Question

bio 227 lab 3, 10 points your name date (c) another form of deafness is caused by a rare autosomal recessive gene. in a mating involving a deaf man and a deaf woman, could some of the children have normal hearing? explain your answer. 4) below is a pedigree of a fairly common human hereditary trait where the boxes represent males and the circles represent females. shading symbolizes the abnormal phenotype. given that one gene pair is involved, (a) is the inheritance pattern x - linked or autosomal, recessive or dominant? (b) give the genotype of each individual in the pedigree. if more than one genotypic possibility exists, present all possible alternatives. 5) in the mouse, gene a allows pigmentation to be deposited in the individual coat hairs, while its allele a prevents such deposition of pigment, resulting in an albino. gene b gives agouti (wild - type fur), while its allele b gives black fur. (a) diagram the cross between a doubly heterozygous agouti mouse mated with a doubly homozygous recessive white mouse. (b) what would be the expected phenotypic ratio in the progeny?

Explanation:

3(c) Step1: Understand autosomal - recessive inheritance

For an autosomal - recessive trait, a deaf person can be homozygous recessive (aa) or a carrier (Aa). If both the deaf man and the deaf woman are carriers (Aa), they can have children with normal hearing.

3(c) Step2: Analyze the cross

When two carriers (Aa x Aa) have children, the Punnett - square analysis gives genotypes AA:Aa:aa = 1:2:1. AA and Aa individuals have normal hearing. So, some of the children could have normal hearing.

4(a) Step1: Check for X - linked inheritance

If it were X - linked recessive, an affected female would need to have an affected father and an affected son (in most cases). Here, the affected female has an unaffected father, so it is not X - linked recessive. If it were X - linked dominant, an affected male would pass the trait to all his daughters, which is not the case here. So, it is autosomal.

4(a) Step2: Determine dominance or recessiveness

Since two unaffected parents can have an affected child, the trait is recessive.

4(b) Step1: Label the genotypes

Let the normal allele be A and the mutant allele be a. The affected individuals have genotype aa. Unaffected parents of affected children must be carriers (Aa). Unaffected individuals with no affected parents or children can be AA or Aa.

5(a) Step1: Determine the genotypes of the parents

The doubly heterozygous agouti mouse has genotype AaBb, and the doubly homozygous recessive white mouse has genotype aabb.

5(a) Step2: Set up the cross

The cross is AaBb x aabb. The gametes from AaBb are AB, Ab, aB, ab and the gametes from aabb are ab.

5(b) Step1: Use the Punnett - square or product rule

Using the product rule:

• For the A - a cross: Aa x aa gives Aa:aa = 1:1.
• For the B - b cross: Bb x bb gives Bb:bb = 1:1.
• The phenotypic ratios are: Agouti (AaBb): Black (Aabb): Albino (aaBb and aabb)= 1:1:2.

Answer:

3(c): Yes, if both parents are carriers (Aa).
4(a): Autosomal, recessive.
4(b): Affected individuals: aa; Unaffected parents of affected children: Aa; Other unaffected individuals: AA or Aa.
5(a): Cross: AaBb x aabb with gametes AB, Ab, aB, ab (from AaBb) and ab (from aabb).
5(b): Agouti:Black:Albino = 1:1:2.