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- #Biology
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A plant of the genotype AaBb is selfed. The two genes are linked and are 50 map units apart. What Proportion of the progeny will have the genotype aabb?
1/2
1/4
1/8
1/16
Answers (1)
The correct proportion of progeny with the genotype aabb can be determined by understanding the principles of genetic recombination and linkage.
When two genes are linked, their alleles tend to be inherited together more often than expected by chance due to their physical proximity on the same chromosome. The degree of linkage is measured in map units or centimorgans (cM), and it represents the frequency of recombination between the genes.
In this case, the genes A and B are 50 map units apart. A 50% recombination frequency corresponds to a genetic distance of 50 map units. This means that there is an equal probability of recombination and non-recombination occurring between the two genes during meiosis.
The genotype AaBb can produce four possible gamete combinations: AB, Ab, aB, and ab. Each of these gametes has an equal chance of being produced.
When these gametes combine during self-fertilization, the resulting progeny genotypes and their respective proportions can be calculated using a Punnett square:
A a
-----------------
B | AB aB
b | Ab ab
From the Punnett square, we can see that 1 out of the 16 possible combinations results in the aabb genotype (lower right corner).
Therefore, the proportion of progeny with the genotype aabb would be 1/16 or approximately 0.0625, which is approximately 6.25%. Hence, the correct answer is option 4.
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