Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
Test crosses of three linked genes, M, N and O, give these recombination frequencies:
Linked genes lie on the same chromosome. In a test cross of a double heterozygote with a double recessive, the recombinant offspring are the small classes. The recombination frequency is the number of recombinants divided by all the offspring, in per cent, and one per cent is one map unit. Of three linked genes, the two that are farthest apart are the end genes.
Have you tried it?
Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Given. M to N , N to O , M to O .
Plan. The largest distance is between the end genes. It must equal the sum of the other two.
Step 1. (a) The largest value is , between M and N, so M and N are the ends. Check: . The gene O lies between them, and the order is M, O, N.
Step 2. (b) The frequency for M and N is : offspring.
Step 3. (c) The smallest distance is , between M and O. These two are the most tightly linked.
Answer: (a) O is in the middle; (b) offspring; (c) M and O.
Check. The order M, O, N gives the distances , and , as in the data.
Watch out. The order M, N, O would need M to O to be , and that is not the case.
Log in to join the discussion.
Loading…
With Standard, you also open 12 more problems, 2 Python labs and 1 extra course. Everything you have used here stays free, always.