Question

Albinism in humans can be caused by recessive mutations in different genes that cause a total lack of melanin pigment. A complementation test would be available when two people with albinism mate to produce offspring. If the two people carry mutations in different genes that each cause albinism, what frequency would albinism occur in their children?

Answer 1

none (zero frequency)

Step-by-step explanation:

The complementation test is widely used in genetics for assessing whether two mutations associated with a phenotypic trait represent two different gene variants of the same gene (i.e., two alleles of the same gene) or if they are mutations associated with two different genes. In a recessive trait, the offspring must receive one recessive allele from each parent in order to show the recessive phenotype (in this case, albinism). Moreover, in heterozygous individuals for a trait that exhibits complete dominance, recessive alleles always are masked by dominant alleles. Albinism is a multigenic (polygenic) recessive trait, thereby individuals who carry recessive mutations causing albinism on different genes will have zero probability (none) of having offspring with albinism.

Answer 2

albinism frequency ---> 3/4 = 75% = 0.75 of the progeny will be albino.

Step-by-step explanation:

Available data:

• Albinism can be caused by recessive mutations in different genes that cause a total lack of melanin pigment.
• Two people with albinism mate to produce offspring
• The two people carry mutations in different genes that each cause albinism

Let us say that genes A and B are involved.

• Gene A: AA and Aa individuals will express pigments, while aa individuals will be albinos
• Gene B: BB and Bb individuals express color, while bb are albinos.

Cross: two people with albinism that carry the mutation in different genes

Parentals) Aabb x aaBb

Phenotype) Albino both parents

Gametes) Ab, Ab, ab, ab

aB, aB, ab, ab

Punnett square) Ab Ab ab ab

aB AaBb AaBb aaBb aaBb

aB AaBb AaBb aaBb aaBb

ab Aabb Aabb aabb aabb

ab Aabb Aabb aabb aabb

F1) 12/16 will be albino

4/16 will express color

4/16 will be Aabb

4/16 will be aaBb

4/16 will be aabb

12/16 = 3/4 = 75% = 0.75 of the progeny will be albino.