Try it on paper first. A hint, the key idea and the worked solution are here when you need them.
A plant is heterozygous for two linked genes, with the alleles C and D on one chromosome and c and d on the other. It is crossed with a plant that is ccdd. The cross gives offspring:
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. CcDd, ccdd, Ccdd and ccDd, out of .
Plan. The big classes, CcDd and ccdd, have the parental combinations. The recombinants are Ccdd and ccDd.
Step 1. (a) offspring.
Step 2. (b) .
Step 3. (c) One per cent of recombination is one map unit, so the genes are map units apart.
Answer: recombinants; ; map units.
Check. The parental classes hold offspring, and .
Watch out. Do not divide by the number of parental offspring. The frequency is a share of all the offspring.
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