Question

. (4 pts.) The mouse autosomal genes B and S are linked and 38 map units apart. We cross mouse strains BBSS and bbss to produce an F1. That F1 is the test crossed to bbss.

Predict the number of progeny from the test cross in each of the following phenotype groups:

B_S_ =

B_ss =

bbS_ =

bbss =

Answer 1

parent: BBSS x bbss

F1: BbSb

Test: BbSs x bbss

F1:

male/ female gametes	bs
BS	BbSs
Bs	Bbss
bS	bbSs
bs	bbss

since the map unit is 38, rate of recombination is 38%.

Hence, progeny possessing parental phenotype/ genotype is 62% (100-38= 62).

so, BbSs and bbss are parental and together constitute 62% of the progeny. Similarly, Bbss and bbSs are recombinants and constitute 38%. So logically looking at it, each of BbSs and bbss have 62/2= 31% occurance and Bbss and bbSs have 38/2= 19% occurance each.

Hence,

phenotype	no. of progeny for every 100 progenies	fraction of progeny population possessing the phenotype
BbSs	31	0.31
Bbss	19	0.19
bbSs	19	0.19
bbss	31	0.31