Question

Mr Crabs suffers from semi-sterility due to a pericentric inversion on chromosome #10. His normal chromosome #10 has the sequence of genes as follows (the dot represents the centromere): ABCD·EFGH. His inverted #10 has the sequence: AFE·DCBGH.

a. What proportion of Mr Crab’s gametes will be viable if no crossing over occurs within the inverted segment? b. What proportion of Mr Crab’s gametes will be viable if a cross over occurs between genes F-E?

c. What is the sequence of the genes on the chromosome in the two gametes that are recombinant due to the cross over between genes F-E?

Answer 1

a. Out of 4 gametes, all the 4 gametes formed will be viable if no crossing over occurs within the inverted segment.So, the proportion is 4/4 = 1

b.Out of 4 gametes, 2 gametes formed will be viable if crossing over occurs between genes F-E.So, the proportion is 2/4 = 0.5. This is because a cross over in the F-E region will form the two gametes that will be with inverted gene sequence while the other two gametes formed will be with duplication or deletion.

d.The sequence will be :

                     AEF.DCBGH

c.