Question

Question 1 (10 pts). Ilne genetic linkage among 3 gene loci in norses is being stuaiea. ine 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 progen Black, tall, long tail Black, tall, short tail Black, short, short tail Black, short, long tail White, short, long tail White, short, short tail White, tall, short tail White, tall, long tail 250 2 3 4 58 7 6 8 total 63 290 56 800 a. 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) d. Calculate the recombination frequency between the tail locus and the height locus. Show your calculations. (2 pt) e. Calculate the recombination frequency between the coat color and the height locus. ShowW your calculations. (1 pt) f. What is the expected frequency of double crossovers? Show your calculation. (1 pt) g. What is the estimate of interference for these data? Show your calculation. (1 pt) h. If the 3 loci were segregating independently of each other, how many children would be expected in the genotypic class, "BbbhTt? (1 pt)

Answer 1

A T 13 uT y Gendistaruュ No.ohnsmimant Coo-DO) Tula) no ん→58 BT ん

h. If the genes are showing independent assortment the ratio will be 1:1:1:1:1:1:1:1, so the proportion of each genotype is 1/8. Then the freequency of BbhhTt will be = 1/8 *800 = 100.

G3 bt btH_うち5 63 + 55 13S SDO ん Taue ↑00 77 18 100 O 032 J& The valuehere should be qual to less than 1, as here the value is 1.08 please verify the data and let me know...