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noname [10]
2 years ago
10

If A/A ⋅ B/B is crossed with a/a ⋅ b/b and the F1 is testcrossed, what percentage of the testcross progeny will be a/a ⋅ b/b if

the two genes are (a) unlinked; (b) completely linked (no crossing over at all); (c) 10 m.u. apart; (d) 24 m.u. apart?
Biology
1 answer:
Alex787 [66]2 years ago
5 0

Answer:

a) 25%; b) 50%; c) 45%; d) 38%

Explanation:

<h3 /><h3>a) Unlinked genes</h3>

A/A B/B X a/a b/b

F1: A/a B/b

Testcross A/a B/b X a/a b/b

The homozygous recessive individual only produces <em>ab</em><em> </em>gametes.

The F1 produces four types of gametes (each of them with a frequency of 1/4):  <em>AB</em>, <em>Ab</em>, <em>aB </em>and <em>ab</em>.

25% of the progeny will be a/a b/b.

<h3>b) Completely linked genes</h3>

AB/AB X ab/ab

F1: AB/ab

Testcross AB/ab X ab/ab

The F1 produces only two types of gametes, the parentals (each of them with a frequency of 1/2):  <em>AB</em> and <em>ab</em>.

50% of the progeny will be ab/ab.

<h3>c) 10 m.u. apart</h3>

AB/AB X ab/ab

F1: AB/ab

Testcross AB/ab X ab/ab

The F1 produces four types of gametes, the parentals <em>AB</em> and <em>ab </em>and the recombinants <em>Ab</em> and <em>aB</em>.

Since the genes are 10mu apart, 10% of the produced gametes will be recombinant and 90% will be parentals. Since there are two types of parental gametes, each of them has a frequency of 45%.

45% of the progeny will be ab/ab.

<h3>d) 24 m.u. apart</h3>

This is very similar to c).

Since the genes are 24mu apart, 24% of the produced gametes will be recombinant and 76% will be parentals. Since there are two types of parental gametes, each of them has a frequency of 38%.

38% of the progeny will be ab/ab.

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