On my honor, this is my work GENETICS 310

EXAM I June, 9, 2017

I. Fill in the missing information in the table below. Assume the original parents are homozygous unless the legend or results dictate otherwise.

KEY: GR = genotypic ratio, PR = phenotypic ratio, L = Legend Crosses and results Legends

a) P1 T”T X T”T Parents phenotype both tailless progeny GR 2 T’T : 1 TT Pregeny PR 2 tailless: 1 normal

T”T” lethal T”T tailless TT normal

b) P1 red, serrate X orange, smooth F1 PR all red smooth GR all Rr, Ss F2 PR: 9 red, smooth 3 red, serrate 3 orange, smooth 1 orange, serrate

color L R_ red rr orange leaf edge L S_ smooth ss serrate (or equivalent)

c) P1 White feathers X Black feathers F1 mix of black and white feathers F2 GR 1 B”B” :2 B”B : 1 BB F2 PR 1 white :2 mix : 1 black

Color L B”B” white feathers B”B mix black and white BB black feathers

d) Black X Brown P1 Genotypes AA, BB X aa, bb F1 PR all black F1 GR All Aa, Bb F2 PR: 15 black: 1 brown

Color L A_, B_ black A_, bb black aa, B_ black aa, bb brown (one example)

Which, if any of cases a-d above provide example(s) of Incomplete dominance (a ok) epistasis d co-dominance c

I

III. In mice: YLYL -Lethal (embryo)

YL Y -Yellow

Y Y -Gray

O _ -slender

o o -obese

H _ -hyperactive

h h -normal

P _ -Plain coat

p p -spotted

1) A yellow, slender, normal activity, plain coat female is mated to a yellow, slender, hyperactive, plain coat male. All possible phenotypes for each trait are seen in the progeny. a) Give the genotype of the female: YLY, Oo, hh, Pp b) Give the genotype of the male: YLY, Oo, Hh, Pp c) How many different genotypes can occur in the sperm? 2 X 2 X 2 X 2 = 16 d) What fraction of the progeny would be expected to have the genotype: YLY, OO, Hh, pp? 2/3 X ¼ X ½ X 1/4 e) What fraction of the progeny would be expected to have the phenotype: Gray, Slender, normal activity, spotted coat? 1/3 x ¾ X ½ X 1/4 2) Two plain coat mice, each of which had one spotted parent are mated. If there are 6 pups born: a) what is the probability that 2 will be plain and 4 will be spotted. 6! (3/4)2(1/4)4 2! 4! b) What are the odds that at least one will be spotted? 1-(3/4)6 IV This pedigree is from a single gene trait. What is the probability that each individual is heterozygous. Mom 1 Dad 1 Daughter 1 2/3 Daughter 2 2/3 Son 1 2/3 Son 2 0

V. a) In rabbits, 4 alleles that affect coat color, listed in order from most to least dominant are: C, brown; Cch, chinchilla; Ch, Himalayan, c, albino. Two brown rabbits are crossed. How many phenotypes could possibly occur in the progeny and what are they. 4: all those listed b) A number of mutations in different genes have been associated with CMT diseases in humans and patients exhibit a wide range of symptoms, including no apparent expression in individuals who then pass the trait on. Based just on this information, what terms associated with ‘extensions’ of Mendelism can be associated with CMT? Genetic heterogeneity, lack of penetrance, variable expressivity VI. Three men are accused as being the father of 2 babies each by two mothers. Mom 1 is A, MN, Rh+ and her babies are 1-1 A, N, rh- and 1-2 AB, M, Rh+ Circle any blood group(s) of babies 1-1 and 1-2 that says the baby is not hers. (underline & blue used for circle) Same for Mom 2 who is O, M, rh- and her babies are 2-1 O, N, Rh+ and 2-2 A, MN, rh- Prospective father 1 is: O, MN, Rh+ Circle the blood groups of the babies below that he could not possibly father: 1-1 A, N, rh- 1-2 AB, M, Rh+ 2-1 O, N, Rh+ 2-1 A, MN, rh- Prospective father 2 is: A, N, rh- Circle the blood groups of the babies below that he could not possibly father. 1-1 A, N, rh- 1-2 AB, M, Rh+ 2-1 O, N, Rh+ 2-1 A, MN, rh- Prospective father 3 is: AB, M, Rh+ Circle the blood groups of the babies below that he could not possibly father: 1-1 A, N, rh- 1-2 AB, M, Rh+ 2-1 O, N, Rh+ 2-1 A, MN, rh- Now do the same using both prospective parents (Mom 1 for 1-1 and 1-2, etc.) : X dad1 1-1 A, N, rh- 1-2 AB, M, Rh+ 2-1 O, N, Rh+ 2-1 A, MN, rh- X dad 2 1-1 A, N, rh- 1-2 AB, M, Rh+ 2-1 O, N, Rh+ 2-1 A, MN, rh- By dad 3 1-1 A, N, rh- 1-2 AB, M, Rh+ 2-1 O, N, Rh+ 2-1 A, MN, rh-

VII. Very briefly, tell why it is so difficult to find a perfect match for an organ transplant. There are many alleles for each of the HLA genes that serve for self identification: any allele in a tissue that is different from those in the recipient will lead to rejection. VIII. In his PhD research, Rehman crossed two true-breeding chickpea varieties to create an F1 hybrid that was then self pollinated so that many F2 progeny could be evaluated. One parental variety matured quickly, setting seeds 60 days after planting, while in the other, average seed set occurred at 80 days. Out of 1,024 F2 plants raised in ideal conditions, 4 set seeds in 50 days and 4 others took 90 days; all others fell between these extremes, a) Predict average ‘days-to-seeding’ in the F1 70 and F2 70 progeny. b) How many ‘days-to-seeding’ QTL genes were heterozygous in the F1? 4 (4/1024 = 1/256 =(1/4)4 c) How many t ‘days-to-seeding’ phenotypic classes were there in the F2 9 d) Was transgressive segregation observed? yes Explain F2 range was from 60 to 90 so was lees than low parent and more than high parent. e) How many days are added by each contributing allele? 5 f) Give potential ‘days-to-seeding’ genotypes for the 60 and 80 truebreeding parents: 60 A’A’, BB, CC, DD 80 AA, B’B’, C’C’, D’D’ g) Describe an experiment that would provide data that would allow you to calculate heritability of days to seeding in these chickpeas. Measure variance in each truebreeding parent or F1 where all individuals have the same genotype to estimate Ve. Measure variance in F2 to get an estimate of Vt. Subtract Ve from Vt to get Vg. Then H2 = Vg/Vt