Question

Duplication of genes is an important evolutionary mechanism. As a result, many, cases are known in which a species has two or more identical genes.

Suppose there are two genes - A and B - that specify production of the same enzyme. These two genes show autosomal recessive inheritance. An abnormal phenotype results only if an individual does not make any of that enzyme. Only one of these two genes needs to be functional in order to make the enzyme.

What ratio of normal to abnormal progeny would result from a mating between two parents of genotype AaBb where A and B represent alleles that specify production of the enzyme, while a and b do not? Write your ratio like this - e.g 1:2

Suppose now there are three genes - A,B and C that specify production of this enzyme. Again, these traits show autosomal recessive inheritance. As before, only one single functional gene is sufficient for a wild-type phenotype. How many of the possible phenotypes would be normal? Write your answer as a number

number of normal phenotypes-

Answer 1

In a dihybid cross, between the heterozygous parents will give a phenotypic ratio of 9:3:3:1

9-when both genes are dominant

3 when only 1 is dominant either A or B

1 when its recessive for both A and B

As only one gene needs to be functional for the enzyme production, only homozygous recessive for both A and B will be unable to form the enzyme .

So the ratio will be 15:1(15 will have normal phenotype and 1 will have mutant phenotype)

In the trihybrid cross when three genes are involved the ratio is 27:9:9:9:3:3:3:1

27 - when all the 3 genes are dominant

9 - when any 2 genes are dominant while 1 is recessive

3 -only 1 dominant gene is there

1- no dominant gene is there

In this also only 1 which is homozygous recessive for all the 3 genes will be unable to form the enzyme.all the others have 1 or more then 1 dominant allele for the gene which can form the enzyme

So the ratio will be 63:1(63 will have normal phenotype and 1 will have mutant phenotype)