Question

Short Answer 27. An individual has the following genotype. Gene loci (A) and (B) are 20 map units apart, which determines their probability of recombination. 1) Indicate all the possible gametes that this individual can produce (with crossing over), 2) Label the gametes that are recombinant and non- recombinant. 3) Calculate the recombination frequency of the recombinant and non-recombinant gametes. The genes are not COMPLETELY linked. bb BB 28. Now imagine the genes are completed linked in #27. 1) What are all possible gametes?

Answer 1

27.

1. The possible gametes if crossing over takes place are Ab, aB, ab and AB.

2. Parental gametes- Ab, aB. Explanation- Those gametes which have genotype combination same to that of their parents are called parental gametes.

Recombinant gametes- AB, ab. Explanation- Those gametes which have genotype combination different than that of their parents are called recombinant gametes.

3. Recombination frequency = Map distance

So, percentage of recombination = 20 %

Recombination frequency of AB gamete = 20 ÷ 2 = 10%

Recombination frequency of ab gamete = 20 ÷ 2 = 10%

Recombination frequency of Ab gamete = (100 - 20) ÷ 2 = 80 ÷ 2 = 40%

Recombination frequency of aB gamete = (100 - 20) ÷ 2 = 80 ÷ 2 = 40%

28.

1. The possible gametes are aB and Ab. Explanation- If the genes are completely linked, then there will be no recombination and hence, only parental type gametes are produced.

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