name the product of step 1

ii. Outline what occurs at step 2.

c. Name the event that occurs at structure 3.

d. i. Name the structure at 4.

ii. Outline the function of the structure you named in d.i.

4. The first six amino acids of the P34H protein areMet Pro Lys Tryp Val cysA mRNA sequence that corresponds to this amino acid sequence is AUG CCA AAG UGG GUA UGCWrite dowm the double stranded DNA sequence corresponding to this section of the gene.

Explain if this is the only base sequence possible for this section of this gene.

5. Question 15Translation of the genetic code occurs in the cytosol of a cell. The following diagram is one representation oftranslation.

In the model presentedA. M represents a ribosome.B. N represents messenger RNA.C. O represents transfer RNA.D. P represents an amino acid.

CHAPTER 8 DNA STRUCTURE/ GENE REGULATION PART 7DNA Structure and Discovery Watson/Crick, Rosalind Franklin, Chargraff’s bases, Chromatin, Histones,GENE REGULATION – Gene switched on/off, Promoter, Operator, regulatory genes ,Questions1. What is the basic building block of DNA?2. What are the 3 major components of a DNA nucleotide?3. Two of the above components are constant. Which?4. The third component varies. Name the four variations.5. What are ‘Complementary bases’?6. What type of chemical bond holds the base pairs together?7. How can the above bonds be broken?8. Do all genes make proteins which will be used outside the nucleus?

1. (Q128 Q13 2001)The base composition of the DNA of a particular bacterium was analysed. It was found that 18 per cent of the bases were adenine. You could reasonably conclude that A the DNA comprised 18 percent adenine-thymine pairs.B 32 percent of the bases were cytosineC the DNA comprised 32 percent of cytosine-guanine pairs.D 36 percent of the bases were thymine

2. (Question 8)In DNA, the number ofA. phosphate groups equals the number of nitrogen bases.B. adenine nucleotides equals the number of cytosine nucleotides.C. phosphate groups equals twice the number of sugar molecules.D. guanine nucleotides equals the number of uracil nucleotides.

The following nucleotide sequence forms part of the template strand of a gene coding for protein X.T G G A T G A C DNA template strand *3. (Question 9)The complementary base found at the fourth nucleotide (marked *) in a sequence transcribed from this sequence would beA. C B. G C. T D. U

4. (Question 10)In a double-stranded molecule formed from this DNA template strand (shown above) the number of deoxyribose sugar units you would expect to find isA. 4 B. 8 C. 16 D. 32

5. (Question 11)Reverse transcriptase catalyses the production of A. DNA from an mRNA template.B. DNA from a protein template.C. mRNA from a DNA template.D. tRNA from a DNA template.

6. (Question 20)One section of a polypeptide has the amino acid sequence Ala . Cys . Lys . Ile . AsnThe codons for these amino acids are- Ala - GCA GCC GCG GCU

Cys - UGC UGU Lys - AAA AAG Ile - AUU AUC AUA Asn - AAC AAU

The sequence of DNA coding for this section of the polypeptide could beA. CGTACGTTTTATTTG B. CGTTCGTTTTATTTGC. CGTACTTTTTACTTG D. CGAACATTCTATTTT7. (Question 7)Organisms can regulate the expression of their genes in a number of ways.a. Suggest why an organism regulates the expression of its genes.

One example in bacteria is the regulation of the expression of a gene which produces an enzyme (enzyme X)involved in the metabolism of the amino acid tryptophan. Enzyme X is only produced when tryptophan is in highconcentration. This gene regulation involves several genes. Two of the genes include a gene for the production ofenzyme X and an operator gene. If a protein, called a repressor protein, binds to the operator gene, transcriptionof the gene for enzyme X is stopped. If no repressor protein is bound to the operator, transcription of the genefor enzyme X occurs. A summary of this regulation is shown in Figure 1.

b. The gene coding for enzyme X is not transcribed when the repressor protein binds to the operator gene.What enzyme is prevented from functioning during this binding?

When tryptophan binds to the repressor protein, the repressor protein can no longer bind to the operator gene.

c. When tryptophan binds to the repressor protein what will happen to the production of enzyme X?

d. Based on Figure 2, suggest how tryptophan prevents repressor protein function.

CHAPTER 8 MUTATIONS PART 8MUTATIONS – Chromosome (Crossover during meiosis, Deletions,Inversions, TRANSLOCATION, Duplication ANEUPLOIDY (Non-Disjuction-Trisomy, Klinfelters, Turners) PolyploidyGene- Substitution, Inversion, Insertion/deletion, Frameshift, Mutagens, Mutation rates, Somatic or Germ line

Questions1. Gene mutations are alterations to the base sequences in genes. Describe the three major types of base mutations – Substitution, Insertion, Deletion and there effects.2. Explain why mutations are usually harmful. Which of the 3 above types is the least harmful. Why?3 Distinguish between germ line and somatic mutations.4 What sort of mutation can be beneficial to an organisms long term survival?5. What is Translocation6. Can genes on different chromosomes become linked? (Think)7. What is non-disjunction.

8. Make a list of examples of a condition that arises as a result of non-disjunction. Explain what must have happened during meiosis and at what stage it happens.9. What is polyploidy. How does it help sterile hybrids?10. List some causes of mutation.

1. (Question 8)The origin of new alleles at a gene locus is the result ofA. migration.B. mutation.C. recombination.D. independent assortment.

2. (Question 13)Achondroplasia is an autosomal dominant trait in humans that results in a form of dwarfism. In some cases a child with achondroplasia is born to parents who have normal height.The most likely reason for the appearance of the child with achondroplasia is thatA. the parents are carriers and the child has inherited the mutant allele from each parent.B. a mutation has occurred in a gamete of either the mother or the father.C. a mutation has occurred in a somatic cell of one of the parents.D. a mutation has occurred in the tissues of the child.

3 (Question 6)Cabbage (Brassica oleracea) and radish (Raphanus sativus) both have a diploid number of 18. However theydo not naturally hybridise with each other.

a. How many chromosomes would be expected in the gametes of the cabbage?

In the laboratory, the two species can be forced to mate and produce offspring. The offspring are sterile.b. i. What would be the diploid number of the hybrid?

ii. Explain why the hybrid of the cabbage and radish is sterile.

An occasional spontaneous event produces a doubling of each chromosome set in the hybrid. The new plants are able to grow and produce fertile offspring.

c. What term is used to describe cells with more than two sets of chromosomes?

d. Explain, with reference to the events of meiosis, why the new plants are fertile.

4 (3.) Genetic ‘accidents’ can occur to chromosomes. One type of accident is when one chromosome becomespermanently attached to another. This kind of arrangement is called a translocation.A scientist observed a translocation involving chromosomes 9 and 18 in somatic cells of a male cat. Figure 8shows chromosomes 9 and 18 in a normal male cat and their arrangement in the cat carrying the translocation.Note that the centromere of the translocated number 18 chromosome has been lost.

In some cases translocations lead to abnormalities, but that was not the case here. The cat had a normal phenotype.

b. i. What is meant by the phenotype of an organism?

ii. The cat with the translocation had only 37 chromosomes in each somatic cell. Explain why it still had a normal phenotype.

The cat’s reproductive tissue also contained the translocation, and investigation showed that he produced fourdifferent kinds of sperm with respect to chromosomes 9 and 18. Two of these four types are shown in Table 2.

c. Complete Table 2 by naming the chromosome make-up of sperm types 3 and 4 respectively.

One type of sperm produced by the male carrying the translocation does not survive.

d. i. Explain which type this is most likely to be.

ii. The male cat with the translocation is mated to a normal female. What is the chance that he will fathera kitten with the same kind of translocation involving chromosomes 9 and 18, and a normal phenotype?

5. Streptococcus pneumonia is a bacterium which causes pneumonia in humans and may show resistance to antibiotics.a. How would antibiotic resistance have first occurred in the Streptococcus pneumonia population?

6. In some Australian insects, new species have arisen through changes that occurred to chromosomes in an ancestral species. Such changes may involve the joining together of chromosomes, the loss of whole or parts of chromosomes, and rearrangement of the genetic material within chromosomes.One ancestral species has the following haploid set of chromosomes.

As the changes in chromosomes accumulate, a number of different species can result from a single ancestral species.

Three species that have evolved from the ancestral species shown above have the haploid sets of chromosomes shown below.

6. (Question 25)The most likely order of evolution of these species isA. ancestral species, species Z, species G, species M.B. ancestral species, species G, species M, species Z.C. ancestral species, species M, species G, species Z.D. ancestral species, species G, species Z, species M.

Below is the DNA sequence from the beginning of a gene coding for an enzyme involved in photosynthesis. The upper line of bases (in bold) represents the template strand.

a. Write the mRNA sequence that would be transcribed from this DNA sequence.

7. (b. i. )The 6th base on the template strand of the sequence above is substituted by C. What type of mutation is this?ii. Explain the effect this mutation will have on the amino acid sequence of the protein produced

c. The 11th base pair of the sequence is deleted. Explain the effect that this mutation will have on the amino acid sequence of the protein produced.

CHAPTER 9 GENETIC ENGINEERING PART 9Tools – Restriction enzymes(sticky/blunt ends), Recognition site.DNA ligase, Recombinant DNA , Vectors- plasmids, viruses, Transgenic organisms, Transformations, Antibiotic resistance markers, PCR

QuestionsWhat is the aim of genetic engineering?What are restriction enzymes and how are they used?Given that geneticists can cut and rejoin pieces of DNA, how do they transport DNA from source to target cells?How do they detect if the gene has been inserted into the new organism?Explain the steps of the PCR (what does it stand for?) and give an example of its use.How is it similar to DNA replication in cell division?

1. (Question 11)Incubation of this length of DNA in a tube containingA. Spe I would result in three pieces of DNA.B. Hin dIII would result in two pieces of DNA.C. Spe I and Eco RI would result in five pieces of DNA.D. Bgl II and Hin dIII would result in four pieces of DNA.

2. (Question 4)Genes can be transferred from one species to another in different ways. One method is to use plasmids, circular pieces of DNA found in some bacteria.In this method, a plasmid is cut and a piece of foreign DNA inserted. The foreign piece of DNA usually contains more than one gene. The process is shown in Figure 5 below.

Many copies of the new plasmid are then incubated with bacteria.a. What is the name given to a plasmid that is used to transfer DNA from one organism to another?

b. What is used to cut the DNA of a plasmid?

c. What is used to join the inserted piece of DNA to the plasmid?

One of the foreign genes inserted into the plasmid, codes for resistance to a particular antibiotic.d. Explain why it is important to include a gene for antibiotic resistance in the plasmid produced.

Bacteria containing plasmids, that are constructed in the way outlined in Figure 5, can be used in a variety of settings with plants and animals. For example, some plants are resistant to attack by insects. The plants produce a protein that poisons the larval stage of some insects that feed on them. The production of the protein is under genetic control.A particular species of crop plant was genetically engineered to contain this gene. Such plants are referred to as GM (genetically modified) plants.

e. Explain why a farmer might choose to grow a crop that was genetically engineered to be resistant to insects, rather than spray the crop with insecticide.

Some plants are resistant to particular herbicides, chemicals that are used to kill plants. This trait is also under genetic control. The gene that confers herbicide resistance has also been incorporated into some GM crop plants. This enables a farmer to spray his GM crop with a herbicide that will not harm the GM crop but does kill weed plants growing within the crop.f. Suggest one advantage for a farmer to be able to spray his crops with a herbicide.

Two farmers have properties next door to each other. They grow the same cereal crop.• Farmer X wishes to grow GM crops that are resistant to herbicide.• Farmer Y wishes to continue to grow non-GM crops.Farmer Y was concerned, and suggested to farmer X that pollen from the GM crop could fertilise the non-GM plants.g. Explain why farmer Y might be concerned about the possibility of his crop being fertilised by pollen from farmer X’s crop.

The farmers agreed to carry out field trials to establish whether leaving a gap between crops reduced the likelihood of cross-pollination. A number of trials were planted so that the results of one trial did not interfere in any way with the results of another. The percentage of seeds produced at various positions as a result of cross-pollination was measured for each trial. The outline of these trials and the results gathered are shown in the following table.

h. From the data, what conclusions can be drawn about cross-pollination and the gap between crops?

3. (Question 7)Sickle cell anaemia is a serious inherited blood condition. It leads to tiredness and kidney or heart failure and without treatment children usually die before the age of 10. Sickle cell anaemia is due to a change in the gene which codes for

beta haemoglobin. There are two alleles for the beta haemoglobin gene; HbA coding for normal beta haemoglobin and HbS coding for the changed haemoglobin. An individual with two copies of the HbS allele will develop symptoms of sickle cell anaemia.It is now possible to genetically test people to see if they carry the HbS allele. This test uses PCR, the restriction enzyme MstII and gel electrophoresis.MstII is a restriction enzyme that recognises the 7-base sequence in DNA,C C T G A G GG G A C T C Cand cuts it between the C and the T to produce

a. What term is used to describe the ends of the fragments produced by MstII?

4 (Question 4)a. Describe the appearance of a bacterial plasmid.

A bacterial plasmid was modified in the laboratory so that it contained a gene for an enzyme which provided resistance to the antibiotic tetracycline. Bacterial cells, which in their natural environment were sensitive to the antibiotic tetracycline, were mixed with the modified plasmid. The bacterial cells were treated so that they could take up the plasmid.

b. What is the name of the process in which a bacterial cell takes up a plasmid and expresses the genes ofthe plasmid?

The outcome of an experiment is shown below.

With respect to the growth of bacteria the results of plates A and C are shown. On plate A there is a continuousgrowth of bacteria over the surface of the agar. On plate C the colonies are distinguishable from each other.

c. i. What result would you expect on plate B with respect to the growth of the bacteria?ii. Explain your answer to c.i.

d. Explain why there is a difference in the way the bacteria have grown on plates A and C.

Question 16The following diagram indicates the cutting sites of three different restriction enzymes on a particular bacterialplasmid.

If the plasmid was incubated with the restriction enzyme Eco R1, the number of pieces of DNA obtained wouldbeA. two.B. three.C. four.D. seven.

CHAPTER 9 GENETIC ENGINEERING PART 10 and 11Gel electrophoresis, P.C.R, DNA sequencing, Human Genome project,DNA profiling(fingerprinting)-STR’s and VNTR’s ,Genetic Testing- Restriction length polymorphisms (RFLP), Gene Probes,

Artificial evolution – Selective breeding, Cloning(Embryo splitting, nuclear transfer) Stem cells, Transformation, Genetic Screening, Gene therapy

QuestionsWhat is the purpose of Gel electrophoresis. How is this accomplished?What do the letters PCR stand for?What is the purpose of P.C.R. How does this work including enzymes required and the role of primers.What does it mean to sequence DNA?What was the Human genome project and why is this information important.How could knowing the sequence of a worm’s genome help understand human development?What do the initials STR and VNTR’s stand for? Why are these regions studied when comparing individuals?What does the phrase Restriction length polymorphisms mean? How can it help detect for some genes of a disease?What is a gene probe? How are they used? Mention autoradiography (radioactive) in your answer.What is the difference between Natural Selection and Artificial Selection.What is a clone? How is embryo splitting and nuclear transfer different?What are Stem cells and why are they useful?What is trying to be accomplished in gene therapy?

1. Q158 2002There were three suspect in an assault case. A forensic scientist found blood other than the victim’s at the site. DNA was extracted from five blood samples: the victim, the site and the three suspects. PCR was used on the extracted DNA.a) A DNA polymerase enzyme is involved in the PCR process. Explain the role of the polymerase enzyme in PCR.

One of the regions used in the forensic analysis was a short tandem repeat (STR) sequence of 4 bases, called HUMTH01. This sequence, located on chromosome 11, has many alleles which differ from each other by the number of times the sequence AATG is repeated. It was this region of chromosome11 which was amplified using PCR.The amplified samples were loaded onto a gel and electrophoresis was performed to separate the fragments of DNA.

b) Name the 2 properties of the DNA fragments which allow them to be separated from each other during the gel electrophoresis process.

c) Why is there only one band in lane 2 but two bands in lanes 3,4,5 and 6?

d) How many different alleles at the 1 HUMTHO 1 locus are represented on the gel in individuals 2,3,4,5,6?e) Which piece of DNA, A or B, has the greater number of the 4 base repeat sequence?f) Which of the suspects appears to have committed the assault ? Explain.2. Question 11Amplification of DNA in the polymerase chain reaction requiresA. nucleotides of uracil. B. DNA polymerase. C. amino acids. D. ribose sugar.

3. Molecular studies have shown that the sickle cell allele differs only by one base pair from the normal allele. This base change occurs in a 7-base sequence that is recognised by the restriction enzyme MstII. This is the only MstII site found within the region of the gene that is used in the genetic test.

The PCR products are digested using MstII. The resulting fragments undergo gel electrophoresis.b. How is the action of the MstII enzyme affected by the HbS mutation?c. Mark on the picture of the gel below the banding patterns you could expect to see for someone who has each of the following genotypes.

i. HbA/HbS

ii. HbS/HbS

Question 5CC. for Carbon Copy is the name of the Þ rst cloned kitten born in 2001. The nucleus of a cat.s egg cell was removed. It was replaced by a nucleus from a somatic cell of a donor female cat. Once development commenced the egg cell was transferred into a surrogate female.a. What is meant by the term cloning?The diploid number of a cat is 38. b. i. How many chromosomes would have been in the nucleus that was removed from the egg cell?ii. Is CC male or female? Explain.

To determine if CC was in fact a true clone, studies were made of specific variable regions in the DNA of the donor, CC and surrogate. The results are shown in the table.

c. For each region of DNA there are two values, for example, 164/164. Suggest a reason for this.d. In the case of DNA variable region 4 in the donor DNA, why are the pairs of values different?e. From the data it was concluded that CC was a true clone. Explain the evidence in the table that supports this claim.

6. Selective breeding has been used to improve the milk yield of cattle herds in Australia.e. Identify a key difference between selective breeding and random mating in a herd of cattle.

f. What is the impact of selective breeding on genetic variability in a herd of cattle?

The quality and yield of milk in cattle has been improved by artificial insemination in which semen from aselected bull is used.

g. Explain how the use of artificial insemination may intervene in the evolutionary process.

7. Question 18A restriction enzyme cutting site may be present or absent in a particular 200 kb region of human chromosome 1.

CHAPTER 10 FOSSILS PART 12 Fossilisation process, Sedimentary rock, Trace fossils, Dating- absolute (radioisotopes), Comparative (Index fossils) Transistional fossils

Biogeography Pangea/Gondwana, Wallaces line

Questions Is it always true to say that fossils are the remains of dead organisms?Under what environmental conditions are fossils formed? What conditions reduce the chances of good fossils forming?What information can fossils provide about the lifestyle of once living organisms?Why is it difficult to identify different species using only fossil evidence?Describe the difference between absolute dating (radioactive isotope dating) and relative dating (stratigraphy).Different radioactive isotopes have different half-lives. What does this mean?Different radioactive mentods are used to date fossils of different ages. List some commonly used isotopes and state their use. Is it always the fossil that is dated?What are “Transistional fossils” like Archaeopteryx and Tiktaalik and the recent Ida.

Use the following information to answer Question 12.Radioisotopes may be used to establish the age of rocks and fossils. Potassium–40 is an isotope which decaysto argon–40. The half-life of potassium–40 is 1.3 million years.Figure 4 shows the proportion of potassium–40 (%) remaining in rock over a period of millions of years.

1. Question 12A rock was found to have 25 per cent potassium–40 remaining.The approximate age of the rock isA. 0.65 million years.B. 1.3 million years.C. 2.6 million years.D. 3.9 million years.

2. Question 18During a volcanic eruption molten material called magma comes out of the volcano, cools and solidifies on the surface of the earth forming basalt. Volcanic ash, also from the eruption, is deposited near the basalt and may contain well preserved fossils. The surrounding basalt can be useful to date fossils in the strata formed by the ash becauseA. eruption dates of volcanoes are known from historical data.B. organic remains are baked and preserved in the basalt.C. radioactive elements within the basalt can be accurately dated.D. the basalt may contain an index fossil.

3. Question 19There is little fossil evidence of the earliest forms of life because the organismsA. decayed quickly in the oxygen-rich atmosphere.B. did not have hard parts which would fossilise easily.C. evolved so quickly that they left few remains.D. lived in water and were not preserved.

4. Question 21Fossils of soft-bodied organisms are relatively rare because theyA. were never common in the environments in which they lived.B. lived in environments where sedimentation did not occur.

C. are generally small in size.D. readily decompose.

5. Trilobites (Figure 6) are an extinct group of marine arthropods. They are a very well-studied group due to the abundance of fossils. Trilobites had a tough exoskeleton and bodies and legs divided into segments. They were distributed worldwide and occupied a range of habitats. They existed for almost 300 million years before becoming extinct around 250 million years ago.a. i. What structural feature improves the chances of a trilobite becoming a fossil?ii. Suggest one other reason why trilobite fossils are abundant

6. (Question 9)Naracoorte Caves are located in southeast South Australia. These limestone caves contain the greatest number, most diverse and best preserved fossils of the Pleistocene Epoch (1.8 million years to 10 000 years ago) in Australia. For more than 300 000 years sediment and animals fell into one particular cave through an opening in its ceiling, forming an enormous cone-shaped pile. Animals that fell in through the hole were unable to escape and died. The pile of sediment and bones eventually grew up to the ceiling and blocked the hole about 15 000 years ago.

The diagram above represents the cone-shaped pile within the cave.a. In which layer (1 to 5) would you expect to find the oldest group of fossils?

b. Scientists sometimes use an index fossil to date a rock layer. Describe two features of a fossil that would make it useful as an index fossil.

c. Outline one other method that the scientists could use to date the fossils in the oldest layer within the cave.

d. This cave has been a good environment for the formation and preservation of fossils. Give one reason why fossils have been well preserved in this cave.

To date 118 species of vertebrate animals have been found. One of these is the now extinct Thylacoleo carnifex, referred to as a marsupial lion because of the cat-like nature of its skull and its carnivorous habit. The scientists have only found hard parts of these extinct animals.e. Describe a particular hard part that would be useful to the scientists to determine that Thylacoleo carnifex was carnivorous.

After studying the fossilised hard parts of the Thylacoleo carnifex, the scientists have decided that the animal when alive weighed around 120 kilograms and had a very muscular body. Drawings have been made to indicate the animal’s appearance.

f. How do the scientists reach a conclusion about the animal’s appearance when only the hard parts of the extinct animal are available?

Question 25The diagrams below represent sedimentary rock strata from two different palaeontological sites.

Based on these diagrams it would be reasonable to conclude that fossils inA. strata B are the same age as fossils in strata E.B. strata A are younger than fossils in strata E.C. strata D are younger than fossils in strata F.D. strata I are older than fossils in strata G.

CHAPTER 10 EVOLUTION PART 13 Adaptation, Recent common ancestors,

Divergent evolution- Homologous features, Convergent-Analogous features, Parallel, Phylogenetic tree (cladogram) Comparisons- Structural, (Homologous features, Embryology) Biochemical(DNA, Nuclear/ mitochondrial, Hybridisation,Molecular clock, proteins-cytochrome C, haemoglobin), Biogeography. Extinction.

QuestionsWhat is a ‘recent common ancestor’?What is a phylogenetic tree?What is evolution? Is it a process or a result? ExplainApart from fossils what other evidence supports evolution.Distinguish between convergent, parallel and divergent evolution.Can an individual organism become extinct?

1. (Question 24)The scales of reptiles and the feathers of birds are considered to be homologous structures because theyA. have arisen as a result of similar selection pressures.B. have a common evolutionary origin.C. are both forms of skin covering.D. serve a similar function.

2. (Question 22)The most recent common ancestor of Homo and Kenyanthropus is represented on the diagram atA. ZB. YC. XD. W

3. (Question 13)Populations of bacteria can evolve rapidly in response to changes in the environment.One factor which contributes to this is that bacteriaA. have a single chromosome.B. have a short generation time.C. are single-celled organisms.D. have a low rate of mutation.

4. (Question 15)From this diagram it would be reasonable to conclude that

A. white-lipped snakes and black snakes lack a common ancestor.B. tiger snakes share more characteristics with brown snakes than with swamp snakes.C. death adders are more closely related to golden-crowned snakes than they are to tiger snakes.D. broad-headed snakes share a more recent common ancestor with white-lipped snakes than they do withswamp snakes.

5. (Question 17)The shingleback lizard is closely related to the bluetongue lizard. The bluetongue has a long pointed tail andsmooth scales. The shingleback has a short stumpy tail and enlarged rough scales.The evolution of these characteristics in the shingleback is an example of which type of evolution?

A. cultural B. parallel C. Divergent D. convergent

A comparison was made between human, rabbit, mouse and chimpanzee of the• DNA coding sequence of the globin gene • DNA sequence in the introns of the globin gene• Amino acid sequence of the globin polypeptide.

6. (Question 18)It is possible to conclude from this data thatA. a human is more closely related to a mouse than to a rabbit.B. the variation between chimpanzees and humans occurs in a region of the globin gene which would codefor amino acids.C. the variation in the intron sequence between human and mouse would account for some of the differencesin the amino acid sequence.D. the comparison between rabbit and human indicates that the differences in their DNA did not always makea difference to the amino acid produced.

7. (Question 19)The common evolutionary ancestry of many organisms is reflected by a geographic distribution consistent withthe former supercontinent Gondwanaland.An example of this is the distribution ofA. parastacid crayfish in South America, New Zealand, Australia and New Guinea.B. bears in North and South America, Europe and Asia.C. flying foxes in Australia, Asia, Africa and Europe.D. mockingbirds in South and North America.

8. (Question 15)Evidence for evolution includes data from comparative anatomy and embryology. When comparing two species such evidence would be obtained fromA. chromosome numbers. C. shared habitats.B. mitochondrial DNA. D. the presence of gill slits during development

9. (Question 19)The linking of the present-day distributions of organisms with past movements of continental plates is referred to asA. continuous variation. B. biogeography. C. genetic drift. D. biodiversity..Trilobites are thought to be closely related to three other groups of fossil arthropods; helmetids, tegopeltids and naraoids. The tegopeltids and helmetids are the two most closely related groups. These two groups are more closely related to trilobites than they are to naraoids. The diagram below illustrates the evolutionary relationships between these four groups.b. Write the names of these four groups in the boxes at the top of the diagram below so that the evolutionary relationships between them are consistent with the information provided.

c. Suggest one reason why all species of trilobites became extinct.

10. (Question 16)Convergent evolution may produce similar structures in two different species.This may lead toA. analogy. B. homology. C. divergence. D. embryology.

11. (Question 17The shark (a fish) and the dolphin (a mammal) are an example ofA. convergent evolution. B. allopatric speciation. C. divergent evolution. D. species radiation.

12. (Question 25Comparisons of the amino acid sequences of the α-globin polypeptide have been made between humans and a number of other vertebrates. The number of differences is shown in the table below.

Based on the information provided, the correct placement of each animal on the figure to show the evolutionaryrelationship is

A. V = cow, W = kangaroo, X = newt, Y = carp, Z = sharkB. V = shark, W = carp, X = newt, Y = kangaroo, Z = cowC. V = carp, W = shark, X = kangaroo , Y = newt, Z = cowD. V = kangaroo, W = cow, X = newt, Y = shark, Z = carp

Thylacinus cyanocephalus (Tasmanian tiger) was the largest living marsupial carnivore in Australia at the timeof European settlement. The thylacine is believed to have become extinct on 7 September 1936 when the lastcaptive thylacine died in the Hobart Zoo. There are thylacine fossils found in Tasmania and mainland Australia, but when Europeans arrived in Australia living thylacines were only found in Tasmania.

d. Suggest why thylacines were not found in mainland Australia at the time of European settlement.

Since 1936 there have been many reported sightings of thylacines in Tasmania and along the southern coast of Victoria.e. Explain why scientists still believe thylacines are extinct.

The dingo is a eutherian mammal and the thylacine is a marsupial mammal. Scientists regard these two carnivores as an example of convergent evolution.f. Explain why scientists would regard the thylacine and the dingo as an example of convergent evolution.

SECTIONCHAPTER 11 NATURAL SELECTION PART 14Variation, Selection pressure, Gene Pool, Wild-type allele, Polymorphism, Mutation, Genetic drift, Bottleneck effect, Founder effect,Darwin, Lamarck, Wallace, - Speciation, Isolating mechanisms (pre-reproductive and post reproductive) Allopatric speciation,

QuestionsWhat is the modern version of evolution by Natural selection? (refer to answer checkpoints pg158 3-35)What is meant by variation when looking at members of a population? These can be listed under 3 main headings. What is the most important source of variation –recombination during meiosis or mutation? Justify your opinion explaining the significance of each.Distinguish between genotypic and phenotypic variation.How are Natural selection and Evolution related?If Natural Selection is to occur, what must exist in a population?Is selection at the phenotypic or genotypic level? Explain what sorts of individuals are most likely to survive in a particular environment.If an individual is to pass on its alleles to the next generation how long must it live?What is meant by the terms gene pool and allele frequency?In terms of the gene pool and allele frequency , what is the difference between gene flow and genetic drift?When will the two processes have the greatest affect on allele frequency?What is a species? (What is the most important test to find members of the same species)What is allopatric speciation? What are some geographic barriers.Name some mechanisms which could prevent interbreeding between closely related species.

1. Question 24The following statements (not in correct order) summarise the steps in natural selection.1. Some individuals are better suited to a particular environment.2. Over time there is an increase in particular characteristics in the population.3. There is variation within a population, some of which is genetic.4. Individuals better suited to the environment are more successful at survival and reproduction.The order of statements which best describe natural selection areA. 1, 3, 2, 4B. 3, 1, 4, 2C. 2, 3, 1, 4D. 1, 2, 4, 3

2. Question 17Two types of bird were originally thought to be different species. However, recently a group of biologists has agreed that they are the same species. The biologists must have found out that the two types of bird

A. look alike enough to be thought one species.B. are separated by a geographical barrier.C. successfully interbreed in nature.D. live in the same habitat.

3. Question 14Many frog species inhabiting tropical rainforests have evolved green skin colour.It would be reasonable to conclude that the main selection pressure responsible for the evolution of green skincolour isA. predation. B. climate. C. reproduction. D. infection by pathogens.

4. Question 16Genetic drift isA. most evident in large populations. B. the gene flow between populations.C. random changes in the gene pool of a population. D. the gradual change in phenotypic frequency resulting from natural selection.

5. Question 20For a species living in an unchanging environmentA. there are no selection pressures.B. the selection pressures remain constant.C. the only selection pressure is genetic drift.D. all individuals are equally suited to the selection pressures.

6. Question 25The variation in allele frequencies between several isolated populations can be due to genetic drift. Genetic drift is likely to be observed whenA. there is gene flow.B. the mutation rate is high.C. there are strong selective pressures.D. a population is reduced to a few individuals.

7. Question 7Warfarin is a poison used to control rat populations. Figure 13 shows changes in the proportion of rats resistantto warfarin in a particular population over a period of about 4 years. High levels of warfarin were used on thispopulation during Year 2 but poisoning stopped at the end of this period. Rats are reproductively mature at anage of three months and can breed about every three weeks.

a. Explain the process which led to the increase in the percentage of resistant rats during Year 2.

b. Using the data in Figure 13, explain what can be concluded about the selective advantage to a rat of beingwarfarin-resistant compared to being nonresistant in an environment without warfarin.

8. Question 8Figure 14 shows the natural distribution of a mammal, the red-necked wallaby, Macropus rufogriseus.

a. Give two reasons why populations of this species in Tasmania have not evolved into a separate species despite being geographically isolated by the waters of Bass Strait.

b. Another mammalian species common in Tasmania is the Eastern Quoll, Dasyurus viverrinus. This species,which is about the size of a domestic cat, was widely distributed in south-eastern mainland Australia untilabout 50 years ago. It is now believed to be extinct in Victoria and possibly over the rest of its formerrange in mainland Australia.Give two possible reasons for the extinction of this species in mainland Australia.

Figure 4 shows the results of a breeding experiment with the vinegar fly Drosophila melanogaster. In each of the first 25 generations the smallest flies were selected to produce the next generation. After 25 generations the selection was reversed. From generation 25–35 the largest flies were chosen to breed the next generation.

9. (Question 21)With respect to the genes controlling body size, the results of the experiment suggest thatA. after 25 generations there was no genetic variation.B. for the first 25 generations there was no genetic variation.C. selection between generations 25 and 35 had a significant effect on average body size.D. if selection for small body size continued after generation 25, average body size would continue to fall.

10. (Question 16)Geographical isolation is important in assisting the process ofA. adaptation. B. gene flow. C. speciation. D. fossilisation.

11. (Question 17)Snakes and legless lizards evolved separately from ancestors with legs. The lack of legs in these reptiles is an example ofA. analogy. B. divergence. C. founder effect. D. polymorphism.

A single gene locus with two alleles determines the colour of snapdragon flowers. A flower may have one of 3 phenotypes. phenotype genotype red CRCR

pink CRCW

white CWCW

A student investigated the frequency of each of these phenotypes in a population of 100 snapdragon plants. The frequency of each phenotype and genotype is shown in the table.

12. (Question 23)From this data it can be concluded thatA. the number of CW alleles in this population is 20.B. the number of CR alleles in this population is 120.C. the proportion of CR alleles in this population is 0.4.D. the total allele pool for this locus in this population is 100.

In a species of marine snail the colour of the shell is controlled by a single gene with 2 alleles, G and g. Allele frequencies for this locus were determined in six populations of snails. These populations were located close to each other, but not able to interbreed. The frequencies of the G allele are shown in the table below.

13, Question 24The frequency of the g allele in population 5 isA. 0.15 B. 0.3 C. 0.35 D. 0.7

14. b. The incidence of antibiotic resistant Streptococcus pneumonia has increased in the last 15 years. Approximately 40% of infections by this bacterium are resistant to commonly used antibiotics.

i. Explain how the increase in bacteria resistant to antibiotics has occurred.

ii. What is the selective agent associated with the increase in antibiotic resistance in Streptococcus pneumonia?

15. Question 10All the alleles in a population are referred to as theA. phenotypic family. B. proteome. C. gene pool. D. genotype.

16. Question 16The founder effect and bottleneck are examples ofA. gene flow. B. speciation. C. genetic drift. D. selection pressures.

17. Question 14The frequencies of the phenotypes of the MN blood group were measured in a European population. Of 100 individuals, 40 were blood type M, 20 were blood type MN and 40 were blood type N.From this data it is possible to conclude thatA. the frequency of the LN allele is 0.3.B. the frequency of the LM allele is 0.5.C. there are 60 LM alleles in this population.D. there is a total pool of 100 alleles at this locus for this population.

16. (Question 15)Natural selection acts upon an organism’sA. habitat. B. Genotype C. phenotype. D. environment.

17. (Question 23)In some autosomal recessive conditions in humans, the homozygous recessive genotype results in death before reproductive age. Despite this, the allele for the recessive trait is maintained in the population. Maintenance of this allele in the population is most likely the result of A. mutation.B. migration between populations.C. the heterozygote being biologically fitter than either of the homozygous genotypes.D. individuals with the homozygous dominant genotype leaving more offspring in each generation.

18. Rock wallabies, Petrogale lateralis pearsonii, on Pearson Island off the coast of South Australia have had no genetic contact with mainland rock wallabies since they were isolated by rising sea levels at the end of the last glacial period, around 10 000 years ago. Scientists have taken blood samples from the wallabies and compared the distribution of unique DNA sequences called microsatellites, which are scattered across the wallabies’ chromosomes. These microsatellites give a measure of the population’s genetic diversity, or lack of it. In this case the microsatellite data showed that the Pearson Island population has low genetic diversity.The scientists concluded that the Pearson Island population of rock wallabies has been through a genetic bottleneck. A genetic bottleneck is an example of genetic drift.a. Explain how a genetic bottleneck may lead to a decrease in genetic diversity.

Despite the Pearson Island rock wallabies’ lack of genetic diversity, the population size has been maintained over many generations. In fact, the wallabies appear to be thriving.b. Suggest one reason for the wallabies’ success despite the lack of genetic diversity within the population.

c. The population of rock wallabies on Pearson Island is most closely related to small populations of rock wallabies in southern Western Australia. Some scientists argue that some individuals from the southern Western Australian populations should be released onto Pearson Island. Give one reason for this suggestion.

18. Question 8The Isthmus of Panama is a narrow strip of land that joins North and South America.

The land bridge formed approximately 3 million years ago.

Snapping shrimps, genus Alpheus, can be found on either side of the land bridge. The two groups arephenotypically similar. However when the males and females from either side of the land bridge were brought together they snapped aggressively at each other and would not mate. They are now considered to be twodifferent species.a. Why is the inability to mate sufficient evidence to call the two groups different species?

b. What type of speciation has occurred in the snapping shrimp?

c. Explain how the differences between the shrimp on either side of the land bridge could have arisen.

Mesosaurus was a giant reptile that lived about 270 million years ago. The average Mesosaurus measured aboutone metre in length, had webbed feet, a long tail and numerous sharp teeth. Fossils from Mesosaurus have beenfound in only two places; the eastern side of South America and the west side of South Africa.19. Question 23Mesosaurus was most likely to beA. a land animal that ate plants.B. an aquatic animal that ate plants.C. an aquatic animal that ate small fish.D. a land animal that ate small animals

20. Question 7The graph below shows changes in allele frequencies at a single locus with two alleles B and b in two verylarge populations. The phenotype resulting from allele B is the dominant phenotype. In one population thereis selection against the homozygous recessive phenotype, and in the other population there is selection againstthe homozygous dominant and heterozygous phenotypes.

a. In which population, X or Y, is selection ocurring against the homozygous recessive phenotype?Explain.b. Describe what has occurred in population Y by the 60th generation.c. Given the results in population Y, why has a similar pattern of events not occurred in population X?CHAPTER 12 HOMINIDS AND HOMININS PART 15 Primates, Apes and monkeys, Characteristics- fossil structure, Bipedalism, opposable thumb, Large brain,

Australopithecus(afarensis,africanus), Paranthropus, Homo habilis, Homo erectus, Homo neanderthal, Homo florensis, Homo sapiens, Cultural evolution, tools, fire, painting, rituals, language,

QuestionsList the major characteristics that distinguish mammals from other animals.List the major characteristics that distinguish primates from other animals.Why are opposable thumbs important to primates?Why do netballers and footballers require good stereoscopic vision?What features distinguish hominins from other primates.Describe the location changes of the foramen magnum.What changes in jaw and tooth structure have occurred during human evolution? Describe how these changes may be linked to an increase in brain size.Skull shape and size have changed during human evolution. Describe their changes and their significance.What has been the limiting factor to a further increase in skull size?What selective pressures might have led to an increased brain size?Is it true to say that man has evolved from chimps and gorillas. If not , rephrase the statement.Biologists talk of ‘physical evolution’ and ‘cultural evolution’. Many would argue that the latter is now more important. What is the definition of culture?

1. Question 23The skull of Kenyanthropus was discovered in 1999 in sediments between 3.2 and 3.5 million years old. Theskull is unusual in that it shows the combination of a small braincase, flat face and small teeth.The following characteristics would be seen in the skull of Homo.A. a small braincase and small teethB. a small braincase and a flat faceC. a flat face and small teethD. a flat face and large teeth

2. Question 6The table below shows the number of nucleotide differences between a region of mitochondrial DNA in humans, chimpanzees and a Neanderthal.

a. Based on the data in the table, which individual is most closely related to the Neanderthal?

b. The differences between the mitochondrial DNA recorded are the result of base substitutions. There are 77 nucleotide differences between Human 1 and Chimpanzee 1. Explain why 77 nucleotide differences is a minimum number of base substitutions.

c. The Neanderthal DNA was extracted from a fossil approximately 25 000 years old. i. What other type of information obtained from the fossil could be used to assist in determining the evolutionary relationship of Neanderthals with humans and chimpanzees?

ii. What method would be used to estimate the absolute age of the Neanderthal fossil?

iii. What method could be used to determine the relative age of the Neanderthal fossil?

3. a. i. Which of the two skulls is the more ancient? ii. List one characteristic of this skull in support of your choice.Although extensive searches have occurred, fossils classified in the genus Australopithecus and other early hominid genera have been found only in Africa. Assume that these fossils in fact only exist in Africa.b. What is a possible explanation for these fossils being limited to Africa alone?

4. Question 18Homo neanderthalensis lived in Europe and Western Asia from approximately 250 000 to 28 000 years ago. In comparison to modern day humans the Homo neanderthalensis possessed relativelyA. smaller brains. B. smaller noses. C. higher foreheads. D. more prominent brow ridges.

5. Question 19Evidence of hominid cultural evolution can be found in the fossil record.This evidence would includeA. position of the attachment of the spine to the head. B. length of arm bones in comparison to leg bones.C. number of teeth present in the skull. D. presence of stone tools.

6. Question 21 Hominids are believed to have evolved in Africa becauseA. the oldest hominid fossils have been found in Africa. B. the most hominid fossils have been found in Africa.C. monkey fossils were found in Africa. D. Africa is the oldest continent.

7. Question 22 Consider the following diagrams of skulls.

The skull most likely to be that of a chimpanzee is A. W B. X C. Y D. Z

Question 23There is evidence that Homo sapiens and Homo neanderthalensis coexisted in Europe more than 30 000 yearsago. Both of these species left signs of cultural evolution from this period. An example of evidence which would show that cultural evolution was occurring in these groups at this time isA. animal remains close to a Homo skeleton . B.. male and female skeletons in the same area. C. drawings and carvings on rocks D. Homo sapiens and Homo neanderthalensis skeletons in the same area.

Question 22Consider the following diagrams of Hominid skulls.

The correct sequence of evolution, from oldest to youngest, of the Hominid species shown isA. 2, 3, 1 B. 3, 1, 2C. 1, 2, 3 D. 2, 1, 3

CHAPTER 7 STRUCTURE OF CHROMOSOMES PART 11. B 2. B Each rabbit only has a pair of homologous chromosomes so it can only have 2 alleles. 3. D 1, 1 1,2 1,3 1,4 2,2 2,3 2,4 3,3 3,4 4,4 10 combinations 4. B 5. 19 6.C Phenotype= Genotype+Environment 7.D 8.C 9.D 10.A 11.A 12.C 13.D 14.C 15. B 16.aA bB17 a. The gene locus is the position of the gene along the chromosome. b. The sequence of bases/nucleotides would be different.

CHAPTER 7 CELL DIVISION PART 21. B 2.C 3. D 4. C, E, D, B, F, A4 b Any two of• increases variation in the gametes or offspring• the recombining of maternal and paternal alleles• exchange of alleles or genetic material between homologous chromosomes or between non-sister chromatids resulting in new combinations of alleles.Some students incorrectly stated that the homologous chromosomes exchanged genes. Homologous chromosomes havethe same genes; they may have different forms of a gene. Therefore, they may exchange different forms of a gene. Thedifferent forms of a gene are called alleles.Ci E Cii In anaphase one chromosomes are double stranded. The chromosomes shown moving to the poles in E are doublestranded.5. D6. a The diploid chromosome number is the number of chromosomes found as homologous pairs in a cell (organism) or thenumber of chromosomes in a cell (organism) that have two copies of each specific chromosome.Responses that were not awarded marks included ‘number of chromosomes in a somatic cell’ and ‘2n = 16’.b. Each daughter cell would contain 15 chromosomes.C i. Each gamete of a female will contain eight chromosomes.ii. Half of the male gametes will contain eight chromosomes and the other half will have seven chromosomes.Students needed to have both answers to be awarded full marks. Some incorrect answers included seven and a halfchromosomes and 15 chromosomes.7. a. 53 b. Meiosis c. Students were expected to be able to accurately show the following: • separation of homologues in meiosis I (at the end of the first divisions, showing one chromosome in each cell as double- stranded) • separation of chromatids in meiosis II (at the end of the second division, showing chromosomes in four cells (two of each type) as single-stranded) • the process of division involving cytokinesis, the spindle, etc. d. Chromosomes are not homologous and therefore will not pair during meiosis.

CHAPTER 7 CROSSES PART 3

1 D 2 A 3 D (A or b could be true but you can’t tell because you don’t know which alleles are linked on the same chromosome, C isn’t true because there will be more of one than the other) 4 B 5. D 6. C 7 A8. The presence of four phenotypes in the F2 generation indicates that two genes are involved. One of the genes determinescoat colour. One allele for coat colour codes for black coats, the alternative allele codes for white coats. The gene forcoat texture has two alleles; one coding for rough texture, the other allele codes for smooth texture.OR there are four phenotypes in the F2 generation with a ratio of 9:3:3:1 indicating the parents are heterozygous for twogene loci each with two alleles.bi RR BB rough, black rr bb smooth, white bii Bb Rr biii BBrr or Bbrr

9. D (1,1 1,2 1,3 2,2 2,3 3,3 6 genotypes A, AB, B, O 4 Phenotypes) 10. C 9:3:3:1 11. B (AWac x aSac 4 genotypesAWaS, AWac, acaS, acac White, White, Sable, copper 3 phenotypes . ) 12. C 13. B (could be AO x BO) 14. B 9:3:3:1= 16 into 1600 so 100 for 1. 15. C 16. D (1,1 1,2, 1,3 2,2 2,3 3,3 6 genotypes) 17.C 18. Y= Yellow y= grey Genotype ratio 1 YY: 2Yy : 1 yy Phenotype ratio: 3 Yellow: 1 grey. They ratio is not this because the YY is lethal and these offspring die.

CHAPTER 7 MONOHYBRID CROSSES X LINKED PART 41. C 2. B 3 A 4 B 5 DdxDd , dd 6.Cross 2 because two resistant plants are crossed and yet 12 offspring who are sensitive are produced which indicates this trait was carried hidden in the parents (recessive). R=resistant r = sensitiveRr x rr = ½ Rr : ½ rr 7. B 8. A Patterned is dominat because patterned parent frogs produced non patterned offspring indicating this characteristic was carried by the parents (recessive). b) P=patterned p= non-patterned PpxPp 1PP:2Pp:1pp 3 patterned : 1 non patterned is the expected outcome.

CHAPTER 7 PEDIGREES PART 51. Autosomal dominant. Dominant because two parents produce a child with a different phenotype to each of them which means they must have been carrying the normal allele. Autosomal because if it was X-linked a father who had the dominant allele gives it to his daughters so any girls should have psotosis.2. ai Pedigree A ii) It can be autosomal dominant or recessive as long as the parent with the dominant phenotype is heterozygous. It can’t be X-linked because there is a mix each of the sexes with and without the trait. bi Pedigree B is dominant because two parents with the trait have had offspring without it indicating that the normal trait was carried (recessive). ii. It is not X linked because a father with the trait gives the X to his daughters so all daughters would be affected. 3. A 4 A 5. Rr x Rr because II 1 is liver coloured which requires rr and Bb x Bb because II 3 is red with requires bb. They each have one R and B because of their own colouring. 5b i) Rr Bb ii) rrBb a cross would result in all the phenotypes being possible 6a) I1 XHXh I2 XhY b) 100% c) 50% d) I1- Gg I2- Gg II1 gg II2 GG or Gg

CHAPTER 8 PROTEIN SYNTHESIS PART 51 C 2 D 3a) Transcription b) PremRNA c) the pre-mRNA has the introns removed and the exons are joined back together to form the final mRNA d) Translation e) i t RNA ii) The role is to bring in the correct Amino Acid to join the polypeptide chain based on the codon on the mRNA fitting the anticodon on the tRNA.

ATG CCA CCG TGG GTA TGC 5 BTAC GGT GGC ACC CAT ACG

CHAPTER 8 DNA STRUCTURE/ GENE REGULATION PART 61. B 2. A 3. D 4.C 5.A 6 A 7 a) Organisms regulate the expression of their genes so that the correct proteins are present in the cell for it to carry out its role in the organism. They are also only produced when needed so that energy and resources are not wasted on producing proteins which can’t be utilized. B) RNA polymerase c) The production of enzyme X will commence as the gene is free to be transcribed because the RNA polymerase is no longer blocked. D) The tryptophan binds to the repressor protein and changes its shape so that it can no longer bind to the DNA.

CHAPTER 8 MUTATIONS PART 71. B 2. B 3a)9 b)i. 18 ii. The chromosomes aren’t homologous so they don’t pair up during meiosis (prophase 1) c) polyploidd) The new plants now have a pair for each chromosome so they canpair up at meiosis and form gametes with one copy of every chromosome. 4. B. The phenotype are they characteristics expressed by the organism. ii. It had two copies of every chromosome it’s just that two of the copies were stuck to each other. This didn’t alter their functioning. c) Sperm 3- 9/18 by itself Sperm 4 – A 9 by itself 5. d) The 9 by itself doesn’t have any information for the 18. e) ¼ chance to get the same as the father . The would have to be the one with the 9/18 by itself.5. A mutation would have occurred which changed the phenotype so that it was no longer susceptible to the antibiotic 6. D 7a. AUG AAA UUC UCG AAU AGC bi. A point mutation called a substitution. Ii. The DNA code changes from TTT to TTC so originally the mRNA was AAA not it is AAg. The original amino acid was Asn and its Lys. This change of an amino acid may have no affect on the poly pepetide produced or it may make the protein more or less efficient. c) All of the code from then on will be read out of order so it will affect eery amino acid from then on. This is called a ‘frameshift’ mutation. AGC TTA TCG will now be ACT TAT and only CG left, so the codes have been totally changed and reduced to 2 codons rather than three.

CHAPTER 9 GENETIC ENGINEERING PART 9 1. C 2. a) A recombinant plasmid (A vector). B) Restriction enzyme C) DNA ligase D) To identify the bacteria which have taken up the plasmids you can include an antibiotic resistance gene and then grow the bacteria on a plate containing that antibiotic. Only the bacteria with the plasmid will grow. E. There are many reasons why a farmer might want to do this – normally crop spraying is done by air and the insecticide can be carried by the wind and kill insects over a wide area. The insects may be beneficial in the food chain feeding birds etc. More insecticide in the environment can encourage selection of resistant insects. It costs more in time and insecticide to spray. F. He can kill the weeds easily even when the crop is growing. He doesn’t have to introduce heavy machinery amongst the crop to kill the weeds or hire costly labourers. G. Consumers in many countries are concerned about using crops that have been genetically modified and some countries are banning the import of GM foods. If the farmer is selling to those markets he would be unable to if his crops where fertilised by the GM crops h. When there is no gap between the crops there is a 10% contamination rate on the edge of the crop and contamination even within the crop. Moving the crops 5 metres or even 7 metres apart, lowers the contamination rate but it doesn’t totally eradicate it. So the farmer could still not claim his crops where GM free at 7 metres apart and once one plant grew with the transgene it could pollinate others the next year and increase in the crop.3.a) Sticky ends 4a a small, circular chromosome. b) Transformation c) i. No growth ii. The bacteria are sensitive to the antibiotic tetracycline and therefore do not grow. D) Plate A: The bacteria are able to grow as there is no tetracycline present. Plate C: Only those bacteria which take up the plasmid can grow in the presence of tetracycline. (It was important to only some of the bacteria take up the plasmid) 5. C

CHAPTER 9 GENETIC ENGINEERING PART 2 PART 10 AND 111a) The polymerase enzyme catalyses the production of a new strand of DNA or is involved in making multiple copies of DNA or amplification of DNA AND DNA polymerase replicates the DNA by extending from the primer or by complementary base pairing or by using the original DNA as a template. (there were 2 marks awarded for this question so 2 separate remarks needed to be made)b) A DNA fragment will move according to its charge and molecular weight (size) or DNA is negatively charged and moves to the positive pole; smaller DNA fragments move further or faster than larger fragments.c) There is only one band in lane 2 because individual 2 is homozygous, the others on the gel are heterozygous or the two fragments of DNA are the same size or the number of repeats in the two fragments is the same.d) There are 5 different alleles at the HUMTHO1 locus represented on the gel.e) DNA piece A has the greater number of the 4 base repeat sequence. The greater the molecular weight of the sample the smaller distance the sample will move from the loading well.f) The bands on the gel for suspect 5 match the sample of blood found on the victim, which was not the victim’s blood (lane 3).2. B 3. b) The enzyme does not cut the DNA, or the enzyme no longer recognises the sequence

C) 5. a) Making an exact genetic copy of an organism. b) 19 ii) female . The donor cat was female so the genetic material would be female c) There are two homologous chromosomes for each region studied. This part was not well answered. Many students incorrectly thought the two numbers were representative of the two strands of DNA. D) The values are different because the regions have slightly different lengths 154/160 One is 154 nucleotides long and the other 160. Most students attempted this question, however, many students again related the numbers incorrectly to the two strands of DNA or made reference to the occurrence of mutations having occurred, which was also incorrect. E) All of CC DNA which was studied was the same length as the donors DNA but did not match any of the regions in the surrogates DNA. This implies that it was a clone of the donor rather than and offspring of the surrogate. 6. Selective breeding means that humans control which bull will mate with which cow. In random mating the animals themselves choose. F. Selective breeding means that the traits found in the herd will be those considered desirable by the breeder. This will reduce the variability in the herd to these characteristics.g. The animals may reach the stage where they are unable to reproduce themselves. They may lose the behaviour involved in mating or some other factor of the mating process.7. A

CHAPTER 10 FOSSILS PART 12Q1. C 2.C 3. B 4 D 5.a) They had a tough exoskeleton. b) Trilobite fossils may be abundant for any one of the following reasons: • were widely distributed geographically • able to survive in a wide range of habitats • existed for 300 million years• lived in an environment where fossilisation can occur readily.Q6 A) Layer one b) Any two of the following answers were acceptable:• an index fossil must be distinctive and easily recognisable • an index fossil should be abundant • an index fossil lived a short range through time • an index fossil lived in a wide geographic distribution.Many students defined an index fossil rather than describing the features of a fossil that would make it useful as anindex fossil. C) One of the following was an acceptable method:• radioactive isotopes (for example, potassium/argon, argon/argon, lead/lead, uranium/lead)• stratigraphy, which involves a study of the relative ages of the surrounding rock layers.Carbon dating was not acceptable, as it only dates to within 30 000 – 50 000 years.d) The remains have been undisturbed from both scavenging animals and the weather. Students were not awarded a mark if they stated the cave protected the fossil and gave no further information as to what it was offering protection from.e) The teeth or jaw would be used to determine that the animal was carnivorous. F) The scientists would compare the hard parts of the extinct animal to living species. They would look for points of muscle attachment to animals living today and for indications on the fossil bones where ligaments might have been attached to gain insight into musculature, or compare the sizes and types of bones of the fossil to animals living today to help establish the weight of the fossil. 7. C

CHAPTER 10 EVOLUTION PART 131. B 2. B 3 B 4.D 5.C 6.D 7. A 8. D 9. B b)Trilobites, tegepeltids, helmetids, naraoids. C) The environment altered and there were no variations within the gene pools which could cope with it.A global climate change must have occurred to make all species of trilobites extinct. Some students mentioned a climate change without any indication of the global nature of the change and therefore were not awarded a mark. 10. A 11. A 12. B 13.d) They had become extinct on the mainland due to not adapting to a new selection pressure (maybe the arrival of the aboriginals? e) Sightings are not concrete enough evidence so show that they still exist. You would need some physical evidence – an actual animal. f) Mammals and marsupials diverged a long time ago into a all the different forms of today. These two have similar features because they are both carnivores and similar features (large teeth, strong jaws etc) would have been separately selected for to arrive at a similar function.

CHAPTER 11 NATURAL SELECTION PART 141. B 2.C 3.A 4.C 5.B 6.D7. a)The process that led to the increase in the percentage of resistant rats included warfarin resistant rats existing in the population before the use of warfarin and when warfarin is used non-resistant rats are killed and warfarin resistant rats survive to reproduce, and pass on the allele for resistance or warfarin resistance is inherited and so is passed on to next generation and over several generations the proportion of warfarin resistant rats increases.Students who could clearly express their ideas in a logical way were more likely to be awarded full marks. Students must be encouraged to formulate answers to questions that require a detailed account of a concept/s.b) From the graph it can be seen that the percentage of resistant rats decreases when the use of warfarin is discontinued (years 3 and 4) therefore resistant rats are at a disadvantage in a non warfarin environment or resistant rats are less fit in a non warfarin environment or non-resistant rats are at a selective advantage in a non warfarin environment.The question asked students to use the data in figure 13. Therefore, students were expected to explain how the data was used in arriving at their conclusion, for example more successful answers specifically mentioned that the number of rats decreased in years 3 and 4.8a) The populations of red-necked wallabies in Tasmania have not been isolated long enough from the populations of red-necked wallabies on the mainland for sufficient genetic differences to accumulate and the populations of the red-necked wallabies occupy similar habitats so similar selection pressures or a specific example of a selection pressure.Students could score 1 mark if they gave a specific example of a similar selection pressure but were not awarded 2 marks if they gave two examples of similar selection pressures. b)The Eastern Quoll may be extinct on the mainland of Australia because a disease may have spread through the mainland populations and killed all quolls or a predator may have been introduced on the mainland which killed all of the quolls or the quoll habitat may have been destroyed when humans cleared much of the mainland for farming.One-word answers such as ‘hunting’ or ‘predators’ are unlikely to be awarded marks. The space provided for the answers indicated the detail needed in the response.9 A 10.C 11. A 12.B 13. B 14. B. When antibiotics are used to treat an infection if there are any resistant bacteria present and the immune system does not destroy these they will multiply. The more times the same antibiotics are used the more resistant bacteria are selected for and so their numbers and percent of the bacterial population increases. C) The Antibiotics which are used to treat the infections. 15. C 16. C 17. B 16. C 17. C 18. A genetic bottleneck results in a severe reduction in population size and as a consequence allele frequencies may change by chance, which will decrease genetic diversity.b) There were many acceptable answers to this question. Some of these included: • the wallabies are well adapted to the environment • there have been no changes in the selection pressures acting on the wallabies • the wallabies are protected from new predators as they are on an island C). The scientists want to increase the genetic diversity of the wallabies, giving the population a better chance of survival if the environment changes. 18 (again!) a) Inability to mate is not enough evidence; it is the inability to produce viable offspring which is important. (Stupid question!) b) Allopatric speciation Incorrect answers included divergent evolution and founder effect. C) There were different selection pressures in the two environments, which allowed differences in allele frequencies to develop. •Over time there is an accumulation of genetic differences which changes a trait, such as mating behaviour. This question was poorly answered and was generally not recognised as a ‘natural selection’ question. Students needed to provide both of the above points for two marks 20a) Population X has selection against the homozygous recessive phenotype, as the frequency of the b allele is decreasing over time. b) Any of: • the allele b has become fixed • the frequency of allele b = 1 • only one allele in population Y. Some students misread the graphs for parts a. and b.Students should be encouraged to take greater care when reading the axes and applying this knowledge to the question asked. c) In population X there is selection against bb, but there are b alleles still present in the heterozygotes that survive and will pass on these alleles to the next generation. This question was not well done. Many students did not realise that the b allele would be present in heterozygotes and would therefore persist in the population.

CHAPTER 12 HUMAN EVOLUTION PART 151. C 2. a) Human 1 b) Mutations are reversible and some substitutions may reverse an earlier change making 77 nucleotides the minimum number of base substitutions. C) Skeletal structure/morphology may be used to assist in determining the evolutionary relationship of Neanderthals with humans and chimpanzees. ii) Carbon dating can be used because the fossil is approximately 25 000 years old. iii)To determine the relative age of the fossil, an index fossil may be used. Stratigraphy which states that the oldest stratum is at the bottom and younger layers lie above can also be used 3 a) Skull one b) Skull one has a more rounded jaw line or large jaw relative to overall skull size, or skull one has more prominent eyebrow ridges or skull one has a smaller brain case. Students need to be reminded that when a question asks for one characteristic, they should only give one characteristic. If they include more and one is an incorrect response they may be penalised. C) The genera may have become extinct before the migration out of Africa

could have occurred OR there were no selection pressures placed on the genera forcing them to move, so they remained in Africa until extinct. 4. D 5. D 6. A 7. B 8. C 9. A