Question

Princess Beauty comes from a very noble family, and has inherited an X-linked Hemophilia A mutation from her mother. Her father King Chuck is red-green color blind. She has recently married Prince Brüno, who has no family history of X-linked conditions (what kind of royal is that?!). What is the likelihood that their first child will be a son who will not be hemophilic or color blind? The two mutations are 20 map units apart. a. 0.05 b. 0.1 c. 0.20 d. 0.25 e. 0.

Answer 1

Answer:

a). 0.05

Explanation:

Hemophilia A = Xh

Colorblindness= Xc

Princess (XhC/XHc) x (Bruno) XHCY ----Parents

Distance between genes (mu) = Recombination frequency (%)

XhC/XHc genotype produces 4 types of gametes as below:

Recombinant gametes (20%):

XHC = 10%                                    

Xhc = 10%

Non-recombinant gametes (80%):

XHc = 40%

XhC = 40%

XHCY genotype produces only two types of gametes XHC (50%) & Y (50%).

	XHC (50%)	Y (50%)
XHC = 10%	XHC / XHC = 10% * 50% = 0.05 (wild type for both traits)	XHC / Y= 10% * 50% = 0.05 (wild type for both traits)
Xhc = 10%	Xhc/ XHC = 10% * 50% = 0.05 (wild type for both traits)	Xhc/ Y = 10%* 50% = 0.05 (hemophilia and colorblind)
XHc = 40%	XHc/ XHC = 40% * 50% = 0.2 (wild type for both traits)	XHc/ Y = 40%* 50% = 0.2 (colorblind only)
XhC = 40%	XhC/ XHC = 40% * 50% = 0.2 (wild type for both traits)	XhC/ Y = 40%* 50% =0.2 (Hemophilia only)
	Female	Male

The likelihood that their first child will be a son who will not be hemophilic or color blind = XHC/Y = 0.05