Question

Draw a Punnett square for a cross between a woman who has hemophilia (X-linked recessive inheritance) & a man who is unaffected by hemophilia. Then, answer the following questions about your results. a. what percent of their offspring are likely to be affected males? b. What percent of their offspring are likely to be affected females? C. What percent of their offspring are likely to be normal (unaffected) males? d. What percent of their offspring are likely to be normal females?

Answer 1

Haemophilia is a x linked recessive trait. A gene 'h' is responsible to cause haemophilia in human.A woman who is haemophilic(very rare) will have "h" gene in his both chromosome i.e. XhXh. A male has not been a carrier for haemophili either he is normal or haemophilic. The genotype of a normal male is XHY0.

Hence in Y chromosome there is no haemophilic gene so we take it as Y0.

A normal X chromosome will be taken as XH

Draw a Punnet sruene for a cross brw a woman who has haemophilia (x-linked recessive a) 8 a mom who is unaffected to by hemophilia pro → Now howmophilia is caused in x chromosome oney. Now here we din identify the haemophilic chromosome with Xu Normal chromosome with XH in female o and in male Normal chromose with XH 8 Yo a ha as Haemophilia is x linued Yo is taken disease. Female with Haemophilia Normal male хи Хи XH Yoblino ہے XXH xu Yo XHXH Carrier gigt XnYo Haemophilic on og a carries gir haemophilic poy ,

Punnet stuare O XH I Yo XHxh Xhyo Xh xH xh xu yo a. There will be 100.1. of the male offspring will have haemophilia. b. 100 % of girls will be carrier for haemophilia. C. → ot. of male offspring will be normas Do - ot of girl be normal. Explanation - As the mother is haemophilia so both of his x chromosome viu be have haemophilic gene h. so all the offspring will be affected. 501 5. & out of a child. (2m 82F) Dood & Female will be carries & two am will be haemophilic. If one gene of mother is non haemophilic then the mother coin be carrien not haemophilic.