Tutorials: focused on problemsProblem sets: due at the beginning of the classSHOW WORK FOR PROBLEM SETS

CHAPTER2

Is the science of heredity•Variation w/in & btw species•

Genetics: science that deals w/ the structure & function of genes & their transmission from 1 generation to the next

Transmission geneticsMolecular geneticsPopulation geneticsQuantitative genetics

Branches of Genetics:

Chapter 2: Mendelian genetics

AlleleCharacter(trait)CrossDiploid(2n)F1 generation (1st filial generation)F2 generation (2nd filial generation)Gamete: mature reproductive cells specialized for sexual fusion; haploid cellsGene: factors; determinant of a characteristic of an organismHaploid (n)Locus (loci)True-breeding/ pure-breedingZygote

Terminology:

Homozygous (SS or ss)-

Heterozygous (Ss)-

Dominant (SS or Ss)-

Recessive (ss)-

Genotype - Genetic constitution of an organism

Phenotype - observable/physical set of traits (structural or functional)

Phenotypic ratio: 3:1 (smooth: wrinkled)Yield genotypic ratio: 1:2:1 (SS:Ss:ss)-

Monohybrid cross: cross b/w organisms that differ in a single trait

Mendel's Principle #1: Segregation

S__ x ss (smooth x wrinkled)-

If offspring are all smooth, then unknown parent's genotype is SS-

If offspring show ratio of 1 smooth: 1 wrinkled, then unknown parent genotype is Ss-

Confirm Principle #: Test-cross (**cross with homozygous recessive**): cross of an individual of unknown genotype (usually showing dominant phenotype) with a homozygous recessive individual to determine the unknown genotype

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Lectures Page 1

Dihybrid cross - cross between organisms for 2 pairs of alleles at 2 different lociMendel's law of independent assortmentIf crossed with itself, always 9:3:3:1

Dihybrid crosses and principle of independent assortment

Trihybrid crosses: 27:9:9:9:3:3:3:1

Product rule ("and") - take probability of 2 independent events occurring simultaneously•Sum rule ("or") - probability of either one or the other of 2 mutually exclusive events occurring at the same time

•

Principles of Probability: ratio of the number of times a particular event is expected

Conditional probability

p(success) and q(failure)

Use 2 variable:○

p = frequency of 1 outcome○

q = frequency of other outcome○

Exponents: # of events○

Use Pascal's triangle○

Expands the binomial expression to reflect the # of outcome combos & probability of each

Binomial probability - involve predicting the likelihood of a series of events (for which there are two outcomes each time)

Take note of all diseased and what type of disorder it is and how it's transmitted (dominant, recessive, somatic, etc.)

AaWw, aaWw○

Ans: a plant that is genotypically AaWW has 2 types of gametes; AW & aW (notice that W is a symbol for one allele, not 2)

○

In a testcross, this plant is crossed to one homozygous for recessive alleles at both the colour gene & the waxy gene (aaww) The gametes of this plant are all aw

○

Genetic Cross: aaww x AaWW○

In maize, a dominant allele A is necessary seed color, as opposed to colorless (a). Another gene has a recessive allele (w) that results in waxy starch, as opposed to normal starch (W). The two genes segregate independently. An AaWW plant is test-crossed. What are the genotypes of offspring?

1.Practice problems:

1/4○

In cattle, the polled (hornless) (P) is dominant over the horned (p) phenotype. Another gene recessive (c) that results in white cattle, as opposed to spotted cattle (C). A hornless and spotted cow (heterozygous for both alleles) is test-crossed. What proportion of the offspring will be horned and spotted? (both recessive)

2.

~33%○

In jimsonweed, purple (P) flowers are dominant to white (p). Self-fertilization of a particular purple-flowered jimsonweed produces 68 purple-flowered & 22 white-flowered progeny. What proportion of the flowers will be true breed?

3.

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½ x 1 x ½ x ¼ x 1 = 1/16○

What is probability of a cross between an individual of genotype AaBBccDdee and an individual genotype AAbbCcDdee producing an offspring of genotype AaBbccDDee?

4.

In mimes, a rare autosomal recessive mutation of a particular gene results in an diamond shape around the eyes (dd) rather than the normal lash lines (Dd or DD). A dominant mutation on another autosomal gene results in plain black shirts (BB or Bb) as opposed to the usual stripes (bb). A pure breeding diamond-eyed, black-shirt mime is crossed with a lash-line eyed, striped shirt mime (pure-breeding).

5.

Answer: ddBB x DDbb○

F2 generation: 100% DdBb (all lash-line, black-shirt)○

What are the possible genotypes and phenotypes of the offspring in the F2 generation?

Answer: 9:3:3:1 --always this outcome for DdBb x DdBb○

D-B- (lash line & black shirt): 9/16○

D-bb (lash line & striped shirt): 3/16○

ddB-: (diamond-eye & black shirt): 3/16○

ddbb (diamond-eye & striped shirt): 1/16○

Intercross the F1 generation, what are the phenotypic ratios?

Show your work to get marks

Tracks phenotype of a specified human trait over several generations-

The affected individual through whom the pedigree is discovered called the proband-

Pedigree analysis

Autosomal dominant-

Autosomal recessive-

X-linked recessive-

X-linked dominant-

Know characteristics of inheritance for traits/diseases:

Review: Chi Square test

A man whose father expresses a recessive trait marries a woman who has a brother expressing the same trait. What is the probability that the two individuals will produce a child expressing that recessive trait?

1.

1/6○

Hint: begin by doing a pedigree

Bob and alice are married and thinking of having children. Alice has a brother with galactosemia (a rare autosomal recessive disease). Bob's aunt also has galactosemia

2.

Draw pedigree for this family. Identify the genotypes of all individuals in the pedigree.•

2/3 (2/3)(1/2) (1/2)(1/2)p(child is gg) = p(Alice is Gg)p(Bob is Gg)p(both pass g to child)○

4/72 = 1/18○

Determine the probability that Bob and Alice will have a child that has galactosemia.•

Determine probability that Bob and Alice will have a child that is a carrier for the recessive allele

•

+ p(Alice is GG)p(Bob is Gg)p(Bob passes g) etc.p(child is Gg) = p(Alice is Gg)p(Bob is GG)p(Alice passes g)

Assume homozygous normal unless otherwise indicated when someone is marrying into a family

•

Gaucher disease is caused by a chronic enzyme deficiency. A man has a sister affected with the 3.

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4/144 = 1/36○

Gaucher disease is caused by a chronic enzyme deficiency. A man has a sister affected with the disease. His parents, grandparents. And three siblings are not affected. Discussions with his wife's family reveal that the disease is not likely present in her family, although some relatives recall that the brother of the wife's paternal grandmother suffered from a similar disease. Determine the higher p that, if this couple have a child, the child will be affected (what is the chance of the worst-case scenario occurring?)

3.

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NOTE that students should be able to do ALL of the problems at the end of Chapter 2 but the following may be helpful preparation for Problem Set 1.*Brief review terminology & expected Mendelian ratios (e.g, #8, 16,17)*Determining parental genotypes based on offspring phenotypes(e.g., #29)*Basic Probability (e.g., #35)

CHAPTER3&4Chi Square test checks for goodness of fit between the expected and observed traitsDue to chanceExample:

154 smooth; yellow-

124 smooth; green-

144 wrinkled; yellow-

146 wrinkled; green-

If n = 568, we expect that each class will contain 1/4 of the total progeny or...○

568 x 1/4 = 142 progeny expected in each phenotypic class○

Expect proportion of each phenotypic class to be 1/41.

Next, subtract each observed number by its corresponding number (o-e)2.Let d = (o-e), square each d3.Divide each d2 by its corresponding expected values4.Add all these numbers5.

Degrees of freedom is n-1○

Determine your degrees of freedom - n of classes you are looking at (4 phenotypic classes)6.

SsYy x ssyy - observed progeny data from dihybrid cross (total 568)

Anything below reject null hypothesis-

.05 use as tolerance - anything above we fail to reject null hypothesisUse chi-square distribution table

Chi Square test cannot tell us that hypothesis is correct or incorrect

One gene is going to mask another geneNo new phenotypes produced

Recessive epistasis: 9:3:4Dominant epistasis: 12:3:1Complementary gene interaction/ duplicate recessive epistasis: 9:7Duplicate (gene interaction) dominant epistasis: 15:1Dominant suppression: 13:3

We can have different types

Epistasis

Problem:Barranca Mesa Flying Lizard. The results suggested that the wing shape for the lizard is due to a recessive epistatic system (9:3:4). To test this, the researcher performed a dihybrid cross, where he crossed two breeding lines of lizard, one with dumpy, and second with no wings. The resulting F1 progeny were all long-winged, the F1 progeny were intercrossed and the following F2 progeny were produced: 1025 no wing; 845 dumpy; 2465 long-winged

Carry 3 decimal places

Chapter 3: chromosomal basis of inheritance*Read non-disjunction, lethality

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Lectures Page 1

Sex chromosomesAutosomesHeterogametic - not always maleHomogametic - same sex chromosome, not always femaleHemizygous

Y present - male○

Y absent - female○

Y-chromosome - testis determining factor

XX males - a small fragment of Y is translocated to an XXY females - have a deletion of the same region of Y

*Read non-disjunction, lethality

X-chromosome: autosome balance systemY chromosome has no effect on sex determination

XXY-female XO-sterile male

(opposite from humans)

In Drosophila..

ZZ-malesZW-females

In birds and moths..ZW-system

What are the genotypes of the parentals?a.What are the phenotypic proportions of the F1 moths?b.If a light F1 female moth is mated with the father, what are the phenotypic proportions of the F2 moths?

c.

Sample question: in particular species of moth, body colour is determined by the d+ and d alleles. The wild-type, dark allele (d+) is dominant to the light (d) allele. A dark female moth is crossed to a dark male (heterozygous)

'+' = wild type - most prevalent in natureDominant and recessive is symbolized by upper-case and lower-case

Male v+/Y OR Xv+Y○

Female: v+/v+ or Xv+Xv+○

If genes were located in the X-chromosome

Drosophila gene symbols

Sample question:

Red is wild-type (w+) to white (mutant; w)-

Gene is on X-chromosome-

What are the gametes produced by each parent?a.What phenotypes and genotypes F1 generation? The proportions?b.F1 generation are intercrossed, what phenotypes and genotypes are observed in F2? The proportions?

c.

We do not obeserve, 2 different ratios for the 2 sexes○

Perform a reciprocal cross. Do we still observe the same phenotypic and genotypic ratio?d.

ALWAYS the same if the genes are located on AUTOSOMES○

What would we expect if the gene was autosomale.

True-breeding red-eyed female and a true-breeding white-eyed male

Problem:You are genetic counsellor, and a man and woman have come to you to discus their family history and asses their risk of having a child affected with Tay-Sachs or haemophilia A. the couple indicates that the woman's maternal grandfather had haemophilia A. no other people in her family are known to have had haemophilia. The mans' paternal grandfather's brother had Tay-Sachs, and the woman's paternal uncle had Tay-Sachs. No other people in the family are known to have had Tay-Sachs. Assume that these mutations are very rare in this population. Tay-Sachs is an autosomal recessive

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Draw pedigreea.P child has Tay-Sachs? 1/72b.P carrier for Tay-Sachs? 8/36c.P boy with haemophilia A? 1/8d.P girl carrier for haemophilia A? 1/8e.

Assume that these mutations are very rare in this population. Tay-Sachs is an autosomal recessive disorder, and haemophilia A is an x-linked disorder.

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CH 3,4 (continued) & gene mapping

Review:CH2 must do - 8, 16-18, 23, 24, 28, 29, 33-35CH3 must do - 12, 13, 14, 23CH4 must do - 18, 19, 30, 31 (ABO system)

CH3: Non-disjunction and Drosophila Practise QuestionCH4: Lethal alleles, Gene interactions (continued)

Sections 5.1-5.2Start CH5: Introduction, Two and three point testcrosses

Fail to segregate properly-

Meiosis I, homologous chromosomes fail to segregate-

Meiosis II (and Mitosis), sister chromatids fail to segregate-

Can occur with autosomes or sex chromosomes-

In humans, resulting progeny (XXX; XO female) (XXY male)-

Drosophila, XXX and YO offspring die!-

In fruit flies, XO male, XXY female-

Non-disjunction of X chromosomes

Show genotypes of the parentals.○

What re the genotypes and phenotypic proportions of F1○

Assume that an F1 female undergoes non-disjunction of the sex chromosomes in meiosis I when producing her all eggs. What will the surviving progeny of a cross between this female and an F1 male look like?

○

In drosophila, vestigial (partly formed) wings (vg) are recessive to normal long wings (wild-type), and the gene for this trait is autosomal. The gene for the white eye trait (w) is on the X chromosome (red eyes are wild-type). Suppose a homozygous white-eyed, long-winged female fly is crossed to a homozygous red-eyed, vestigial winged male

-

Non-disjunction question:

Affects an essential gene (mutated genes that result in a lethal phenotypes)-

There are dominant and recessive lethal genes-

Mice (two copies of lethal allele dies) - example of recessive lethal-

Lethal allele

Probability of cross between two white bunnies producing a brown bunny? 2/3 Ff white and 1/3 FF brown

-

What can we conclude about the fur allele? Allele F is recessive lethal-

Question: you know that gene F codes for fur colour in bunnies. Bunnies that are hetero (Ff) have white fur. Bunnies that are homo for dominant (FF) are brown. ff condition is lethal.

Two mice ratty a lethal allele which has dominant effect of colour, making them yellow rather than the wild type brown colour. The mice mate multiple times and 100 pups are produced. How many pups are expected to be yellow?~66%

a+ b+ c+ d+ = Independently assorting controlling black pigment. The alternative alleles that give abnormal functioning of these genes are a, b, c, d. black individual of genotype a+/a+, b+/b+, c+/c+, d+/d+ is crossed with a colourless individual of genotype a/a, b/b, c/c, d/d to produce all black F1. Two F1 progeny are crossed.

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Lectures Page 1

What proportion of F2 progeny are colorless? 37/64-

What proportion of F2 progeny are brown? 27/256-

Two F1 progeny are crossed.

What proportion of the F2 progeny is colorless?-

What proportion of the F2 progeny is black?-

What proportion of the F2 progeny is red?-

Black can be produced only if both the red pigments are present and thus the c+ converts both pigments together into black.

Genes A, B, and C are independently assorting and control the production of a black pigmentSuppose instead that a diff pathway is utilized. In it, the C allele produces an inhibitor that prevents the formation of black by destroying the ability of B to carry out its function.A colorless A/A B/B C/C indv is crossed with a/a b/b c/c, a colorless giving colorless F1. F1 are selfed.What is the ratio if colorless to black in the F2? Only colorless and black phenotypes are observed

For this pathway to produce black, ind must have A and B functions BUT NOT the inhibitor function provided by C

-

A/- B/- c/c (black) = 3/4 x 3/4 x 1/4 = 0.1406 -

Colorless = 55/64-

Lectures Page 2

Please note that you are responsible for ALL end-of-chapter problems for Chapters 3 and 4. However, the following is a list of concepts and textbook questions may be helpful preparation for Problem Set 2.

Chapter 3:*ZW sex determination (e.g., #23)*Pedigree Analysis with sex-linked traits (e.g., #13)*Determining parental genotypes w/ sex-linked traits (e.g., #14)-students are responsible for ALL end-of-chapter problems.

Chapter 4:*Chi-Squared Analysis (e.g., #30)*Genetic Pathways (e.g., #18)*Review gene interactions, epistasis, and modified Mendelian ratios (…lots of possible questions; also see table 4.19 on page 127)-students are responsible for ALL end-chapter problems.

Ch5 - Gene Mapping (Part I)Questions do not do: 18, 20, 28, 29, 31

Emphasis on Sections 5.1 to 5.3•Introduction - "Lingo" for gene mapping•2 & 3 point test crosses•Chi-Square analysis - genetic linkage•

Today:

Several genes are located on same chromosomes = linked genes-

Two parents w/ different alleles are crossed (e.g. white eyes/ mini wings x wild-type eyes and wings)

-

Parental combination of alleles = parentals (white eyes/ mini wings x wild-type eyes and wings)

-

Non-parental combination of alleles = recombinants (white eyes/ wild-type wings x wild-type eyes and mini wings)

-

Gene mapping

Linked genes are always SYNTENIC and are always located near one another on a chromosome-

Genetic linkage leads to more frequent parental phenotypic classes than recombinant-

Frequency of CO is roughly proportionate to the distance b/w genes -> relationship that allows for genes to be mapped!

○

CO is less likely to occur when linked genes are closer together on a chromosome than b/w genes farther apart on the chromosome

-

% recombinants below 50% suggest genes are located on same chromosome (Linked genes)1.% recombinants above 50%, genes are on different chromosomes (genes are not linked)2.Max recombinant is 50%3.

Fundamentals to genetic mapping

Two arrangements of alleles:○

a+b+/ab or a+b/ ab (both have same phenotypes)○

Coupling (CIS) configuration: when wild type alleles are on one chromosome and mutant

In heterozygote of two linked genes-

Concept of genetic mapping

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Coupling (CIS) configuration: when wild type alleles are on one chromosome and mutant alleles are on another chromosome

○

Repulsion (TRANS) configuration: when one wild type and one mutant allele are on each chromosome

○

Can use recombinant frequencies as a quantitative measure of the genetic distances b/c 2 genes are on a genetic map

-

mu (map units) or cM (centimorgan)-

1 mu = 1 cM = 1% recombination frequency-

Be careful in using cM, just use mu to be safe-

Crude approximations, but good enough for genetic instructions!-

Genetic distances

Analyze testcross data and determine whether a deviation is "significant"-

Used because it is not possible to predict phenotype frequencies produced by linked genes

○

Genes assort independently in meiosis

If 2 genes are not linked, a testcross should be 1:1 ratio of parentals: recombinant○

Ho (Null hypothesis) - genes are not linked-

Expected values - based in the assumption that the genes in question are assorting independently

-

Chi-square and gene linkage

In drosophila, b is recessive autosomal mutation that results in black body color, and vg is a recessive autosomal mutation that results in vestigial (short, crumpled wings). Wild-type flies have grey bodies and ling, non-crumpled wings (normal)

-

Cross: true-breeding black, normal wings x true-breeding grey, vestigial wings-

P: b/b vg+vg+ x b+b+ vg/vg-

F1: all progeny flies -> grey with normal wings (b+/b vg+/vg)-

Test cross F1 female x black body, vestigial wing (b/b vg/vg)-

Males cannot be used for these studies (they do not go over crossing over, only female does!)-

Testcross: F1 female x black body, vestigial wing (b/b vg/vg)-

Grey, normal wings - 283 (non-parental or recombinant)○

Grey, vestigial wings - 1294 (parental)○

Black, normal wings -1418 (parental)○

Black, vestigial wings -241 (non-parental or recombinant)○

Phenotype Observed Expected (O-E)2/E

Grey, normal wings 283 809 341.998

Grey, vestigial wings 1294 809 290.760

Black, normal wings 1418 809 458.443

Black, vestigial wings 241 809 398.793

Testcross progeny (3236 total)-

Chi-square = 1489.994; df = 3; reject null hypothesis-

NO, since p value is less than 0.001. There is a significant deviation.Is it likely that the observed numbers are the result of 2 independent assorting genes?-

Genes are linked!○

Genes are (241+283)/3236 x 100% = ~16% or ~16 mu apart ○

Alternate hypothesis explain our observed data-

Sample chi-square test - linkage

What have we seen up to this point-

Remember: a+b+/a+b+ x a b/a b cross-

Results in the following two point test cross: a+b+/ab x a b/a b-

Two point test crosses

Lectures Page 2

Results in the following two point test cross: a+b+/ab x a b/a b-

rf= (number of recombinants/number of testcross progeny) x 100%○

Frequency of recombinants gives an estimate of genetic distance in map units-

But the number of recombinants does not change○

Number of phenotypes of recombinants changes when arrangement of F1 (cis or trans) changes

-

There is large number of frequency of parentals if there is linkage-

This method can be considered accurate as long as the genes are relatively close together-

This method is not accurate when genes are far apart-

(figure from textbook)

Used to determine linkage and rf's of three genes-

a+b+c+/a b c x a b c/a b c○

Three-point testcross:-

There is 8 phenotypic classes (see figure from textbook)-

Three-point testcrosses

p = purplep+= yellow (WT)r = roundr+ = elongatedj = juicyj+= dry

p+r+j+ 179 parental

prj 173 parental

p+rj 46 recombinant

pr+j+ 52 recombinant

p+rj+ 22 recombinant

pr+j 22 recombinant

p+r+j 4 recombinant, double crossover (fewest progeny)

prj+ 2 recombinant, double crossover(fewest progeny)

Total progeny = 500○

Classify phenotypic classes in progeny: of the 8 classes, 2 are parentals, 4 are recombinants.1.

Double crossovers are rare; hence, the recombinants associated with the fewest progeny arose due to double crossovers!

○

In recombinants, determine which arose due to a double crossover2.

We see that only the j gene is changed. While the other two remain in their parental arrangement

○

J is in the middle. Therefore, the order of the genes: p j r or r j p○

Determine order of genes: after a double crossover, only one of the genes will be exchanged between the homologous chromosomes, leaving the other two unchanged (i.e. parental arrangement)

3.

pr+j+ = 52 (purple, elongated, juicy fruit)Results: p+jr = 46 (yellow, round, juicy fruit)

First single crossover○

pj+r = 2Results: p+jr+ = 4

Second single crossover○

Determine which SCO progeny is produced in the chromosomes4.

Calculate rf between p and j:Calculate the recombinant frequency for two genes at a time5.

Steps in genetic mapping

Lectures Page 3

Calculate rf between p and j:○

Single crossover recombinants p and j = 46+52 = 98Double crossover recombinants p and j = 4+2 = 6Rf = (single crossover progeny + double crossover progeny / total progeny) = 20.8 muCalculate rf between r and j:○

Single crossover recombinants r and j = 44Double crossover recombinants r and j = 6Rf = (single crossover progeny + double crossover progeny / total progeny) = 10.0 muCombine data:○

Single r and j = 44Double r and j = 6Single p and j = 98Double p and j = 6Rf = 30.8 mu

Draw you genetic map! Remember your distances and units!6.

Calculate coefficient of Interference and coincidence (if necessary)7.If each single crossover is considered an independent event. Then we can say that the double crossover (2 single crossovers occurring simultaneously) probability is the product of each single crossover

-

Prob/freq for 1st single crossover = 0.208, 2nd single cross = 0.10-

Expect the double crossover to be = Product rule -> 0.208 x 0.10 = 0.0208 -

Observed is 6/500 = 1.2% . Expected is higher than observed because of interference.-

I = 1 - coefficient of coincidence○

Interference = one single crossover interfering with another single crossover-

Coeff of 1 = DCO occurred that were expected on the basis of 2 independent events (no interference)

○

Coeff of 0 = none of the expected DCO occurred (total interference, with one CO completely interfering a second crossover in that region)

○

Coefficient of coincidence = (observed/expected freq) = 0.012 / 0.0208-

Examine testcross progeny numbers and indicate which are parentals and recombinants. Determine gene confirmation (i.e. cis or trans)?

1.

Indicate your double crossover progeny (look for the lowest number of progeny)2.Determine the order of your genes via DCO3.Determine which SCO progeny is produced from CO between regions 1 & 24.Calculate your recombination frequency for each region5.Draw your final genetic map6.Calculate coefficient of coincidence and interference (if necessary or required by question)7.

Generating your genetic maps: summary of the 7 steps to creating your genetic map are

Lectures Page 4

Finish Chapter 5 (up to 5.3)Students are responsible for the following end-of-chapter questions for Chapter 5:1-17, 19, 21-28, 30-32....However, some questions/concepts that may be helpful preparation for Problem Set 3 include:Three-point test crosses (e.g., #15, #16, #23, 26)Basic recombination (leading to more challenging recombination)(e.g., #17)X2 with linkage (e.g., #13, #32)

TAs may also:-Revisit complementation analysis (Chapter 4, Section 4.4).-Start Chapter 10.

CH 5 - Gene Mapping (Part II)

In soy bean plants, there are three linked genes coding for pod color, pod texture, and pod size. The wild-type phenotypes are green, smooth, and big. The mutant phenotypes are turquoise (t), wrinkled(w), petite(p). A triple heterozygote is test-crossed producing the ff. progeny types:

1.Questions:

F2 genotype Number a.) Recombinant or parental? b.) None, SCO, or DCO types?

w++ 476 P none

++p 375 R SCO (t-w)

+tp 496 P none

wtp 88 R DCO

wt+ 385 R SCO (t-w)

w+p 204 R SCO (w-p)

+t+ 249 R SCO (w-p)

+++ 66 R DCO

Solution: c.) What is the gene order? twp or pwt

'w' is middle gene

Parents: DCO

w+++tp

+++wtp

Rf(t-w) =

Rf(w-p) =

d.) draw the genetic map showing the correct order and relative distances (including units) between the genes

t---39.08mu----w---25.95mu----p p---25.95mu---w---39.08mu----t

C.O.C= observed DCO/expected DCO = 0.6492I = 1- 0.6492 = 0.3508

e.) Coefficient of coincidence and interference values

f.) What does the above data tell you? Most of the DCO are occurring 35.07%

Three mutant phenotypes are observed: brown kernels (b); wide leaves (w); short (s). The wild-type (+) phenotypes are yellow, narrow leaves and tall, respectively. All mutant alleles are recessive to wild-type. A triple heterozygote corn plant is testcrossed to produce the ff. progeny types:

2.

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Lectures Page 1

heterozygote corn plant is testcrossed to produce the ff. progeny types:

F2 phenotype Number a.) F2 Genotype b.) Recombinant or parental? c.) None, SCO, or DCO types?

Brown, short 18 bs+ R DCO

brown 279 b++ P None

short 82 +s+ R SCO (s-w)

wide 12 ++w R DCO

Wild-type 127 +++ R SCO(b-s)

Wide, short 251 +sw P None

Brown, wide, short 143 bsw R SCO(b-s)

Brown, wide 88 b+w R SCO(s-w)

Solution:d.) What is the gene order? bsw or wsb

"s" is the middle gene

Parentals DCO

b++ bs+

+sw ++w

Rf=(b-s) = 30.0mus-w = 20.0mu

b--30.0mu---s---20.0mu----w

COC=0.50

What does our interference value indicate? Most of our expected DCO are occurring although there is 50% interference indicates that 50% of DCO to occur are not occurring due to in of DCO event w/

I=0.50 or 50%

3. A cross was made between an aaccFF female and an AACCff male. The F1 males were testcrossed and produced the ff:F1 testcross -> AaCcFf x aaccffF2 progeny

Phenotype Observed

ACf 394

aCF 402

acf 90

aCf 102

AcF 98

ACF 110

Acf 398

acF 406

Assume A, B, C are dominant phenotypes to a, b and c.

If a pair of genes is linked, then we should expect to see more parental types than recombinant types. That is parental recombination would be greater than 50% and rec would be less than 50%

-

If a pair of genes are not linked, then we should to see equal numbers of parental to recombinant types (i.e. independent assortment -> 1:1:1:1)

-

A and C C and F A and F

A C -> 394 + 110 = 504; 504/2000=0.252 = ~25%a c = ~25%A c= ~25%

C F -> 402 + 110 = 512; 512/2000=0.256 = ~25%cf= ~25%Cf= ~25%

A F -> 98 + 110 = 208 ; 208/2000=0.104 = ~10% << 25% (recombinant)af=~10%Af=~40% (parental)

Which pair of genes is linked> A and C, A and F, F and C?a.

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A c= ~25%a C= ~25%

Cf= ~25%cF= ~25%

Af=~40% (parental)aF=~40% (parental)

C and F are unlinked Therefore, A and F are linked and are on the same chromosome and the C is on a different chromosome

Calculate the distance between ANY linked genes. Draw the associated genetic map(s).b.Rf (a-f) =total number of recombinant progeny/total progeny = (208+192)/2000=~20% or 20 mu

A----20mu---F ----C----

4.

h i+ x h i

h+ i h i

Genes h and i are linked and are 43 map units apart. Given the ff test cross:

What is the probability of the dihybrid parent producing a gamete with genotype h i+? 28.5%h and i = 43 mu apart = 43% rfParentals = 57% rfIn 2 point testcross you can get:

5.

t+ m; i+ x t m; i

t m+; i t m; i

Genes t and m are linked and are 30 map units apart. Gene I is on a different autosome. Given the ff test cross:

t+/t and m+/m loci are linked 30 mu apart - crosses involving these loci will have 70% parental types and 30% recombinant types The i+/i locus assorts independently

PT = 35% t+m and 35% tm+ RT = t+m+ (15%) tm (15%)

If our autosome was gene i (homozygous mutant) than all progeny types would have iParental (35% each)Recombinant (15% each)Since our second autosome heterozygous; we would need to account for the possible outcomesParental: 17.5% eachRecombinant: 7.5% each(t+ m+; i+) (t m; i) = 7.5%

6. A----23mu---B---11mu---C------22mu-------D--6mu--ECalculate the frequency of b c gametes from an individual with the genotype bC//Bc

Parentals (57% each w/ 28.5%) Recombinant(43% each w/ 21.5%)

h i+ h+ i+

h+ i h i

- It has to be a DCO genotype- ANS: 0.99%- Solution: based on map distances; one expects 39 rf between b and e; 33% rf between b and d and 6% rf

between d and e

Calculate the frequency of B d E gametes from an individual with the genotype b d e // BDE (assume that there is no interference)

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Chapter 4 (Section 4.4 – Complementation Analysis):-Try the example in figure 4.22 (you can also try to do question 22 at the end of the chapter but there are multiple typos in the table).

Chapter 10:-Students are responsible for all end-of-chapter problems.-Question 15 is similar to the question in the problem set.-Students should know the difference between Southern, Western and Northern blot techniques and know how different size/different polarity molecules will travel through gels. You should also know some basics about restriction enzymes (e.g., what sized bands would be produced for a particular restriction digest) and molecular probes.-You need to be able to determine genotypes and/or phenotypes using these techniques and then apply stuff they learned in earlier chapters (e.g., calculating probabilities).

Chapter 12:-Students are responsible for questions: 14, 15, 16, 18, 27, 30, 32-Question 14 is similar to the question in the problem set (mutation rates.-Students should understand most of the basic concepts from the chapter (which should be review from BIO206 and/or BIO215).

Chapters: 4(Complementation), 10(SCD), 13(Chromosomal Aberrations)

Chapter 4.4 - Mutants and Mutations

How do we determine the number of genes involved in set of mutations with the same phenotype?

1.

How do we know if these mutations affect the same or different genes?2.

Two strains of petunias - Californian and Netherlands; two geneticists identify mutations (independently from each other) that cause petunia flowers to be white (mutant phenotype) instead of purple (wild type)

Performed by crossing two mutants (pure-breeding) with the same phenotype and observe progeny

-

Complementation test determines whether two independently isolated mutations with the same phenotype have affected same or different genes

Complementation (+) = production of WT progeny - mutations are affecting different genes1.No complementation (-) = production of mutant progeny - mutations affect the same gene (i.e. mutant alleles of the same gene)

2.

Two outcomes:

e/e b+/b+ (black body color) x e+/e+ b/b (black body color)--> e+/e b+/b (wild-type body color resulting from complementation of mutant genes)

Sample problem: In drosophila, recessive mutations a, b, c, d, e, f, g all produce the same phenotype (red pigment)

a b c d e f g

g + - + + + + -

f - + + - + -

e + + - + -

d - + + -

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c + + -

b + -

a -

Complementation; mutations on different genes○

All the resulting F1 progeny expresses WT phenotype○

Parental (g/g a+/a+ x g+/g+ a/a) --> F1 (g+/g a+/a)○

What genotypes were crossed to show complementation between mutants g and a? a.

No complementation; mutations on the same gene○

All F1 expresses mutant phenotype○

What genotypes were crossed to show no complementation between g and b?b.

Identify # of complement groups, this is equal to # of genes○

Mutants that fail to complement each other are on the same gene (g and e; c and e; a, f and d)

○

Therefore, there are three genes○

How many genes are present? c.

Mutant that fail to complement each other are on the same gene (g and b; c and e; a, f and d)

○

Which mutants have defects in the same gene?d.

In the nematode worm, recessive mutations l, m, n, o, p, q, r, and s all have the same phenotype (aberrant locomotion). In pair-wise combinations of complementation tests, the following results were produced where (+) indicates complementation and (-) indicates NO complementation.

m and ra.m and sb.o and qc.p and rd.l, p and re.B, C and E are correctf.A and C are correctg.

Which mutants have defects in the same gene?1.

Wild-typea.Mutantb.An intermediate phenotype between s and lc.A 3:1 ratio of WT to mutantd.1:2 ratio of mutant to WTe.None of the offspring will survivef.There is insufficient information providedg.

If worms bearing mutations s and l are crossed, what is the phenotype of their progeny?2.

l m o p q r s

s + - + + + + -

r - + + - + -

q + + - + -

p - + + -

o + + -

m + -

l -

A potentially fatal ,autosomal recessive caused by a structural abnormality in Hb that affects its ability to carry O2

-

This in turn causes shorten life span of RBC (anemia - an abnormal LOW number of RBC)-

Sickle-cell disease (SCD), also sickle cell anemia

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This in turn causes shorten life span of RBC (anemia - an abnormal LOW number of RBC)-

Sickle shape instead of concave shape RBC (drives heterozygous advantage)-

There are hundreds of variant hemoglobin alleles -

Common to Mediterranean, Middle east and African -

Tetramers with two protein chains each of two diff. globin genes (2 units alpha and 2 units beta)

-

SCD is common hereditary anemia caused by single-base pair substitution in the beta-globin geneβsβs is mutant, βAβA is normal, βAβs is carrier

Review: gel electrophoresis ("molecular sieve") -> separation; agarose or polyacrylamide; "RUN TO RED"/Electrophoretic mobility)

-

Molecular genetic techniques for analyzing out SCD alleles, mRNA and proteins that are produced by these alleles

Western blotting

Carrier will have 3 bands-

Mutant will have a band at 1.35-

Normal will have bands at 1.15 and 0.20-

Southern blotting - pay attention to band fragments that are produced

Normal = succumb more easily to malaria-

Mutant = suffer from hereditary anemia-

Carrier = survive and reproduce more dependably that other genotypes in region where malaria is common

-

Malaria endemic environments favors the survival and reproduction of individuals who are heterozygous for beta A and one of the mutant alleles.

Interrupts the developmental cycle of the plasmodium larvae-

Advantage causes populations to evolve a gene pool with high levels of both normal and mutant alleles

-

Advantage of heterozygosity is balanced by the disadvantages associated with homo for either allele - equilibrium frequency

-

Balanced polymorphism - selection against on of its phenotype is balanced by natural selection in favour of the allele of another phenotype

-

Heterozygote advantage = based on the shorter average life span of RBC

Non-disjunction of X chromosomesResults in gametes with abnormal chromosome number (aneuploidy)-

Meiosis I - X chromosomes fail to segregate. Result = 2 X or no X chromosomes-

Chapter 13 - chromosomal aberrations

Non-disjunction problem: in drosophila, the bobbed gene (b) is located in X chromosome. Mutants (b) have shorter, thicker bristles than wild-type (b+) flies. Unlike most X-linked genes, a bobbed gene is also present on the Y chromosome. The mutant allele (b) is recessive to the WT (b+) allele. If a WT F1 female that resulted from primary non-disjunction in oogenesis (in a cross of a bobbed female with a WT male) is mated to a bobbed male, what will be the frequency of the WT offspring?

Examine Parental cross = XbXb (bobbed female) x Xb+Yb+ (wild type male)Female undergoes primary non-disjunction

Parental Xb+ Yb+

XbXb XbXbXb+

diesXbXbYb+

female

O OXb+

maleOYb+

dies

Remember: XXX and YO dies!

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Remember: XXX and YO dies!

XbXbYb+ × XbYb

Examine the F1 bobbed cross: wild type (b+) female x bobbed (b) male

*Non-disjunction in fruit flies occurs 4% of the time (secondary ND in XXY female)Normal disjunction is 96% Possible gametes from XXY female: Xb, XbYb+; XbXb, Yb+

Normal disjunction (96%)

Secondary ND (4%)

Xb (48%) XbYb+ (48%) XbXb (2%) Yb+ (2%)

Xb (50%) XbXb

(viable, bobbed, female 24%)

XbXbYb+

(viable WT female 24%)

XbXbXb

(dies, inviable 1%)

XbYb+

(viable WT male 1%)

Yb (50%) XbYb

(viable, bobbed, male 24%)

XbXbYb

(viable, WT, male 24%)

XbXbYb

(viable bobbed 1%)

YbYb+

(dies 1%)

Frequency (wild-type offspring) 49%/ 98% = 0.5 What is the frequency of bobbed offspring? 0.5Viable, bobbed male? (0.24 or 24%)

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Autosome1: A---20----B----10----C------15-----D--5--E Autosome 2: F-----12-------G-----6----HCalculate the frequency of b c gametes from an individual with genotype bC//Bca.

Parentals (90%) = 45% Bc and 45% bC Recombinants (10%) = 5% BC and 5% bc

Ans: 5%

Calculate the frequency of CdE gametes from an individual with the genotype cde//CDEb.

CdE = DCO typef(DCO) = (0.15) (0.05)/2 = 0.00375 So, f(CdE) = 0.375%

Ans: 0.375%

A homozygous dominant DEF individual is crossed to a homozygous recessive def individual and the resulting F1 is self-crossed. What is the probability of producing a phenotypically deF individual in the resulting F2 progeny?

c.

F belongs to second autosomal linkage group

DE/DE; F/F X de/de; f/f

de;F = (0.475)(0.475)(0.75) = 16.9%

Ans: 16.9%

Use the maps below to answer the ff questions. Both autosomes are from the same organism. Distances are in mu. Capital letters are dominant to lower case letters. Assume that there is no interference.

1.

Three point test cross: Three loci, mitochondrial malate dehydrogenase that forms a and b(MDHa, MDHb), glucouronidase that forms 1 and 2 (GUS1, GUS2), and a histone gene that forms + and - (H+, H-), are located on chromosome #7 in humans. Assume that the MDH locus is at position 35, GUS at position 45, and H at position 75. A female whose mother was homozygous for MDHa, GUS2, and H+ and whose father was homozygous for MDHb, GUS1, and H- produces a sample of 1000 egg cells. Give the genotypes and expected numbers of the various types of cells she would produce. Assume no chromosomal interference.

2.

Pseudodominance: occurs when a recessive allele is "unmasked" by a deletion (and is EXPRESSED in the phenotype) that removes the dominant allele on the homolog

-

It is used to map the position of genes on a chromosome via deletion mapping-

When heterozygous individuals do not show the recessive mutant phenotype= the deletion they carry has not removed the WT copy of the gene

○

When heterozygotes do show the mutant phenotype = the deletion they carry does remove the WT copy of the gene

○

Results of deletion mapping:-

Deletion mapping

Drosophila-

Seven partial deletions-

Determine the location of and order of the genes on the chromosome-

Deletion map and score matric (+/-) that examine phenotype expressed-

Cross: flies HOMOZYGOUS for mutant allele x flies HOMOZYGOUS for partial deletion - F1 progeny that would be heterozygous

-

Question 24 (p 457):

CH5, #19, p181

TUT7March-07-142:18 PM

Lectures Page 1

Chapter 21 (Quantitative Genetics)

Ex. In pea experiments, flowers were either purple or white, and each was associated with particular genotype

-

Discontinuous traits have only a few distinct phenotypes:

In complex organisms, few traits fall into this category due to penetrance, expressivity, epistasis,

Polygene hypothesis: G x E interaction

Additive genes - when genes contribute to polygenic traits, contribute EQUAL amounts/ share to that total phenotypic variance

-

Additive trait - each allele of AG can be assigned a QUANTITATIVE value that indicates its contribution to a polygenic trait

-

Multiple genes and additive gene effects

Why stats?

Focusing on nature vs. nurture-

How much of the variation observed in a quantitative trait is genetically determined and how much is environmentally induced?

Population and samplePopulation: ultimate group of interestSample: a ransom subset

Frequency of distributionNormal distribution

Sample mean (or average)Sample varianceStandard deviationCorrelationCoviariance: allow to quantify gene variation and

Amount of phenotypic variation attributable to genetic factors-

Heritability

Gene-environment interactionGenotype only, no environmental effectEnvironment only, no genotypic effect

Additive genetic varianceDominance varianceInteraction variance: Vg (overall genetic variance)

General environmental effect: factors such as nutrition and temperature which determine final state of phenotype

1.

Special environmental effect2.Family environmental factors3.Maternal effects4.

Environmental variation

Final phenotypic variance equation

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Contributions to phenotypic variance can be expressed as:-

Final phenotypic variance equation

Measures proportion of phenotypic variance that results from genetic difference among individuals in a specific population (as factors may differ between populations)

-

Formula: H^2 = VG/ VP-

0 = 0 none of the phenotypic variation due to genetic diff (but genes are still involved, they just don’t differ)

○

1 = 1 all of the phenotypic variation due to genetic differences○

A high heritability○

This quantity can range from 0 to 1:-

Broad-sense heritability (HB^2) or (H^2)

Interested in make prediction about the resemblance between offspring and○

Looks at how much of phenotypic variance can be attributed to additive genetic variance-

Narrow-sense heritability (HN^2) or (h^2)

Evolution by natural selection occurs because certain traits leave more offspring than others-

Evoluton by artificial selection - superior individuals are used to create agricultural varieties-

Response to selection

Selection response (R) - amount that a given phenotype changes in one generation due to natural or artificial selection

Genetically diverse fly population with mean body weight of 1.3 mg-

F1 mean body weight = 1.3 mg○

In half of this population, individuals are allowed to interbreed randomly -> unselected-

F1 mean body weight = 2 mg○

In other half, only large individuals (mean = 3 mg) are allowed to interbreed - selected-

If genetic variation underlies variation in body size-

Ex: Body weight in drosophila:

In drosophila example: s = 3mg - 1.3 mg = 1.7 mg○

S = measures the DIFFERENCE between the population mean value (for a trait)and the mean trait value for the mating portion of a population

-

Hn^2 = R/ s = 0.7/ 1.7 = 0.4118○

Knowing R (e. g. 0.7 mg) and s(1.7 mg)-

Selection differential(s):

Genetic analysis: P715, 21.1 8 cm to plant height <-- Plant 1(148 cm): AABBCC x aabbcc: plant 2 (12cm) --> Each allele contributes 2 cm to plant height--> height of F1?

F1 = AaBbCc = 24cm + 6cm = 30cm (each dominant contributes 8cm and recessive 2cm)a.F1 x F1 = P (plant 42 cm in length)b.P (being hetero at one locus and homo at other )= P (AaBBCC) + p(AABbCC) + P (AABBCc) > x 3= 1/2 x 1/4 x 1/4 x 3 = 3/32

Two pure-breeding lines of tomatoes P1 and P2 produce fruit with different average weights and are crossed. The means and variances of their F1 and F2 progeny are shown below:

-

Line Average fruit weight (g) Vp

P1 6.5 1.6

P2 14.2 3.5

F1 10.2 2.2

Q#2 on p56

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F1 10.2 2.2

F2 9.8 4.0

What is VE for this trait? VP = VG + VEa.Therefore VP = VE (for genetically identical organisms --> P1/P2 and F1)Therefore average environmental variance is = (1.6 + 3.5 + 2.2)/ 3 = 2.4333gDetermine VG from the F2:b.VG = VP - VE = 4.0 - 2.43 = 1.57g

H2 = 1.57/ 4 = 0.3925

CH21 MUST DO questions: 8, 10, 12, 13, 14, 21, 22

342 people with M type-

500 people with MN people-

187 people with N type-

Sample problem: A worldwide survey of genetic variation reported autosomal codominant MN blood group types in a sample of 1029 people. The sample contained:

Determine the frequency of alleles

Solution: p2 = 342/1029=0.3324 2pq = 500/1029 = 0.4859, or 2 x 0.58 x 0.426 = 0.4942 q2 = 187/1029=0.1817

F(M allele) = 0.332 + (0.5*0.486) = 0.575 F(N allele) = 0.186 + (0.5*0.486) = 0.429

Solution:q2 = 1/2000; q = sqrt(0.0005) = 0.0223606797749979 p = 1 - 0.02236 = 0.97764 F(carriers) = 2pq = 2*0.97764*0.02236 = 0.04372

You know that the frequency of cystic fibrosis in a population is 1/2000. From this can you determine the allele frequency? Frequency of carriers?

421 have type A-

168 have type B-

336 have type O-

75 have type AB-

A total of 1000 members of a population are typed for ABO. In the sample:

Determine the frequency of the allelesSolution:r2 = ii = 336/1000 = 0.336 p2 + 2pr =IAIA + IAi = 421/1000 = 0.421 q2+ 2qr =IBIB + IBi = 168/1000 = 0.168

r = sqrt(0.336) = 0.579655069847578(p + r)2 = p2 + 2pr + r2 = 0.421+0.336=0.757, p= sqrt(0.421+0.336)-0.579=0.291057469366248 --> A allele frequency

2pq = IAIB

p2 + 2pq + q2 + 2qr + r2 + 2pr

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Chapter 22: Population GeneticsFocus on: Hardy-Weinberg, Map of natural selection

What is it?

Both pop and quantitative genetics use Mendelian genetics-

Pop genetics - field of genetics that studies heredity in group of indi for traits that are determined by one or only a few genes;

Genotype frequencies - way to study the genes in particular gene pool

BB = 452Bb = 43bb =2n = 497

Example: Spot patterns on the moth

f(BB) = 452/497 = 0.9095 Genotype frequencies

Allele frequencies

Allele frequency = (# of copies of a given allele) / (total # of alleles in pop)1.When two alleles are present at a locus, we can use the formula: -

p = f(A) = (2xAA) + (# of Aa) / (2xtotal number of individuals)-

p+q=1-

This formula is for traits that are not sex-linked-

Allele (gene) counting2.

P = f(A1) = (2 x A1A1) + (A1A2) +(A1A3)/ (2xtotal number of individuals)○

q=f(A2) = (2 x A2A2) + (A1A2) +(A2A3)/ (2xtotal number of individuals)○

r=f(A3) = (2 x A3A3) + (A1A3) +(A2A3)/ (2xtotal number of individuals)○

Three alleles-

Allele frequencies at an X-linked locus are more complex because one sex will have only one X-linked allele while the other has 2 X-linked alleles

-

p=f(XA) = (2x XAXA) + (XAXa) + (XAY)/[(2x # females) + (# of males)]○

f(XA) and f(Xa) formula:-

Calculating:

Hardy-weinberg law5 assumptions: infinetly large, randomly mating population, free from mutations, migration and natural selectionResult # 1: the frequencies of the alleles don’t change over time where p is the allele frequcy of A and q is the allel fre of aResult # 2: p^2 + 2pq + q^2 = 1

Ask: what is the chance that we would get this magnitude of deviation due to chance?-

Allows to calculate the prob that the diff b/w what we observed and what we expect (under the HW law) id due to chance

○

Chi square test: goodness of fit test-

MM(12 voles)○

MJ(53 voles)○

JJ(12 voles)○

n=77

Example: Transferrin (a blood protein) locus in the red-backed vole-

Testing proportions (Chi-square analysis)

TUT9March-28-142:31 PM

Lectures Page 1

n=77○

Let p = f(M) and q=f(J)○

p=f(M) = (2xMM) +MJ/2x total = 0.5

p+q=1 --> q=0.50

Using the formula○

Calculate allele frequencies for the population-

f(MM) = p^2 = 0.5^2 = 0.25 x 77 = 19.25 ○

f(MJ) = 2pq = 2x0.5x0.5=0.5 x77 = 38.5○

f(JJ)=q^2 = 0.5^2 = 0.25 x 77 = 19.25 ○

Use the calculated alleic fre to calcule expected values for each-

Construct your chi-square test:-

Phenotypes Observed Expected O-E (O-E)^2/E

MM 12 19.3

MJ 53 38.5

JJ 12 19.3

Totals 77 77

Dof = # of phenotyic classes - 2 = 3 - 2 = 1 Chi square = 10.98Given calculated chi -square value and dof, locate p-value-

Therefore, we reject the hypothesis○

Your conclusion: p-value is less than 0.01 (p<0.005)-

HW equilibrium is maintained when there is no evolutionary change-

Natural selection -

Individuals (favoured phenotype) survive to reproductive age at higher rates and/or -

Consequence of natural selection -> leads -

Organims with highest reproductive success --> w = 1.0○

Individuals reproducing less○

W = 1- s○

Common way to quantify differential reproductive fitness -> compare the relative abilities of population members to produce offspring = relative fitness (w)

-

Sometimes natural selection can maintain genetic diversity within pop-

Balanced polymorphism is eventually reached -> in which allele freq are maintained-

Relative to SS = 1- s○

Relative to ss = 1 - t○

S and t - variables used represent natural selection against the homozygous genotypes-

pE = t/ (s+ t)○

qE = s/ (s+t)○

Equilibrium frequency -

Differential reproductive fitness

Heterozygote advantage

Genotype BABA BABS BSBS

Frequency 0.51 0.41 0.080

Relative fitness 0.80 1.000 0.12

Selection co-eff (s) for BA = 1-0.8=0.2 -

Selection co-eff (t) for BS = 1-0.12 = 0.88 -

EQ freq for BA = t/(sum) = 0.81-

EQ freq for B = s/(sum) =0.19-

Genetic profile for sickle-cell disease (SCD) for a particular population in a malaria-prone region. Calculate the equilibrium frequency of the BA and BS alleles

Lectures Page 2

In the absence of mutations or selection, what % of the next generation would be heterozygotes and/or homozygous recessives?

-

AA p^2 = 0.81 p = 0.9

p + q = 1 q = 0.1

f(Aa) = 2pq = 2x0.9x0.1 = 0.18 q^2 = (0.1)^2 = 0.01

In a large interbreeding pop, 81 % are homozygous for a dominant character:

What is the frequency of the t allele in the population?-

Solution

tt 100

Tt 800

TT 1100

Total 2000

f(t) = [ (2xtt) + (Tt) ]/ (2x total) = 0.25 = qf(T) = 0.75 = pChi-square-

Pehotypic class O E O-E (O-E)^2

TT 1100 1125 -25 625 0.555

Tt 800 750 50 2500 3.333

tt 100 125 -25 625 5.000

Chi-square = 8.889Dof = 1p-value < 0.001Reject null hypothesis and that the population is not in HW equilibrium

In a population of 2000 vipers, a genetic difference with respect to venom exists at a single locus. The alleles are incompletely dominant. Individuals with genotypes tt are non-poisonous; Tt are mildly poisonous and TT are lethally poisonous. The population shows 100 individuals that are non-poisonous, 800 that are mildly poisonous and 1100 that are lethally poisonous.

Problem setQ2 hint: Males are hemizygous. Females are going to copy of XA chromosome.Q3 hint: focus on french and jewish; take into consideration 1 in 30 and getting allele from either parents

Lectures Page 3

Hardy-Weinberg law - if population is in equilibrium, allele frequencies do not change

Two forms: Inbreeding, Selection of individual based on phenotypeNon-random mating: alters genotype frequencies

Individuals share a greater proportion-

- Main genetic consequence is increase in one type of genotype (homozygous) and decrease in heterozygous

- This makes sense because the increase in homozygous genotype produces an increase of homozygous and decrease in heterozygous genotype

- Various examples in plant and animal species (i.e. moth, fruit flies, rat)- Sexually reproducing individuals = the effect produced is over a large number of

generation

- Quantifies the probability of producing an inbred individual that is homozygous for an allele obtained from a common ancestor

- Can also estimate the proportion of loci that will be IBD (identical by decent)

Coefficient of inbreeding (F) - a measure of the probability of homozygosity for an allele obtained in identical copies from an ancestor

# of alleles of a gene carried by common ancestors (i.e. one common ancestor = 2 alleles)

1.

# of transmission events required to produce the homozygous IBD genotype2.Probability of transmission of the allele for each event3.

Three key elements for determining F from pedigrees:

1/2 = prob of transmission of an allele

n= # of transmission events to produce an individual that is IBD

General solution for F = (1/2)n

Inbreeding (consanguineous mating) - mating between related individuals

The pedigree in 22.11.IV-1 and III-1 produces an offspring in V-1ANS: F = 4(1/2)^7 = 1/32 for any alleleF = (1/2)^7 = 1/128, for one of the alleles to be homo IBD

Practice problems:

Is IV-I an inbred individual? If so, who is/are the common ancestors? YES, I-II since IV-I could have 2 copies of an allele derived by this individual, making IV-I IBD

2.

What is F for IV-I? F = 2(1/2)^6 = 1/32

Which individuals are inbred? Who is/are common ancestors?3.What is F for any inbred individuals?

In the European land snail, Cepaea nemoralis, multiple alleles at a single locus determine shell color. The allele for black (CB) is dominant to the allele for purple (CP) and to the allele for yellow (CY). The dominance heirarchy among these alelles is CB>CP>CY. In one population sample for Cepaea, the following color phenotypes were recorded: black (325); purple (319); and yellow (220). Assuming that this population is in HW equilibrium (large, randomly mating, and free from evolutionary processes), calculate the frequencies of the CB, CP, and CY alleles.

4.

CB>CP>CY

3 alleles at single locus (variant)-

n=864

TUT10April-04-142:21 PM

Lectures Page 1

3 alleles at single locus (variant)-

Recall: p2+2pq+q2 for 2 alleles -

Expand and represent as trinomial expression-

p2+2pq+q2+2pr+r2+2qr-

Black = CBCB+CBCP+CBCY = 32

An individual heterozygous for the genes (A/a; B/b; C/c; D/d) is testcrossed5.

Phenotype Number

aBCD 42

Abcd 43

ABCd 140

abcD 145

aBcD 6

AbCd 9

ABcd 305

abCD 310

total 1000

Compute a genetic map, genetic distances and interference value

A and B A and D B and C C and D

AB = 445ab = 455aB = 48Ab = 52

No recombination Recombination Recombination

Since A and D show NO recombination, recreate the table, drop out D

Phenotype Number

aBC 42

Abc 43

ABC 140

abc 145

aBc 6

AbC 9

ABc 305

abC 310

total 1000

Gene order: B A CRf (b-a) = 10 muRf (a-c) = 30 mu

Follow through with the rest of the regular three-point testcross analysis:

CH5: Genetic linkage - You self-cross an individual of genotype AaBb and 1% of the progeny are aabb. What is the map distance between the genes A and B?

6.

CH4-Gene interaction7.Using replica plating and nutrient supplementation to examine biosynthetic pathwaysWhat is the order of the compounds in the pathway?

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What is the order of the compounds in the pathway?

Mutant A B C D E G

1 - - + - - +

2 - - + - +

3 - - - - - +

4 - + + + - +

5 + + + + - +

Answer: E A B D C G

The ff corn recessive loci are on one arm of chromosome 9 in the following order. The wild-type alleles are dominant to the recessive/mutant alleles:

8.

c---------42--------bz----28-----wxCalculate the frequency of c + wx gametes from an individual with the genotypes c bz wx// + + + (assume that there is no interference)ANS: 5.88%

= (c-bz) (bz-wx)/2 = (0.42)(0.28)/2 = 0.0588Solution: f(c + wx) = gamete type forms as a result of a DCO with + bz + being second type

A scientist wants to determine the narrow-sense heritability of tail length in mice. He measures tail length among the mice of a population and finds a mean tail length of 13.5cm. He then selects the 22 mice in the population with the longest tails. Mean tail length in these selected mice is 17.7 cm. he interbreeds the mice with the ling tails and examines tail length in their progeny. The mean tail length in the F2 progeny of the selected mice is 15 cm. calculate the selection differential, the response to selection, and the narrow sense heritability for tail length in the these mice.

9.

R = (mean selected F1 body size) - (mean unselected F1 body size)After 1 generation of artificial selection there was a 1.5 cm response*15-13.5 = 1.5 cmS = (mean phenotype selected) - (mean phenotype in pop before selection)*17.7 cm - 13.5 cm = 4.2 cmHn^2 = R/S = 1.5/4.2 = 0.36

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