Question

1.) Choose all of the following pairs that would share 1/2 of their genome (assuming no crossing over and non sex chromosome inheritance).

-Father and child

-mother and child

-Full siblings

-uncle and niece

-grandparent and grandchild

2.) A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height. The man's father was 6 feet tall, and both the woman's parents were of average height. Achondroplastic dwarfism is autosomal dominant, and red-green color blindness is X-linked recessive.

What proportion of their sons would be color-blind and of normal height?

Answer 1

1) The correct options are (a), (b) and (c)

The male gamete, spermatozoa contain 23 chromosomes including sex chromosomes whereas the female gamete, ovum, also contains 23 chromosomes including sex chromosomes. After fertilization, a zygote is formed which contains 46 chromosomes (23 pairs), 50% from each parent. Hence, a child shares 1/2 of his/her genome with mother and father both.

In full siblings, there is about a 50% overlap between the DNA from the mother and the DNA your brother or sister got from that same mother. So full siblings share 50% of 50% of the mother’s DNA that is 25%. Similarly, there is about a 50% overlap between the DNA from the father and the DNA your brother or sister got from that same father and thus they share 50% of 50% of the father’s DNA that is 25%. So the total percentage that full siblings share is 50%.

Options-(d) and (e) are incorrect because they share only 25% of their genome with each other because they are either from the mother's side or from the father's side but not from both. The grandparent or uncle is either from the mother's side or from the father's side, thus sharing 50% of the genome with mother and father. But their child will share only 50% of this shared genome that is 25%.

2) A man who is an achondroplastic dwarf with normal vision marries a color-blind woman of normal height.

Achondroplastic dwarfism = autosomal dominant

Red-green color blindness = X-linked recessive

As the man's father was 6 feet (homozygous recessive), which means the man is heterozygous for Achondroplastic dwarfism as it is an autosomal dominant trait.

Both the woman's parents were of average height means, the woman genotype is homozygous recessive.

So the cross will produce the following offsprings:-

Aa/aa	a	a
A	Aa	Aa
a	aa	aa

50% will be Achondroplastic dwarfism and 50% will be of normal heights.

Now, the male has normal vision whereas the female is color blind.

The genotype of male will be = XY

The genotype of the female will be = X'X' (color-blind)

So the cross will produce the following offsprings:-

XY/X'X'	X'	X'
X	XX'	XX'
Y	X'Y	X'Y

All sons will be color blind whereas all daughters will be carriers for the color blind.

So, the proportion of their sons for color-blind and of normal height would be = 1/2 (as 50% chance of having a son) X 1 (all sons will be affected with color blind) X 1/2 (50% will have normal height)

Therefore, the final probability would be = 1/4