Question 1 (10 pts). The genetic linkage among 3 gene loci in horses is being st
ID: 259028 • Letter: Q
Question
Question 1 (10 pts). The genetic linkage among 3 gene loci in horses is being studied. The loci are part of autosomal chromosomes and they MIGHT be on the same pair of homologous chromosomes. There are two alleles at each locus. At the coat color locus, allele B is for black, and b is for white. At the height locus, allele H is for tall, and h is for short. At the tail locus, allele T is for long, and t is for short. Upper case alleles are completely dominant to their respective lower case allele. There are no interactions among the 3 loci.
A male horse with the phenotype black, tall and long-tailed was mated with a female horse that was white short and short-tailed. Those matings produced 800 progeny as described in the table:
Class #
Phenotypic Class of progeny
Number of progeny
1
Black, tall, long tail
250
2
Black, tall, short tail
13
3
Black, short, short tail
58
4
Black, short, long tail
63
5
White, short, long tail
15
7
White, short, short tail
290
6
White, tall, short tail
55
8
White, tall, long tail
56
total
800
What are the genotypes of the likely parental types of gametes produced by the male? (1 pt)
b. What locus is in the middle on the genetic map for these loci? (1 pt)
c. Calculate the recombination frequency between the coat color and the tail locus. Show your calculations. (2 pt)
Calculate the recombination frequency between the tail locus and the height locus. Show your calculations. (2 pt)
Calculate the recombination frequency between the coat color and the height locus. Show your calculations. (1 pt)
What is the expected frequency of double crossovers? Show your calculation. (1 pt)
What is the estimate of interference for these data? Show your calculation. (1 pt)
If the 3 loci were segregating independently of each other, how many children would be expected in the genotypic class, “BbhhTt? (1 pt)
Class #
Phenotypic Class of progeny
Number of progeny
1
Black, tall, long tail
250
2
Black, tall, short tail
13
3
Black, short, short tail
58
4
Black, short, long tail
63
5
White, short, long tail
15
7
White, short, short tail
290
6
White, tall, short tail
55
8
White, tall, long tail
56
total
800
Explanation / Answer
Answer:
a). Male gametes’ genotypes are BHT, BHt, BhT, Bht, bHT, bHt, bhT & bht
b). Tail locus, T or t
c). The recombination frequency between the coat color(B) and the tail locus(T)= 17.75%d).
d). The recombination frequency between the tail locus (T) and the height locus (H)= 18.25%
e). The recombination frequency between the coat color(B) and the height locus(H)= 29.25%
f). The expected frequency of double crossovers = 3.2%
g). The estimate of interference for these data = -0.09
h). BhT/bht = 100
Because, in independent assortment, the assortment of gametes is equal the all progeny are produce en equal proportion (800/8=100)
Hint:
Parental genotypes are more than any type of recombinant progeny. Hence parental genotype is BHT/bht
Class #
Phenotypic Class of progeny
Number of progeny
Genotype
1
Black, tall, long tail
250
BHT
2
Black, tall, short tail
13
BHt
3
Black, short, short tail
58
Bht
4
Black, short, long tail
63
BhT
5
White, short, long tail
15
bhT
7
White, short, short tail
290
Bht
6
White, tall, short tail
55
bHt
8
White, tall, long tail
56
bHT
Total
800
1).
If single crossover occurs between B&H
Normal combination: BH/bh
After crossover: Bh/bH
Bh progeny= 58+63= 123
bH progeny = 55+56= 111
Total of this progeny = 234
The recombination frequency between B & H = (number of recombinants/Total progeny) 100
RF = (234/800)100 = 29.25%
2).
If single crossover occurs between H&T
Normal combination: HT/ht
After crossover: Ht/hT
Ht progeny= 13+55= 68
hT progeny = 15+63=78
Total this progeny = 146
The recombination frequency between H & T = (number of recombinants/Total progeny) 100
RF = (146/800)100 = 18.25%
3).
If single crossover occurs between B&T
Normal combination: BT/bt
After crossover: Bt/bT
Bt progeny= 13+58= 71
bT progeny = 15+56=71
Total this progeny = 142
The recombination frequency between B & T = (number of recombinants/Total progeny) 100
RF = (142/800)100 = 17.75%
Recombination frequency (%) = Distance between the genes (cM)
b-------18.25cM--------t------17.75cM-------h
Heterozygous genotype = b+ d+ c / b d c+
Double crossover genotype = b+ d c / b d+ c+
Expected double crossover frequency = (RF between b & t) * (RF between t & h)
= 0.1825 * 0.1775 = 0.032 = 3.2%
The observed double crossover frequency = 13+15/800=14/900 = 0.035
Coefficient of Coincidence (COC) = Observed double crossover frequency / Expected double crossover frequency
= 0.035/0.032
= 1.09
Interference = 1-COC
= 1-1.09 = -0.09
Class #
Phenotypic Class of progeny
Number of progeny
Genotype
1
Black, tall, long tail
250
BHT
2
Black, tall, short tail
13
BHt
3
Black, short, short tail
58
Bht
4
Black, short, long tail
63
BhT
5
White, short, long tail
15
bhT
7
White, short, short tail
290
Bht
6
White, tall, short tail
55
bHt
8
White, tall, long tail
56
bHT
Total
800
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