Question

Question 12 0.5 pts The a and b loci are 40 CM apart. An AABB individual mates with an aabb individual. What gametes will the F1 individual of this cross produce and in what proportions? AB 20% Ab 30% aB 30% ab 20% AB 30% Ab 20% aB 20% ab 30% AB 60% Ab 40% aB 40% ab 60% AB 25% Ab 25% aB 25% ab 25%

Answer 1

12. Answer :- AB - 30% , Ab - 20% , aB - 20% , ab - 30%

Explanation :- According to the question it is a typical dihybrid cross where AABB mates with aabb. So here f1 individual is AaBb which produce four different type of gametes which are AB , Ab , aB , ab. According to the question a and b loci are 40 cM apart that means there is 40% probability of these gene to be inherited together so rest 60% are non recombinant.

Here recombinant gamete produce 40% i.e Ab 20% and aB 20% and 60% is non recombinant i.e AB 30% and ab 30%.

13. Answer :- AB 21% , Ab 29% , aB 29% , ab 21%.

Explanation :- It is also a typical dihybrid cross like the question 12 and here also gametes produced by f1 individual remain same i.e AB , Ab , aB , ab but here a and b loci 58 cM apart that means 58% of these gene inherited together and other 42 % are non recombinant.

So recombinant gametes are 58% i.e Ab 29% , aB 29% and non recombinant are 42% i.e AB 21% and ab 21%.

14. Answer :- AB 20% , Ab 30% , aB 30% , ab 20%.

Explanation :- ​​​​​​ It is also a dihybrid cross but here AAbb mates with aaBB so f1 individual is AaBb and it produce 4 different gametes i.e AB , Ab , aB , ab. Here a and b loci are 40 cM apart i.e there are 40 % probability that these genes are inherited together and rest 60% are non recombinant.

Here recombinant gametes are 40% i.e AB 20% and ab 20% and other 60% non recombinant i.e Ab 30% and aB 30% .