Answer all of 3 A cross was carried out between two individu

A cross was carried out between two individuals having the following genotypes: (/calculations for full credit) BbddEeGGHhNnrr times bbDdeeggHhNNRr How many different gametic genotypes can each parent produce? Female _____ Male _____ Assume independent assortment and give the gametic genotypes and gametic ratios for the male parent. What is the probability that these two individuals will produce an offspring having the following genotype? bbDdEcGghhNnrr What is the probability that these two individuals will produce an offspring having the dominant phenotype with respect to all 7 traits?

Solution

BbddEeGGHhNnrr bbDdeeggHhNNRr

Since the female is heterozygous for four genes (B, E, H and N), it will form total 2⁴= 16 types of gametes
Since the male is heterozygous for three genes (D, H and R), it will form total 2³= 8 types of gametes.

B: Gentoype of male parent: bbDdeeggHhNNRr
Gametic genotypes and ratio= 1 bDegHNR: 1 bdegHNR: 1 bDeghNR:1 bDegHNr: 1 bdeghNR: 1 bdeghNr: 1 bDeghNr:1 bdegHNr

C: probability of genotype bbDdEeGghhNnrr
Bb x bb= 1/2 Bb: 1/2 bb
dd x Dd= 1/2 Dd : 1/2 dd
Ee x ee= 1/2Ee: 1/2 ee
GG x gg= 1Gg
Hh x Hh= 1/4 HH: 1/2 Hh: 1/4 hh
Nn x NN= 1/2 Nn: 1/2NN
rr x Rr= 1/2 Rr: 1/2 rr
So, probability of genotype bbDdEeGghhNnrr= 1/2 bb x 1/2 Dd x 1/2Ee x 1Gg x 1/4 hh x 1/2 Nn x 1/2 rr = 1/128

D: probability of genotype bBDdEeGgHH (or Hh)Nn (or NN) Rr
1/2 Bb x 1/2 Dd x 1/2Ee x 1Gg x 3/4 HH and Hh x 1 Nn (NN) x 1/2 Rr = 3/64