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alisha [4.7K]
3 years ago
11

A mutation creates a dominant negative allele of a particular gene. The normal allele of the gene encodes a protein that forms a

trimer within the cell. If one or more of the subunits has the mutant structure, the entire trimeric protein is inactive. In a heterozygous cell, if the proteins of both alleles are present at the same levels, what percent of the trimers present in the cell will be inactive
Biology
1 answer:
Wewaii [24]3 years ago
8 0

Answer:

87.5% of the trimers in the cell will be inactive.

Explanation:

A monomer is a single subunit. Look at an easy comparison. If a monomer is a ball. For the protein to be functional, it needs to be a trimer .A trimer = 3 monomers joined / bound together with bonds. Therefore, if a monomer is a ball, we need 3 balls, we attach it together and that is the primer . if one ball becomes missing, that protein won't work .

In chemistry, a trimer is regarded as three molecules or ions of the same substance (that is, same type) which are associated together .

Double negative : Each gene has 2 alleles. So, for a positive protein, both allele needs to be same, let's say + means on. So for a x protein , if it's x+/x+ it means the protein is homozygous and active. If for some mutation, it's x+/x- , it's a heterozygous protein. Double negative means when it is x-/x- and this means the protein has lost is function . It's not functional at all.

The heterozygous cell will form one or more subunits has the mutant structure when the gene is expressed. From the question if there is a mutation in dominant negative allele, it will cause the inactivation of one trimer resulting in 87.5% inactive trimers for a heterozygous individual.

There are 87.5 % percent of the trimers present in the cell which will be inactive for a heterozygous individual, If at least one of the subunits has the mutant structure.

The probability for each individual subunit to be a mutant is 1/2

Gene codes for a protein that forms trimer so the percentage of the trimers present in the heterozygous individual is third power of 1/2 =1/8

1/8=0.125 will be active.

1- 0.125 will be inactive.

=0.875 will be inactive.

0.875 × 100= 87.5%

Therefore 87.5% will be inactive.

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