Question

D. 25% orme condition. The male is XX and the female is XY. But X-linked genes have the same effects in mugwumps as in humans. A recessive X-linked allele "n" produces red-green color blindness. The dominant allele "N" produces normal color vision. If a female mugwump with normal color vision mates with a male mugwump, phenotypically normal for color vision but whose mother was color blind, what is the probability that a son from that mating will be color blind? A daughter? b. C. d. e. f. 100%, 0% 75%,25% 50%, 25% 25%,50% 096,50% 0%, 75% Write out:I 1. Traits 2. Cross (2 genotypes) 3. male and female gametes 4. Punnet Square with offspring genotypes 5. offspring phenotypes

Answer 1

Given the male is XX and female is XY.

Recessive x linked gene for color blindness is ‘n’ and dominant gene for normal vision is ‘N’.

The female has normal vision which means its genotype is XN Y. The male is phenotypically normal but it’s mother is color blind which means it’s heterozygous for this trait. It’s genotype is XN Xn.

The possible gametes are XN, Y from female and XN, Xn from male. The possible zygotes are XN XN, XN Xn, Y XN and Y Xn.

The probability of son being color blind is zero, 0%. The probability of daughter being color blind is 1 out of 2 which is 50%. This is option e.

1. The traits are normal colour vision and red green colour blindness.

5. The possible phenotypes of offsprings are (a)norml vision male child, (b)normal vision female child and (c)color blind female child.

2. The genotypes possible are (a) XN Xn heterozygous norml vision male (b)XN XN homozygous normal vision male (c)Y XN normal vision female and (d)Y Xn color blind female

3. The male gametes are XN and Xn. The female gametes are XN and Y

4. The punnet square shows male gametes Xn and XN crossing with female gametes Y and XN. The offspring genotypes are XN XN, XN Y, XN Xn and Xn Y.

Xn

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Xn