Question

In: Biology

1. Female mammals who are carriers of an X-linked recessive trait caused by a non-functional allele...

1. Female mammals who are carriers of an X-linked recessive trait caused by a non-functional allele would express the encoded protein:

a. at the full amount in some cells, but express no protein in other cells

b that displays a novel function in the protein to produce a different dominant trait

c that displays dominance due to haploinsufficiency

d at 1/2 of the normal amount in each cell in a form of the mutant protein

e that would stop function in a dominant negative manner

2. You are following segregation of four genes in a cross – A, B, C, and D Dominant alleles are capital letters and recessive alleles are small case.

You take the cross - aa Bb cc Dd x Aa Bb CC Dd. What is the frequency expected among the progeny for the heterozygous genotype: Aa Bb Cc Dd?

1/4

1/16

1/32

0

1/8

Expert Solution

Answer 1:

Males have single X chromosome, but females have two X chromosome, and to maintain dosage compensation, one of the X chromosome is randomly inactivated in each female cell.

Thus, if a female is carrier for a X-linked trait for non-functional allele, then some of her cells would express comple amount of protein, whereas some cells will not produce any protein.

Suppose that the normal allele is X+ and non-functional allele is X-, which doesn't produces any protein.Thus, carrier female would have the genotype X+X-.

In some cells, the chromosome containing X+ will be randomly inactivated and thus would not express any protein.

Other cells, in which X- would be randomly inactivated, would express the X+ allele and full amount of protein would be made.

Hence, correct choice should be:

(a)	at the full amount in some cells, but express no protein in other cells

Answer 2:

Given cross - aa Bb cc Dd x Aa Bb CC Dd.

Cross at each locus can be shown as :

aa×Aa

Gametes	A	a
a	Aa	aa

Thus, probability of Aa, p(Aa) =1/2

Bb×Bb

Gametes	B	b
B	BB	Bb
b	Bb	bb

p(Bb) = 2/4, which becomes 1/2 on simplification.

cc × CC

Gametes	C
c	Cc

Thus, p(Cc) = 1

Dd×Dd

The cross will be similar to Bb×Bb and p(Dd) will also be 1/2.

We need to find probability of Aa Bb Cc Dd progeny.

Using product rule,

p(AaBbCcDd)

=p(Aa) × p(Bb) × p(Cc) × p(Dd)

=(1/2) ×(1/2) ×(1) ×(1/2)

=1/8

Hence, correct choice should be 1/8.

gladiator answered 4 months ago

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