= disease =color blind pedigree COT +B-colorblind and disease A recessive trait

Question

= disease =color blind pedigree COT +B-colorblind and disease A recessive trait for a late-onset neurological disease is located 8 map units from the recessive deutan color blindness trait, on the X chromosome. Below is a pedigree from a family in which both traits are found. The woman's husband is color-blind, and she has two sons who show the disease but are not color blind. Her daughter shows neither trait, and is well past the age where the symptoms appear. What is the probability Il-2 is a carrier of the disease (number between 0 and 1)? Answer .5 Il-1 and I-2 are expecting a child. The sonogram shows they will be having a son. What is the probability he will develop the disease late in life? Answer: |.17 Il-1 and 1l-2 have their son. At an early age, it is apparent that he is color-blind. What is the chance he will develop the disease late in life? Answer: .92

Explanation / Answer

Q 2 ) Let us denote an X linked trait, 'A' writtern as a superscript on X. Ie;

In Females as they have two X chromosomes there are 3 possible genotypes.They are,

XAXA - Normal

XAXa - Carrier

XaXa - Diseased

Whereas in males the genotypes can be either one among two,

XAY - Normal

XaY - Diseased

Here, as discussed in question number 2, the II-1 and II-2 are expecting a child and it is found that their child is a boy (son). So, he can be of either one of the genotype XAY ( normal) or XaY (diseased) for the trait. Here we have to check the probability tht he gets the diseases, which means the probability to get a genotype XaY. From the pedigree it is sure that the father of the child is normal and therefore his genotype will be XAY( as this is the only genotype that is possible for a normal male in case of an X linked recessive disease). But in case of mother she can be either XAXA ( Normal) or XAXa ( Carrier). It is sure that the son gets the Y chromosome from father and X chromosome from mother. So, for him to get the diseases, he should get Xa allele from mother. For that mother should be a carrier. The probability that mother is a carrier is 0.5.

Therefore the probability of son to get diseased is = XAXa X XAY

( Mother) ( Father )

XaY

( Diseased son)

Probability of getting Xa from mother is 1/2 and there is complete chance of getting Y from father ie; = 1. In addition the mother should be also a carrier and its probability is 1/2 ( AND rule ).

Therefore the total probability = 1/2*1*1/2

= 1/4

= 0.25

Note: AND rule - Multiplication

OR rule- Addition

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