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Harman [31]
3 years ago
14

The amount of enzyme activity in a cell that is homozygous for a mutant allele is 400 units. The amount of enzyme activity in a

cell homozygous for the WT allele is 200 units. The amount of enzyme activity in a heterozygote is 300 units.
What is the dominance relationship between the WT and mutant allele for the phenotype of amount of enzyme per cell?

A) The mutant allele is dominant.
B) The WT and mutant alleles are codominant.
C) The WT allele is dominant.
D) The WT and mutant alleles show incomplete dominance.
E) It is impossible to determine from the information given.
Biology
1 answer:
lutik1710 [3]3 years ago
7 0

Answer:

D) The WT and mutant alleles show incomplete dominance.

Explanation:

Incomplete dominance is an example of intragenic interaction where the two alleles interact with each other. Neither allele is dominant and when they are present together they produce a blended trait i.e. an intermediate trait between the two extremes.

Here, enzyme activity produced by homozygous WT genotype is 200 units. Enzyme activity produced by homozygous mutant genotype is 400 units. Heterozygote ( one mutant and one WT allele ) produces an intermediate enzyme activity of 300 units because it lies between 200 and 400 units. Hence, WT and mutant alleles show incomplete dominance.      

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HELP PLS I WILL GIVE LOTS OF POINTS pls answer good and i will give brainliest and give more points if its correct
Fofino [41]

Answer: .374 and .626

Explanation:

5. total number of alleles = 4712 + 2816 = 7528

       frequencey of B allele = B allele / total number of alleles --> 2816/7528 = 0.374                

             

6. total number of alleles =  4712 + 2816 = 7528

   freq of b allele = b allele/total number of alleles --> 4712/7528 = 0.626

We can confirm our answer as correct because if we add up the answers to 5 and 6 we should get one:

0.374 + 0.626 = 1

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hodyreva [135]
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The cell would lose structure and shape when the ribosomes of a cell were damaged and stopped functioning.

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