Genetics

http://www.youtube.com/watch?v=CBezq1fFUEA&list=PL3EED4C1D684D3ADF

4.1.1

State that eukaryote chromosomes are made of

DNA and proteins.

4.1.2

Define gene, allele and genome.

• Gene- heritable factor that codes for a certain trait– Ex: eye color

• Allele- what will be expressed in that gene– Ex: blue eyes

• Genome- Collection of all of an organism’s genes– usually encoded in DNA– HGP

4.1.3

Define gene mutation.

• Sequence change• Different amino acid possible• Can be beneficial• Mutagens

4.1.4

Explain the consequence of a base substitution mutation in relation to the processes of

transcription and translation, using the example of sickle-

cell anemia.

• 1/655 African Americans

• Single base change• Change beta chain

shape (needles)

4.2.1

State that meiosis is a reduction division of a diploid

nucleus to form haploid nuclei.

4.2.2

Define homologous chromosomes.

Same•Length•Loci for genes•Shape

4.2.3

Outline the process of meiosis, including pairing of

homologous chromosomes and crossing over, followed by two divisions, which results in

four haploid cells.

• http://www.youtube.com/watch?v=qCLmR9-YY7o&list=EC3EED4C1D684D3ADF

Interphase

• Chromosomes not condensed• DNA replicates- S1

Prophase I

• Nuclear membrane breakdown• DNA condensing• Spindle fibers• Centrioles move• Move towards equatorial plate• Longest phase of meiosis

Metaphase I• Line up in center• Chromosomes most condensed• Crossing over (Metaphase and Prophase)-

chiasma• Independent assortment

Anaphase I

• Homologous pair movement• “Arrow shape” of pairs

Telophase I

• Sets at opposite sides• Nuclear membrane may reform (species)• Cleavage furrow• End meiosis I

Prophase II

• Nuclear membrane breaks down• Spindle fibers reform

Metaphase II

• Line up center• Independent assortment

Anaphase II

• Spindle fibers contract• One chromatid per pole

Telophase II

• Nuclear membranes form• Haploid• Crossing over variation

4.2.4

Explain that non-disjunction can lead to changes in chromosome number,

illustrated by reference to Down syndrome (trisomy 21).

• Can happen in Meiosis I or II

• Result in too few or too many of one chromosome

• Monosomy-more deadly, trisomy

4.2.5

State that, in karyotyping, chromosomes are arranged in pairs according to their

size and structure.

• Pictures during metaphase• 23 pairs

4.2.6

State that karyotyping is performed using cells

collected by chorionic villus sampling or amniocentesis, for

pre-natal diagnosis of chromosome abnormalities.

• Obtain fetal cells to do a karyotype to find out if baby has disorder like Downs

• Amniocentesis- removal of amniotic fluid containing fetal cells– Centrifuge separates chromosomes into size

• Chorionic villus sampling- samples chorion – Must be after 8 weeks of pregnancy– Uses catheter

4.2.7

Analyse a human karyotype to determine gender and

whether non-disjunction has occurred.

4.3.1

Define genotype, phenotype, dominant allele, recessive allele, codominant alleles,

locus, homozygous, heterozygous, carrier and test

cross.

4.3.2

Determine the genotypes and phenotypes of the

offspring of a monohybrid cross using a Punnett grid.

• 2:2 ratio• Heterozygous

crosses 3:1 • Each fertilization

independent of others

• Larger the population, the closer to expected ratio

4.3.3

State that some genes have more than two alleles

(multiple alleles).

• Creates more phenotypes and genotypes

• Example: Rabbit coat colour(C) has four alleles which have the dominance hierarchy: C > cch > ch> c

• This produces 5 phenotypes, Dark(C_) , Chinchilla( cchcch), light grey (cchch ,cchc), Point restricted (ch ch, chc) and albino (cc)

4.3.4

Describe ABO blood groups as an example of

codominance and multiple alleles.

• I is immunoglobulin• The Allele hierarchy is IA = IB > I• When A and B present, both expressed and

both mask O

4.3.5

Explain how the sex chromosomes control

gender by referring to the inheritance of X and Y

chromosomes in humans.

• 23rd pair• X much longer than Y• X from mom Y from dad in boys• Theoretically 50/50 chance for gender, but

some have predisposition• SRY gene (supposed to be found on tip of Y)

determines gender

4.3.6

State that some genes are present on the X

chromosome and absent from the shorter Y

chromosome in humans.

4.3.7

Define sex linkage.

• Genes on non-homologous region of X• Females spare tire other X, males don’t have

this• More common in males• Boys inherit these from mom

4.3.8

Describe the inheritance of colour blindness and

hemophilia as examples of sex linkage.

• Genes on non-homologous part of X• Males always get affected gene from mother• Males cannot pass on affected gene to sons,

but can to daughters

• Color blindness– Red green color blindness sex linked recessive– More common in males

• Haemophelia – Sex-linked recessive– Cannot produce clotting factor

4.3.9

State that a human female can be homozygous or

heterozygous with respect to sex-linked genes.

4.3.10

Explain that female carriers are heterozygous for X-linked recessive alleles.

• Carriers are heterozygous women• Have both recessive and dominant alleles• Females have 2 long X chromosomes• Dominant overpowers recessive

4.3.11

Predict the genotypic and phenotypic ratios of offspring

of monohybrid crosses involving any of the above

patterns of inheritance.

PTC Test

• Determined by a single gene• Ability to taste=T, inability=t• Practice– If heterozygous parents were crossed, predict the

genotypic and phenotypic ratios– What is the likelihood that the fourth child is a

taster?– What is the likelihood that the first three kids will

be non-tasters?

3 tasters:1 non-taster

75%

.25x.25x.25=.015=1.5%

Roosters

• A rooster with grey feathers is mated with a hen of the same phenotype. Among their offspring 15 chicks are grey, 6 are black and 8 are white.– What is the simplest explanation for the

inheritance of these colors in chickens?– What offspring would you expect from the mating

of a grey rooster and a black hen?

codominance

50% black, 50% grey

Flowers

• In snapdragons, red flower color is incompletely dominant over white flower color; the heterozygous plants have pink flowers.– If a red-flowered plant is crossed with a white-

flowered plant, what are the genotypes and phenotypes of the plants of the F1 generation?

– What genotypes and phenotypes can be produced in the F2 generation?

All Rr, all pink

25% RR red, 50% Rr pink, 25% rr white

• Haemophilia or red-green color blindness for sex linked

4.3.12

Deduce the genotypes and phenotypes of individuals

in pedigree charts.

•Squares male, circles female•Red=affected, blue=unaffected•Horizontal line= mated•Vertical line denotes offspring

• Phenylketonuria (PKU) is a metabolic disorder and a recessive genetic condition.

• The pedigree shows the inheritance through a particular family.• Which individuals can we be sure about their genotype?• Since it was not possible to identify the condition of 12 and 13 suggest

their genotype and phenotype and how the diagram may need modifying?

4.4.1

Outline the use of polymerase chain reaction (PCR) to copy and amplify minute quantities of DNA.

• DNA cloning• Used when DNA source is small• Temperature used to break and reform bonds• Polymerases thermostable so high temps can

be used

PCR steps,• Initialization step: heat the reaction to a

temperature of 94–96 °C (or 98 °C if extremely thermostable polymerases are used) for 1–9 minutes.

• Denaturation step: heat the reaction to 94–98 °C for 20–30 seconds. It causes DNA melting of the DNA template by disrupting the hydrogen bonds between complementary bases, yielding single-stranded DNA molecules.

• Annealing step: The reaction temperature is lowered to 50–65 °C for 20–40 seconds allowing annealing of the primers to the single-stranded DNA template. Stable DNA-DNA hydrogen bonds are only formed when the primer sequence very closely matches the template sequence. The polymerase binds to the primer-template hybrid and begins DNA formation.

• Extension/elongation step: commonly a temperature of 72 °C is used with Taq polymerase. At this step the DNA polymerase synthesizes a new DNA strand complementary to the DNA template strand by adding dNTPs that are complementary to the template in 5' to 3' direction, condensing the 5'-phosphate group of the dNTPs (deoxyribonucleotide triphosphates ) with the 3'-hydroxyl group at the end of the DNA strand. The extension time depends both on the DNA polymerase used and on the length of the DNA fragment to be amplified. As a rule-of-thumb, at its optimum temperature, the DNA polymerase will polymerize a thousand bases per minute. Under optimum conditions, i.e., if there are no limitations due to limiting substrates or reagents, at each extension step, the amount of DNA target is doubled, leading to exponential (geometric) amplification of the specific DNA fragment.

• Final elongation: This single step is occasionally performed at a temperature of 70–74 °C for 5–15 minutes after the last PCR cycle to ensure that any remaining single-stranded DNA is fully extended.

• Final hold: This step at 4–15 °C for an indefinite time may be employed for short-term storage of the reaction.

4.4.2

State that, in gel electrophoresis, fragments of DNA move in an electric

field and are separated according to their size.

• Sample of fragmented DNA is placed in one of the wells on the gel.

• An electrical current is passed across the gel.• Fragment separation based on charge and size• Large fragments move slowly

4.4.3

State that gel electrophoresis of DNA is

used in DNA profiling.

• Satellite (Tandem repeating) DNA are highly repetitive sequences of DNA from the non coding region of DNA.

• Different individuals have a unique length to their satellite regions.

• These can be used to differentiate between one individual and another.

Main uses

• Forensic crime investigations• Parentage Issues• Animal breeding pedigrees• Disease detection

4.4.4

Describe the application of DNA profiling to determine

paternity and also in forensic investigations.

4.4.5

Analyse DNA profiles to draw conclusions about

paternity or forensic investigations.

• The DNA fragments in the child comes from the mother and father.

• A band present in the child must come either from the mother or from the father

• The bands on the child's fragments are either found on the mother or the male1.

Parentage

Forensics

• Compare bands of specimen to suspects

• Not enough to convict someone but can narrow the search

• Match suspect one

4.4.6

Outline three outcomes of the sequencing of the

complete human genome.

• It is now easier to study how genes influence human development

• It helps identify genetic diseases• It allows the production of new drugs based

on DNA base sequences of genes or the structure of proteins coded for by these genes

• It will give us more information on the origins, evolution and migration of humans

4.4.7

State that, when genes are transferred between species, the amino acid sequence of

polypeptides translated from them is unchanged because the genetic code is universal.

• All known organisms use the same genetic code

• Transferring a gene from one species to another would result in the production of the same protein

• Some prokaryotes differ slightly, but not dramatically

4.4.8

Outline a basic technique used for gene transfer involving

plasmids, a host cell (bacterium, yeast or other cell), restriction enzymes (endonucleases) and

DNA ligase.

Stage 1: obtaining the gene for transfer

• Restriction enzymes are used to cut out the useful gene that is to be transferred.

• Note the 'sticky ends' of unattached hydrogen bonds.

Stage 2: Preparing a vector for the transferred gene

• Plasmids are small circular DNA molecules found in bacteria• These can be cut with the same restriction enzyme before• This leaves the same complementary 'sticky ends' in the

plasmid• The plasmid can be cut at particular sites. These are called

restriction sites and some are named in the diagram

Stage 3: Recombinant DNA

• (a) plasmid that will be the vector• (b) plasmid cut at restriction site Pstl• (c) Source DNA cut with same restriction enzyme as

plasmid to (d)• (e) Recombinant DNA• (f) unaffected plasmid

Stage 4: Isolation of transformed cells

• Recombinant DNA is introduced into the host cells• Many cells remain untransformed• Some cells are transformed to contain the recombinant

DNA.• These transformed cells must be separated from

untransformed

Stage 5: Product manufacture

• The transformed bacterial cells are isolated.• They are introduced into a Fermenter to be cloned.• The bacterial population grows by asexual reproduction.• The Recombinant DNA is copied along with the rest of the

bacterial genome.• In a fermenter the conditions for growth and reproduction

are controlled.• Once the bacteria express the transformed gene the

product is produced.• The next (long ) step is to isolate and purify the product.

This is called downstream processing.

4.4.9

State two examples of the current uses of genetically modified crops or animals.

Factor IX• produced by genetically modified sheep• expressed in milk from which it must be isolated before use by

haemophiliacs • A ewe is treated with fertility drugs to create super-ovulation.• Eggs are inseminated.• Each fertilised egg has the transgene injected.• A surrogate ewe has the egg implanted for gestation.• Lambs are born which are transgenic, GMO for this factor IX gene.• Each Lamb when mature can produce milk.• The factor IX protein is in the milk and so must be isolated and

purified before use in human.

Herbicides• Weeds growing near a crop use up soil nutrients that would

otherwise be used by the crop plant• This competition of resources reduces the productivity of the crop

plant and therefore the efficiency of farming• Herbicides can be used prior to crop planting to kill weeds.• The herbicide cannot be used after crops have been sown as they

will also kill the crop.• However, Cotton, Corn and Soybeans have been genetically

modified to contain an enzyme that breaks down glyphosate• This makes these crops resistant to the herbicide• Herbicide can then be use after the crop has grown to prevent the

reoccurrence of weed competition

Retinol Rice

• Retinol deficiencies can cause stinted growth or blindness• Third world countries• Rice does not contain retinol or beta-carotene (used to make

retinol), but contains molecule normally used to make beta-carotene

• The gene and enzymes to manufacture are missing from rice• Source of the gene is either Erwinia bacterium or the

common daffodil• The transgenic rice is usually yellow in color because of the

accumulation of beta-carotene• This transgenic rice is then crossed with local strains of rice

4.4.10

Discuss the potential benefits and possible harmful effects of one

example of genetic modification.

Adding Bt toxin to corn crops

• Used to keep insects from eating crop• Benefits– Higher crop yield– Less land needed for more crops– Reduce use of pesticides

• Costs– Unsure of consequences to human health– Kill other insects or animals on crop or surrounding plants

(contain toxin in pollen)– Other plants obtain the advantage and crowd out those who

do not have it

General

• Benefits– Reduce spoilage– Increase crop or animal byproduct yields– Increase sources of essential dietary needs

• Costs– May be unsafe for human– Considered “unnatural”– “Contaminate” other organisms

4.4.11

Define clone.

• Clone: a group of genetically identical organisms or a group of cells derived from a single parent

4.4.12

Outline a technique for cloning using differentiated

animal cells.

Dolly

• Udder cell removed from parent sheep• Remove egg cell nucleus of second

sheep• Cells are fused• Egg cell can still divide by mitosis• Cell grown IV until contains 16 cells-

implanted in surrogate• Normal gestation period• Sheep 1 and 4 genetically identical

4.4.13

Discuss the ethical issues of therapeutic cloning in

humans.

Benefits

• Can replace tissues/ organs saving lives and stopping pain

• Cells can be taken from embryos no longer developing so they would have died anyway

• Can be taken when embryos cannot feel due to lack of nerve cells

Costs

• Embryos should be given the chance to develop

• Often too many embryos are produced and are simply killed

• Embryonic stem cells may become cancerous

Genetics HL

10.1.1

Describe the behavior of the chromosomes in the phases of meiosis.

• Prophase I– Chromosomes coil tightly– Homologous chromosomes pair up– Crossing over– Nuclear membrane dissolves

• Metaphase I– Spindle fibers attach to centromeres– Homologous line up at the middle– Chromosomes most condensed

• Anaphase I– Spindle fibers shorten– Chromosomes pulled in opposite directions

• Telophase I– Chromosomes uncoil– Nuclear membrane may reform– May enter an interphase period

• Meiosis II the same but nucleus reforms and there are half the number of chromatids

10.1.2

Outline the formation of chiasmata in the process of

crossing over.

• Interphase – DNA replicates

• Prophase I– Molecules (cohesins) hold the pairs

together– DNA is exchanged

• Metaphase I– Homologous pairs repel at centromere

but are held together at chiasma– Cohesins break down

• Anaphase I– Pulling apart breaks chiasma

10.1.3

Explain how meiosis results in an effectively infinite genetic variety in gametes through crossing over in prophase I and random orientation in

metaphase I.

Crossing over

• Exchange maternal and paternal chromosome info

• Chromatids with new combinations• Different from both parents- recombinants• Can occur at any locus and more than one

could form

Random assortment• Homologue from either parent can go in

either direction• Over 8 million possible orientations (2 )• Combined with crossing over, the possibilities

are extremely large

23

10.1.4

State Mendel’s law of independent assortment.

• The transmission of genes to each child is independent of the others

10.1.5

Explain the relationship between Mendel’s law of independent assortment

and meiosis.

• Orientation of chromosomes at prophase I random

• Direction the chromosome is facing is also random and independent

10.2.1

Calculate and predict the genotypic and phenotypic

ratio of offspring of dihybrid crosses involving

unlinked autosomal genes.

• Crossing the F1 generation of homozygous dominant and recessive P1 for true cross (RRYYxrryy or RRyyxrrYY)

• Two characteristics controlled by two different genes

• 16 square Punnet square• 9:3:3:1 ratio phenotypes in

double heterozygous (RrYy)

Test Cross for Heterozygotes

• Cross expected heterozygote with double homozygous recessive

• Homozygous dominant can only produce dominant offspring

• 1:1:1:1 phenotype ratio should be found if heterozygous

10.2.2

Distinguish between autosomes and sex

chromosomes.

• Sex chromosomes determine gender (X and Y)– Not always same size (X much larger

than Y)• Autosomes- other 22 pairs

10.2.3

Explain how crossing over between non-sister

chromatids of a homologous pair in prophase I can result

in an exchange of alleles.

10.2.4

Define linkage group.

• A set of genes on a chromosome that tend to be inherited together

10.2.5

Explain an example of a cross between two linked

genes.

• Will follow the phenotypic ratio for a monohybrid because they are inherited together

• Recombinants will only occur if crossing over in prophase I

• 3:1 rather than 9:3:3:1

Flower color and pollen length

10.2.6

Identify which of the offspring are recombinants in a dihybrid cross involving

linked genes

• Combinations not found in parents

• Lower frequency

10.3.1

Define polygenic inheritance.

• A single characteristic determined by two or more genes

• Follow a bell curve generally• Phenotype combinations increase-

continuous

10.3.2

Explain that polygenic inheritance can contribute to

continuous variation using two examples, one of which must be human skin color.

Human skin color

• Determined by amount of melanin present• At least four genes involved• One allele codes for melanin the other for

nonproduction

Finch beak depth

• A= add depth (1 unit)• a= no depth added• B= add depth (1 unit)• b= no depth added• C= add depth (1 unit)• c= no depth added

Wheat grain color

• Varies from white to dark red• Controlled by three genes• Pigment producing vs no pigment alleles