Question

Hello, please read the question and answer the 3 parts of this question (Q's are bolded)

Merfolk have three pairs of autosomes and one sex chromosome; males are X0 and females are XX.

In merfolk the genes controlling tail fin shape and scale colour are on different autosomes. Solid tail fin (Ff) is dominant to forked tail fin (ff), and the F allele is recessive lethal. There are two alleles of the scale colour gene – CB(blue) and CR (red)– CB and CR show incomplete dominance with heterozygotes being purple.

The silent trait is X-linked and dominant (XS) to normal voice (X+).

Atolla is a male merfolk, he has a forked tail fin, blue scales, and is voiceless (silent). Brachy is a female merfolk with a solid tail fin, purple scales and a normal voice.

What is the probability that a female child of a mating between Brachy and Atolla will be silent (it doesn't matter what tail shape or scale colour the child has)?

Select one:

a. 3/4

b. 1/4

c. 1/2

d. 0

e. 1

What is the probability that a mating between Brachy and Atolla would produce a child with a forked tail and purple scales (doesn’t matter whether they are silent or not)?

Select one:

a. 2/3

b. 1/2

c. 3/4

d. 1/3

e. 1/4

You mate a female merfolk with blue scales who is heterozygote for silent and a carrier of forked tail, to a silent male with blue scales who is also a carrier of forked tail. Which of the following are TRUE statements (you can choose more than one)

For this question ignore the phenotypes that are unmentioned. For example "a blue female" is a blue female who can be any colour, and either be silent or normal.

Select one or more:

a. This mating can produce silent or normal voiced offspring

b. The probability of an offspring from this mating having a forked tail and normal voice is 1/6

c. This mating can produce silent or normal voiced female offspring

d. The probability of a female offspring having a forked tail and blue scales is 1/3

Answer 1

First Question:

The answer will be 1 (Option e).

Explanation: As tail shape or scale color doesn't matter in this case, we will only consider sex chromosomes in this case. Genotype of Atolla will be XS0 as he is silent. Genotype of Brachy will be X+X+ as she has normal voice. The cross between them is shown in below Punnett square.

Gametes	XS	0
X+	XSX+ (Silent female)	X+0 (Male with normal voice)
X+	XSX+ (Silent female)	X+0 (Male with normal voice)

From the above Punnett square, we find that probability that a female child will be silent is 1 (We will ignore X+0 genotype in this case as only female child needs to be considered).

Second Question:

The answer will be 1/4 (Option e).

Explanation: In this case we only need to consider autosomes. Genotype of Atolla will be ff CBCB as he has forked tail fin & blue scales. Genotype of Brachy will be Ff CBCR as she has solid tail fin & purple scales. The only gamete of Atolla will be f CB. The gametes of Brachy will be F CB, F CR, f CB & f CR. The cross between them is shown in below Punnett square.

Gametes	f CB
F CB	Ff CBCB (Solid tail & blue scales)
F CR	Ff CBCR (Solid tail & purple scales)
f CB	ff CBCB (Forked tail & blue scales)
f CR	ff CBCR (Forked tail & purple scales)

From the above Punnett square, we find that probability that a child with a forked tail & purple scales is 1/4.