9. In humans pattern baldness is a sex-influenced trait, whose degree of express

Question

9. In humans pattern baldness is a sex-influenced trait, whose degree of expression is controlled by the sex of the individual, such that it is autosomal dominant in males while autosomal recessive in females. On the other hand, colour blindness and haemophilia are bothrecessive X-linked traits.

B = bald; b = non-bald; C = normal vision; c = colour blind; H = normal blood clotting; h = haemophilia.

a. Give the genotypes for individuals in the P and F1 generations for the matings below. Assume that no crossing over has occurred.

Mating 1

P: bald, non-colour blind, non-haemophiliac female X non-bald, colour blind, haemophiliac male

F1: bald, colour blind, haemophiliac male

Mating 2

P: bald, colour blind, non-haemophiliac female X bald, non-colour blind, haemophiliac male

F1: non-bald, non-colour blind, non-haemophiliac female

b. Suppose that the male and female from the F1s marry, what are the probabilities of the male and female offspring outlined below?

i. haemophiliac but otherwise normal

ii. colour blind and haemophiliac but normal for baldness

iii. colour blind but otherwise normal

iv. bald, colour blind and haemophiliac

Explanation / Answer

a) Mating 1

P: bald, non-colour blind, non-haemophiliac female X non-bald, colour blind, haemophiliac male

F1: bald, colour blind, haemophiliac male

P/Male = bb”c”XY”h”XY” ("c"XY indicates X linked recessive for colour blindness,"h"XY indicates X linked recessive for haemophilia)

P/Female = BBCcHh, BBCCHH (Female would be BB since phenotype is bald and since being a recessive gene, has to be homozygous for B, she can be "Cc" or "CC" for colour blindness, phenotypically she isnt colour blind but might be a carrier, similar point for hameophilia)

F1(Male)= Bb”c”XY”hXY (Bald,colour blind haemophilic male)

Mating 2

P: bald, colour blind, non-haemophiliac female X bald, non-colour blind, haemophiliac male

F1: non-bald, non-colour blind, non-haemophiliac female

P/Male = Bb”C”XY”h”XY” (male is bald, can be BB or Bb since its an autosomal dominant trait for male)

               BB”C”XY”h”XY”

P/Female = BBccHh,BBccHH

F1(Female)= BbCcHh (Female would be Bb since she is non-bald and baldness gene is autosomal recessive for females, for a female to be bald , she needs to be BB)

b. considering Male and Female genotypes from part a as

F1(Female)= BbCcHh

F1(Male)= Bb”c”XY”hXY

i) male : 50% (considering hXY genotype, only mother can pass the “h” to the male offspring)

female : 25% of getting genotype “hh” (1/2 * ½ = ¼)

ii) male : for the genotype “c”XY”h”XY”bb

50% of being color blind; 50% of being haemophiliac , 25% for non baldness i.e ½ * ½ * ¼ = 1/16

Female: 25% of being color blind; 25% of being haemophiliac,non- Baldness, ¾ (Bb or bb)

(1/4 *1/4 * 3/4) = 1/64

iii) Male : 50% colour blind 50% non-haemophilic ,25% non-bald

(1/2 * ½ * ¼)= 1/16

Female: 25% colour blind ,75% non-hameophilic ,75 % non-bald

(1/4 * ¾ * ¾) = 9/64

iv) male : ¾ (bald) * ½ (color blind)* ½ (haemophiliac) = 3/16

Female: ¼ (bald) * ¼* colour blind * ¼ haemophiliac = 1/64

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