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)

Answer 1

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