Genetics

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AFLATPEG

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A geneticist testcrosses a dihybrid (heterozygous for two genes) mouse and notices the double dominant and double recessive phenotypes are present at 8% and 10% frequency, respectively. Which of the following is the best conclusion from these data?


(A) The dominant allele of one gene is linked to the recessive allele of the other, and the two genes are approximately 20 map units apart.
(B) The dominant alleles of the two genes are linked and the two genes are 80 map units apart.
(C) The recessive alleles of the two genes are approximately 20 map units apart and are not linked.
(D) The two genes are linked and 80 map units apart, but it cannot be determined which alleles are being inherited together.



correct answer bolded. explain?
 

gwjib04

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So the double dominant and double recessive should be 6.25% each in an unlinked Mendelian dihybrid cross. But here they're 8% and 10% which is abnormally high.

If the dihybrid was GgYy, it seems to me that the linked alleles (because they're less than 50 map units so they tend to recombine together) should be the GY and gy because there's an increased proportion of homozygous offspring.

The answer I would have chosen is
"The dominant alleles of the two genes are linked and the two genes are 20 map units apart."

But that's not listed.
 

Ibn Alnafis MD

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So the double dominant and double recessive should be 6.25% each in an unlinked Mendelian dihybrid cross. But here they're 8% and 10% which is abnormally high.

If the dihybrid was GgYy, it seems to me that the linked alleles (because they're less than 50 map units so they tend to recombine together) should be the GY and gy because there's an increased proportion of homozygous offspring.

The answer I would have chosen is
"The dominant alleles of the two genes are linked and the two genes are 20 map units apart."

But that's not listed.

How so?

Mendelian dihybrid test cross results in 9:3:3:1 phenotype ratios. You are right that the double recessive ratio will be 6.25%, but shouldn't the double dominant trait be 56.25%?
 

WSHRocks

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The dihybrid is being test crossed, which means that in your example GgYy is being crossed with ggyy (a test cross always crosses the tested organism with a homozygous recessive organism to observe heterozygosity). Setting up the Punnett square GY,Gy,gY,gy versus gy (for all 4 combinations), you find that each combination is 25%, meaning that GY and gy should be 25% each, so they are NOT as frequent in the observed cross. This is due to the increased frequency of Gy and gY due to linkage (one dominant and one recessive of each gene being linked). The distance between the genes is the result of recombination, which in this case is 18% of progeny or 18 map units (so about 20).
 

Ibn Alnafis MD

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The dihybrid is being test crossed, which means that in your example GgYy is being crossed with ggyy (a test cross always crosses the tested organism with a homozygous recessive organism to observe heterozygosity). Setting up the Punnett square GY,Gy,gY,gy versus gy (for all 4 combinations), you find that each combination is 25%, meaning that GY and gy should be 25% each, so they are NOT as frequent in the observed cross. This is due to the increased frequency of Gy and gY due to linkage (one dominant and one recessive of each gene being linked). The distance between the genes is the result of recombination, which in this case is 18% of progeny or 18 map units (so about 20).

Thank you for clarifying that testcross means crossing the organism with one that is homozygous recessive for both traits.

Doing that, according to Independent Assortment, will yield offsprings with the dominant trait (GgYy) and the recessive trait (ggyy) 50%/50%. However, the question stem indicates that only 8% had the dominant phenotype and 10% had the recessive phenotype. Therefore, the frequency of the recombinant offsprings should be 82%. Shouldn't that translate into ~80 map units separation between the genes?
 

WSHRocks

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Thank you for clarifying that testcross means crossing the organism with one that is homozygous recessive for both traits.

Doing that, according to Independent Assortment, will yield offsprings with the dominant trait (GgYy) and the recessive trait (ggyy) 50%/50%. However, the question stem indicates that only 8% had the dominant phenotype and 10% had the recessive phenotype. Therefore, the frequency of the recombinant offsprings should be 82%. Shouldn't that translate into ~80 map units separation between the genes?

I think it's easiest to take a step back first. We notice that there is an increased frequency of heterozygous allele assortments (all non-GY, non-gy, so 82%) as compared to expected Mendelian inheritance via independent assortment. This increased frequency is due to Gy and gY alleles being linked and not GY and gy, otherwise the frequency would be reversed (this is due to the rarity of recombination). With this in mind, GY and gy ARE the recombinant offspring and not Gy and gY, which are non-recombinant. Therefore, it must be 18% recombinant, thus ~18 map units.
 
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