Question

In humans, the RH factor and elliptocytosis genes are located on the same chromosome at 3% from each other. The color blindness gene and the night blindness gene are located 40% apart on the X chromosome. Rhesus positivity and elliptocytes - dominant alleles, and night and color blindness are recessive.

А) A woman who was heterozygous in all her genes and whose ancestors had no crossing over was married to a man who suffered from night blindness and color blindness, with Rh-negative blood and normal red blood cells. Determine the probability of birth in this family of a boy with color blindness, RH positive blood and normal red blood cells.

B) A Heterozygous woman with all genes, whose mother suffered from elliptocytosis and night blindness, and whose father had Rh-negative blood and suffered from color blindness, married a man suffering from night and color blindness, with Rh-negative blood and normal red blood cells. Determine the probability of birth in this family of a girl with normal vision blindness, Rh-negative blood and without elliptocytosis.

Answer 1

(A). The cross is between a woman heterozygous for all the said traits (night blindness, colour blindness, Rh positivity and elliptocytes) and a man who suffered from night blindness and color blindness, with Rh-negative blood and normal red blood cells. Since none of the woman's ancestors had crossing over, her genotype is:

RE/re ; CN/cn - That is genes are in coupling phase. To make it more clear, dominant alleles of a linkage pair are present on one chromosome and the recessive ones on the other.

R and r represents alles for Rh positivity and Rh negativity respectively, R being the dominant allele.

E and e represents alleles for Elliptocytes and normal RBC respectively, E being the dominant allele.

c and C represents X linked alleles for color blindness and no colour blindness respectively, C being the dominant.

n and N represents X linked alleles for night blindness and no night blindness respectively

Y represents normal Y chromosome as it has no allele for X linked traits.

R and E are located 3% apart, hence has a recombination percentage of 3%.

c and n are 40% apart and hence has a recombination percentage of 40%.

Genotype of man : re/re ; cn/Y

We need to find the probablity of birth of a boy with genotype : Re/re ; cN/Y ( Normal RBC, Rh positive, color blind - since father has no allele for Rh positive and mother is heterozygous for Rh positive, the son with Rh positive blood can only be heterozygous ).

The cross :

RE/re ; CN/cn x re/re ; cn/Y

Required offspring : Re/re ; cN/Y

The Y in offspring is from male parent.

X containing cN should come from female parent, which is not possible without recombination.

Male parent has no dominant alleles for R and E, hence, dominant R should come from female parent. Male parent can produce gametes only with re and not Re or RE or rE, as male parent is homozygous recessive. Hence Re chromosome of offspring should be come from female.

The percentage of formation of fenale gamete with Re; cN :

For this to happen, recombination should happen in X chromosome and RE chromosome simultaneously. The probablity of this event = product of recombination percentage between RE and recombination percentage between cn

= 3% * 40% = 12%.

This recombination event produces four types of female gametes viz., Re;cN , Re;Cn , rE;cN and rE;Cn

But we need only the Re;cN gamete.

Probablity of formation of this gamete is one in four of 12% (here, 12% is already found out on previous step)

= 12% ÷ 4 = 3%

Also, males produces two type of gametes viz. re;cn and re; Y. But we need only re;Y as we are looking for male child.

The probablity of production of re;Y male gamete is 50% ( since males produce only 2 type of gametes, each gamete has 50% probablity )

Hence the production of male child with the genotype Re/re ; cN/Y = probablity of production of Re;cN female gamete * probablity of production of re;Y male gamete

= 3% * 50%

= 1.5%

(B) The cross here is between female heterozygous for all four alleles and a man with night and color blindness, night blindness, normal RBC and Rh negetive.

Genotype of female parent:

Her mother had elliptocytosis and father had Rh negetive blood. This means that the female's heterozygous make up for R and E are in coupling phase i. e. Both dominant alleles came from mother's gamete and the recessive from father's gamete.

Her mother was night blind and her father was color blind. Hence, her genes c and n are in repulsion phase, i.e. dominan allele of one trait and recessive allele of other in one chromosome.

Hence, genotype of female parent :

RE/re ; cN/Cn

Genotype of Male parent:

re/re ; cn/Y (Rh negetive, normal RBC, colorblind, night blind).

Child required: re/re ; CN/cn ( girl with normal vision blindness, normal RBC and Rh negetive )

Normal vision blindness means normal for both color and night blindness trait.

But males can produce only type of gametes for a girl child with respect to C and N. Hence, cn is received from male and CN is received from female parent. For CN to be received from female parent, recombination should take place. The probablity of recombination to take place in female between C and N is 40%.

After this recombination, we get four types of female gametes viz. RE;CN , RE;cn , re;CN and re;CN ----- each with equal chance of forming.

Hence, chance for production of re; CN female gamete is one in four of the recombination percentage between C and N.

Chance for production of re;CN gamete = 40% ÷ 4 = 10%.

Also, males produce only two types of gametes viz. re;cn and re;Y in equal proportion. We need only a girl child and for that we need the male gamete re;cn. The probablity of production of re;cn male gamete is 50%.

Hence, probablity of formation of girl child with

re/re ; CN/cn = probablity of production of re;CN female gamete * probablity of production of re;cn male gamete

= 10% * 50% = 5%