![DO EITHER QUESTION 11 OR QUESTION 12. NOT BOTH. We will only mark Question 11 if it is not clear which one you attempted. 11) Hemophilla is a sex-linked trait. A normal woman marries a normal man and they have 1 daughter and 2 sons. One son has hemophilia but the other two are normal. The normal daughter marries a hemophiliac man and they have2 sons and 2 daughters. One son and one daughter have hemophilia but other 2 children are normal. (6 marks] A) Draw a pedigree showing the sex-linked genotypes of all the individuals in the pedigree. 13 marks B) What is the probability that the unaffected granddaughter is a carrier? 11 mark] C) What is the probability that the original couple will have another child that has hemophilia? 12 marks) 14](http://img.homeworklib.com/questions/acc5fa40-c300-11ea-839b-7f2f84dad2b5.png?x-oss-process=image/resize,w_560)
Homework Answers
A) X represents the normal allele while X* represents the haemophilia allele.
B) The
unaffected granddaughter will be Carrier, the
probability would be 1. Since another
granddaughter is affected which indicates that she has both the
haemophilia causing alleles. Therefore both the parents (II-3 &
II-4) must have at least one X* allele. The unaffected
granddaughter received one X from II-3 and another X* from II-4
(because II-4 is affected male and has only haemophilia causing
allele).
C) There are four possible genotypes of offspring of parents: XX, XX*, XY, X*Y.
Out of four only one is haemophilic. Thus the probability of child being affect = 1/4.
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