Inheritance and HEREDITY = Genetics Father of Genetics: Gregor Mendel
Austrian Monk Grew Peas and studied their characteristics (inheritable feature) and their traits (varia- tions of a character) Flower color: characteristic (gene) Pink or white: Trait (allele) Mendels Work: Reasons for Studying Peas Easily distinguished traits
Fast reproduction Self pollinate: develop true breeding varieties - parents and offspring have the same characteristics from generation to generation Controlled breeding of plants
Crossed plants with opposing traits to see what would happen P generation: parent generation - initial cross of a genetic study F1 generation: first filial - first generation of offspring F2 generation: second filial - second generation of offspring Monohybrid study of one gene Dihybrid study of two genes
Two types of Studies: Monohybrid study of one gene Dihybrid study of two genes Two types of Organisms: Pure bred = Homozygous - two of the same allele Hybrid = Heterozygous - two different alleles - typically one dominant and one recessive Results of Mendels Work
Law of Dominance: Tall pure bred X Short pure bred found all F1 were Tall one trait would over shadow the other trait Against the convention of the time blending Took F1 Generation and did a self cross Tall F1 X Tall F1
Law of Segregation: TookF1 Generation and did a self cross Tall F1 X Tall F1 found 75% of F2 were Tall and 25% of F2 were short traits were preserved from one generation to the next and separated from one another during reproduction Law of Segregation Law of Independent Assortment
Tall Purple flowered pure bred X Short white flowered pure bred F1: all tall purple flowered - self cross F2: ratio of 9 tall, purple:3 tall, white :3 short, purple :1 short, white traits were inherited independently of one another and recombined during fertilization Law of Independent Assortment Modern Terminology Factors = genes Alternative traits = alleles
Each characteristic controlled by a gene pair or allele pair Mendels Crosses: How to do genetics Problems
Homozygous Tall Pea plant X Homozygous Short Pea plant Genotype: genetic make up Phenotype: what it looks like if heterozygous: dominant trait appears MONOHYBRID CROSS: the Punnett square P generation: TT x tt
Male Genotype Possible Gametes Female Genotype Possible Gametes Male Genotype TT Possible Gametes Female Genotype T T t Possible Gametes tt t Male Genotype TT Possible Gametes Female Genotype T T t Possible Gametes tt t Male Genotype TT Possible Gametes Female Genotype T T t Tt Possible Gametes Tt tt t Tt Tt Genotypic Ratio: Phenotypic Ratio: Male Genotype TT Possible Gametes Female Genotype T T t Tt Possible Gametes Tt tt t Tt Tt Genotypic Ratio: 4:0 Phenotypic Ratio: 4:0 F1 Generation Cross: Tt X Tt
Male Genotype Possible Gametes Female Genotype Possible Gametes F1 Generation Cross: Tt X Tt
Male Genotype Tt Possible Gametes Female Genotype T t T Possible Gametes Tt t F1 Generation Cross: Tt X Tt
Male Genotype Tt Possible Gametes Female Genotype T t T Possible Gametes TT Tt Tt t Tt tt Genotypic Ratio: Phenotypic Ratio: F1 Generation Cross: Tt X Tt
Male Genotype Tt Possible Gametes Female Genotype T t T Possible Gametes TT Tt Tt t Tt tt Genotypic Ratio: 1:2:1 Phenotypic Ratio: 3:1 Test Cross T___ X tt crossing of a dominant phenotype with an unknown genotype with a homozygous recessive Possible Outcomes T ___ t t
If some of the offspring show the recessive trait = Possible Outcomes Must be recessive if tt appears T t t Tt tt t Tt tt
If some of the offspring show the recessive trait = unknown genotype to be recessive If some of the offspring show the recessive trait = Possible Outcomes T __ t Tt t Tt If no offspring show the recessive = Possible Outcomes T T t Tt Tt t Tt Tt
If no offspring show the recessive = most likely that the unknown genotype is homozygous dominant depends on the number of offspring produced GENETICS AND PROBABILITY
Coin tosses Formation of gametes represents probability Homozygous Dominant: probability of getting dominant trait = 100% Heterozygous: probability of getting dominant trait = 50% Inheritance: Rule of Multiplication: Probability and dependent events Determining the probable genotypes of offspring Heterozygous Parents Probability of getting a gene from mother: Probability of getting a gene from father: Total probability = x = Probability of being heterozygous in a heterozygous cross
Rule of Addition: Probability and independent events Probability of being heterozygous in a heterozygous cross two ways to be heterozygous: find individual probabilities and add them together + = TT Tt Tt tt Probability and the Dihybrid Cross
Homozygous Tall Purple X Homozygous short white P Generation TTPP x ttpp Each gamete must get one of each gene Gametes: TP or TP and tp or tp F1 Genotypes: TtPp F1 Gametes: formed from TtPp Genes separate in meiosis
Independent assortment allows for several combinations TtPp OR TtPp Genes separate in meiosis each gamete gets one of each gene
F1 Genotypes: TtPp Genes separate in meiosis each gamete gets one of each gene Independent assortment allows for several combinations TtPp OR TtPp T P t p T p t P Dihybrid Genotype results in 4 possible gametes. Expands the Punnett to a 4x4 grid TP Tp tP tp TP Tp tP tp TP Tp tP tp TTPP TTPp TtPP TtPp TP TtPp Ttpp TTPp TTpp Tp TtPP TtPp ttPP ttPp tP TtPp ttpp tp Ttpp ttPp Dihybrid Cross Ratios Genotypic Ratio: Phenotypic Ratio: Dihybrid Cross Ratios Genotypic Ratio:
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp Phenotypic Ratio: Dihybrid Cross Ratios Genotypic Ratio: 1:2:2:4:1:2:1:2:1
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp 1:2:2:4:1:2:1:2:1 Phenotypic Ratio: Dihybrid Cross Ratios Genotypic Ratio: 1:2:2:4:1:2:1:2:1
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp 1:2:2:4:1:2:1:2:1 Phenotypic Ratio: Tall Purple: Tall White: Short Purple: Short White Dihybrid Cross Ratios Genotypic Ratio: 1:2:2:4:1:2:1:2:1
TTPP:TTPp:TtPP:TtPp:TTpp:Ttpp:ttPP:ttPp:ttpp 1:2:2:4:1:2:1:2:1 Phenotypic Ratio: Tall Purple: Tall White: Short Purple: Short White 9:3:3:1 Phenotypic Ratio for ANY Dihybrid Cross where both parents are heterozygous for both traits. Extensions Of Mendels Ideas
Not everything in genetics is as clear cut as Mendels idea of dominance.Many traits are controlled by several other types of inheritance patterns. Incomplete Dominance Blending
neither trait is dominant so both are partially expressed Red flower X White flower = pink Different Rules: Key: Use a capital Letter to represent the characteristic Use a capital letter superscript to represent the trait CR = Red CW = White Sometimes its just the R and W or just R and r Red X white Genotypic: Phenotypic: Cross Pink with Pink Codominance Both traits are equally expressed and DISTINGUISHABLE
Ex: Roan Cattle Red Bull x White Cow ROAN Offspring looking closely at the fur = #s of red hairs and white hairs KEY: Function the same way as incomplete dominance CR = red CW = white Codominance Crosses Crosses Red with White Roan with Roan Multiple Alleles instead of just two possibilities, there are more
gives multiple factors to consider: which alleles are dominant? which are codominant? which are incomplete? Ex: Rabbit Fur one of four genes
Full color dominant over chinchilla Chinchilla dominant over Himalayan Himalayan dominant over Albino Key for Multiple Allele Problems
Large Letter for characteristic Superscript for traits - only lower case is the least recessive Full color - CF Chinchilla - CCh Himalayan - CH Albino - c Cross a heterozygous full/chinchilla with a heterozygous Himalayan/albino
Genotypic: Phenotypic: Rabbit Fur Mutations HUMAN BLOOD TYPE Multiple Alleles and Codominance
Importance: Matching Donors and Recipients - Immunoglobin Responses Blood Type Interaction Type Type Type Type Epistasis: One gene at a different locus (location) alters the expression of another gene
Mice Coloration: Black dominant over brown Color dominant over colorless BB, Bb black bb brown HOWEVER: Also dependent upon the gene for color CC, Cc color will be deposited in the hair follicle cc no color despite BB, Bb, or bb - results in an albino Determine Colors: BbCc: BBcc: bbCC: bbcc: Cross heterozygous male with a heterozygous female (not 9:3:3:1) Determine Colors: BbCc: black BBcc: albino bbCC: brown bbcc: albino Cross heterozygous male with a heterozygous female (not 9:3:3:1) Labrador Retriever Genetics Black is dominant to chocolate B or b Yellow is recessive epistatic (when present, it blocks the expression of the black and chocolate alleles) E or e Phenotype Possible Genotypes BBEEBbEEBBEeBbEe bbEE bbEe BBee Bbee bbee Determine the number of chocolate labs produced from a black female and a yellow male (BbEe x bbee) Polygenic Traits: one characteristic is controlled by the cumulative effect of many genes
Ex: Human height and skin color - each controlled by at least three gene pairs T - tallness t - shortness TTTttt - medium height TtTtTt - medium height TTTTTT - tall tttttt- short TtTtTt x TtTtTt - result in many different combinations Human Skin Pigmentation
same basic process - gene codes for activity of melanocytes - all humans have the same number of pigment producing cells in their skin - the level of activity determines skin coloration - enhanced by exposure to UV light - protection against FURTHER UV damage Pleiotropy -a gene affects an individual in more than one way - multiple affects sickle cell anemia: mutated gene causes blood cells to have a sickle cell shape - causes blood clots, weakness, anemia - PAGE 254 Comparison of Polygenic Inheritance and Pleitropy Penetrance Degree to which the gene is expressed in the population of those with the genotype - usually pertains to disease Ex: A disease with an 85% penetrance means that 85% with the gene has the disease Differs from expressivity which is variations in the phenotype of those demonstrating the gene Three individuals with a trait for green hair could have dark green, light green or medium green Environmental Impact on Phenotype
page STUDYING INHERITANCE IN HUMANS
must be multigenerational Use Pedigree Analysis: Pedigree Chart to study one trait at a time and determine the probability of a trait being passed on to an offspring - usually done to predict the likelihood of a child inheriting a hereditary disease: EX: Hemophilia, breast cancer, cystic fibrosis Making a Pedigree Chart
Square - male Circle - female Shaded - demonstrate trait Unshaded - do not demonstrate trait Half-shaded: carrier for a homozygous recessive trait Types of Disorders Recessive Dominant
Multifactoral: based on gene interactions and environmental influence Chromosomal Inheritance
Thomas MORGAN: worked with fruit flies lots of offspring 2 week life cycle only 4 pairs of chromosomes mated fruit flies until one day he noticed a male with white eyes
wild type normal red eyes mutant phenotype - alternatives to the wild white eyes Key: use mutant to indicate trait and mutant with + for wild w+ = wild type w = white eyes gray body vs. black body (mutant) b+ = gray b = black Figure 15.3 Morgan mated white, eyed male with a red eyed female F1 = all red eyed
Crossed F1 F2 = white eyed - expected UNEXPECTED - white eyed were only MALE All females (1/2 of population) were red eyed Half of male (1/4 of population) were white eyed Figure 15.4 EXPERIMENT Morgan mated a wild-type (red-eyed) female
The F2 generation showed a typical Mendelian 3:1 ratio of red eyes to white eyes. However, no females displayed the white-eye trait; they all had red eyes. Half the males had white eyes, and half had red eyes. Morgan then bred an F1 red-eyed female to an F1 red-eyed male to produce the F2 generation. RESULTS P Generation F1 X F2 Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F1 offspring all had red eyes. EXPERIMENT CONCLUSION: Gene for eye color is on the X chromosome Sex determination in fruit flies XX = female XY = male X and Y Chromosomes Y chromosome is structurally different from the X and is missing many of the alleles that are found on the X Result: it only takes one recessive on the X to get the recessive phenotype Cross: Wild type female with mutant type male
F1: All Females: All Males: Cross: Wild type female with mutant type male
Xw+Xw+xXwY F1: All Females: Xw+Xw All Males: Xw+Y Cross F1 female with and F1 male
Females: Males: Cross F1 female with and F1 male
Females: Xw+Xw+or Xw+Xw - all red eyed Males: Xw+Y (red) or XwY (white) CONCLUSION Since all F1 offspring had red eyes, the mutant
white-eye trait (w) must be recessive to the wild-type red-eye trait (w+). Since the recessive traitwhite eyeswas expressed only in males in the F2 generation, Morgan hypothesized that the eye-color gene is located on the X chromosome and that there is no corresponding locus on the Y chromosome, as diagrammed here. P Generation F1 F2 Ova (eggs) Sperm X Y W W+ Implications: Re-evaluate Mendels Ideas What if traits are located on the same chromosome? Law of Independent Assortment Revised: applies to genes on different chromosomes HOWEVER: genes on the same chromosomes only TEND to be inherited together WHY? Crossing over - rearranges gene sequences by mixing parental DNA EVIDENCE FOR LINKED GENES
Heterozygous Wild type (gray body, normal wings) crossed with a recessive mutant (black body and vestigial wings) Genotypes: EVIDENCE FOR LINKED GENES
Heterozygous Wild type (gray body, normal wings) crossed with a recessive mutant (black body and vestigial wings) Genotypes: b+bv+vXbbvv Recombinant (nonparental-type)
Double mutant (black body, vestigial wings) Wild type (gray body, normal wings) P Generation (homozygous) b+ b+ vg+ vg+ x b b vg vg F1 dihybrid (wild type) b+ b vg+ vg TESTCROSS b+vg+ b vg b+ vg b vg+ b+ b vg vg b b vg+ vg 965 (gray-normal) 944 Black- vestigial 206 Gray- 185 normal Sperm Parental-type offspring Recombinant (nonparental-type) RESULTS EXPERIMENT Morgan first mated true-breeding wild-type flies with black, vestigial-winged flies to produce heterozygous F1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F1 dihybrid females with black, vestigial-winged males, producing 2,300 F2 offspring, which he scored (classified according to phenotype). CONCLUSION If these two genes were on different chromosomes, the alleles from the F1 dihybrid would sort into gametes independently, and we would expect to see equal numbers of the four types of offspring. If these two genes were on the same chromosome, we would expect each allele combination,B+ vg+ and b vg, to stay together as gametes formed. In this case, only offspring with parental phenotypes would be produced. Since most offspring had a parental phenotype, Morgan concludedthat the genes for body color and wing size are located on the same chromosome. However, the production of a small number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the linkage between genes on the same chromosome. Figure 15.5 b+v+ b+v bv+ bv bv bv bv bv Genotypic Ratio: Phenotypic Ratio: b+v+ b+v bv+ bv bv b+bv+v b+bvv bbv+v bbvv bv bv bv Genotypic Ratio: Phenotypic Ratio: b+v+ b+v bv+ bv bv b+bv+v b+bvv bbv+v bbvv bv bv bv Genotypic Ratio: 4:4:4:4 Phenotypic Ratio: 4:4:4:4 b+v+ b+v bv+ bv bv b+bv+v b+bvv bbv+v bbvv bv bv bv
Genotypic Ratio: 4:4:4:4 = Parental Type same as parents Phenotypic Ratio: 4:4:4:4 = Recombinants differ from parents Expected Cross: Genotypic Ratio: 4:4:4:4 Phenotypic Ratio: 4:4:4:4 If 2300 offspring are produced: Parental Types (like parents) wild type (b+bv+v) = 575 mutant (bbvv) = 575 Recombinants: (different from parents) gray, vestigial (b+bvv) = 575 black, normal (bbv+v) = 575 ACTUAL RESULTS: Parental types: wild: 965 mutant: 944 Recombinants: gray, vestigial: 206 black, normal: 185 HUGE DIFFERENCE: WHY?? Traits Linked on Chromosomes WHY RECOMBINANTS?? Crossing over How much cross over? RECOMBINATION FREQUENCY - frequency that two genes of sister chromosomes cross over in prophase I Figure 15.6 = Testcross parents Gray body, normal wings
(F1 dihybrid) b+ vg+ b vg Replication of chromosomes Meiosis I: Crossing over between b and vg loci produces new allele combinations. Meiosis II: Segregation of chromatids produces recombinant gametes with the new allele Recombinant chromosome b+vg+ b vg b+ vg bvg+ bvg Sperm Meiosis I and II: Even if crossing over occurs, no new allele combinations are produced. Ova Gametes offspring b+ vg+ b vg b+ vg b vg+ 965 Wild type (gray-normal) b vg b vg+ 944 Black- vestigial 206 Gray- 185 normal Recombination frequency = 391 recombinants 2,300 total offspring 100 = 17% Parental-type offspring Recombinant offspring Black body, vestigial wings (double mutant) Recombination Frequency
RF = (# of recombinants/total offspring) X 100 ( ) Morgans Cross = _____________ X 100 = 17% (2300) Practice Problems: Pg 291 # 5 Recombination frequencies reflect the distance between genes on a chromosome.
Allows the calculation of relative distances between chromosomes up to a recombination frequency of 50%. 50% or greater frequency eliminates the difference between crossing over and simple independent chromosomes Once gene frequencies are found, gene maps can be constructed from the cross over data GENE MAPPING use recombination frequency to map RELATIVE distances of genes the farther two genes are from one another on a chromosome the more likely there is to be crossing over as you look at the frequency of cross over you can place the genes in order based on the recombination frequencies translate frequency into map units 1% = 1 map unit Recombination frequencies 9% 9.5% 17% b cn vg Chromosome The bvg recombination frequency is slightly less than the sum of the bcn and cnvg frequencies because double crossovers are fairly likely to occur between b and vg in matings tracking these two genes. A second crossover would cancel out the first and thus reduce the observed bvgrecombination frequency. In this example, the observed recombination frequencies between three Drosophila gene pairs (bcn 9%, cnvg 9.5%, and bvg 17%) best fit a linear order in which cn is positioned about halfway between the other two genes: RESULTS A linkage map shows the relative locations of genes along a chromosome. APPLICATION TECHNIQUE A linkage map is based on the assumption that the probability of a crossover between two genetic loci is proportional to the distance separating the loci. The recombination frequencies used to construct a linkage map for a particular chromosome are obtained from experimental crosses, such as the cross depicted in Figure The distances between genes are expressedas map units (centimorgans), with one map unit equivalent to a 1% recombination frequency. Genes are arranged on the chromosome in the order that best fits the data. Figure 15.7 EXAMPLE: three genes X, Y and Z by doing cross over studies: X is 12 map units from Z Y is 7 map units from Z X is 19 map units from Y Relative Gene Order: EXAMPLE: three genes X, Y and Z by doing cross over studies: X is 12 map units from Z Y is 7 map units from Z X is 19 map units from Y Relative Gene Order: X-Z-Y Four genes are studied: ABCD
A & B are15 map units apart B & C are 27 map units apart C & D are 7 map units apart D & A are 5 map units apart A & C are 12 map units apart B & D are 20 map units apart ORDER: Four genes are studied: ABCD
A & B are15 map units apart B & C are 27 map units apart C & D are 7 map units apart D & A are 5 map units apart A & C are 12 map units apart B & D are 20 map units apart ORDER: CDAB or BADC Five Genes: ORDER: Gene Pair Recombination Frequency A & B 5 A & P 2.5
45 B & O 25 B & I 40 B & P O & P 27.5 Five Genes: ORDER: APBOI Gene Pair Recombination Frequency A & B 5
2.5 A & I 45 B & O 25 B & I 40 B & P O & P 27.5 Sex Determination X-Y System: mammals and fruit flies XY = male
XX = female X-O System: grasshoppers and crickets one X = male two X = female Z-W System: birds, some fishes and some insects WW = Male ZW = female
use different letters to represent chromosomes so not to confuse the system with the X-Y system Haploid-Diploid System: bees and ants haploid - male: unfertilized eggs diploid - female: fertilized eggs therefore, all male bees and ants are bastards (d) The haplo-diploid system
Figure 15.9bd 22 + XX X 76 + ZZ ZW 16 (Haploid) (Diploid) (b) The X0 system (c) The ZW system (d) The haplo-diploid system Sex-Linked Inheritance in Humans
Fathers gametes determine the gender of the offspring passing of traits from one generation to the next is determined by linkage X-linked genes: genes found on X chromosome can be passed to offspring by mother and father - father can only pass X-linked to daughters Ex: hemophilia, color blindness Y-linked genes: genes found on Y chromosome - can be passed only to sons by fathers Ex. Hairy ear rims, SRY (sex determining region of the Y chromosome) X-linked recessive traits
More common in males only takes one allele - hemizygous condition Ex: color blindness Can occur in females- need one recessive from each parent A normal visioned woman whos father was color blind has children with a color blind man.
A) What is the probability they will have a color blind son? B) What is the probability they will have a color blind daughter? C) What is the probability they will have a color blind child? D) What is the probability their daughter will be color blind? A normal visioned woman whos father was color blind has children with a color blind man.
A) What is the probability they will have a color blind son? 1/4 B) What is the probability they will have a color blind daughter? 1/4 C) What is the probability they will have a color blind child? 1/2 D) What is the probability their daughter will be color blind? 1/2 A certain recessive gene on the X-chromosome causes death to developing mouse fetuses. If a female carrier mates with a normal male, what fraction of the offspring will be carriers of the trait? Answer: A certain recessive gene on the X-chromosome causes death to developing mouse fetuses. If a female carrier mates with a normal male, what fraction of the offspring will be carriers of the trait? Answer: 1/3 - since the gene only need be present on one chromosome in the males, 1/4 (1/2 of males) will die in utero. Of the remaining population, 1 out of 3 will be a carrier. X-Linked Dominant Traits
Seen in males and females almost equally Ex: hypophosphatemia - low blood phosphate Sex Influenced Traits Traits that are not carried on the sex chromosomes but are enhanced by the sex hormones Ex: Male pattern baldness on Chromosome 15 Dominant trait in males Recessive trait in females Fate of the Double Xs in Female Cells
Embryonic X-inactivation - formation of the Barr body early stages of development, one of the Xs condenses and attaches to the inside of the nuclear envelop - 50/50 chance for either to condense - the resulting cells from that cell all have the same Barr body - can cause differences in female phenotype depending on which chromosome is inactivated Examples: calico cat, sweat glands, partial albinism Figure 15.11 Two cell populations in adult cat: Active X Early embryo:
Orange fur Inactive X Early embryo: X chromosomes Allele for black fur Cell division and X chromosome inactivation Black Figure 15.11 ERRORS IN CHROMOSOMAL INHERITANCE
Separation Problems Non-disjunction: chromosomes or chromatids dont separate properly Results in additional or missing chromosomes Aneuploidy: abnormal chromosome # - type depends on which gamete is received Monosomic: one copy of a chromosome Trisomic: three of one chromosome Trisomy 21: Downs Syndrome Polyploidy: gamete has 2 full sets of chromosomes due to complete non-disjunction:resulting fertilization makes a triploid or tetraploid organism - animals = death (almost always) - plantscan actually benefit - larger leaves, fruits and flowers easier to deal with than aneuploidy because chromosome numbers are equal Nondisjunction of sister chromatids in meiosis II
Gametes n + 1 n 1 n 1 n 1 n Number of chromosomes Nondisjunction of homologous chromosomes in meiosis I Nondisjunction of sister chromatids in meiosis II (a) (b) Alterations in Chromosome Structure:
errors in replication or crossover 4 types: Deletion: lose part of the chromosome - can lose important genetic information 2. Duplication: fragmented chromosome rejoins another chromosome causing a section to be repeated 3. Inversion: fragmented chromosome rejoins in reverse position 4. Translocation: fragmented chromosome rejoins a none homologous chromosome Figure 15.14ad A B C D E F G H Deletion Duplication M N O P Q R
Inversion Reciprocal translocation (a) A deletion removes a chromosomal segment. (b) A duplication repeats a segment. (c) An inversion reverses a segment within a chromosome. (d) A translocation moves a segment from one chromosome to another, nonhomologous one. In a reciprocal translocation, the most common type, nonhomologous chromosomes exchange fragments. Nonreciprocal translocations also occur, in which a chromosome transfers a fragment without receiving a fragment in return. Extranuclear Genes Are genes found in organelles in the cytoplasm
Mitochondrial DNA Trace lineages on mothers side Chloroplasts