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I have two questions pertaining to this pedigree I believe it to be an autosomal recessive trait.

  1. The probability that individuals IV-1 and IV-2 would give rise to an affected individual would be: A. 1/4 b. 1/2 c. 1/8 d. 1/12 e. 1/24

  2. Suppose individual IV-2 was unaffected. The probability of his marriage giving rise to an affected child would be: A. 1/4 b. 1/2 c. 1/8 d. 1/12 e. 1/24

I worked on this problem for the last hour or so and figured up question 1 to be 1/2 and question two to be 1/4. I don't feel this is correct. I feel like I am missing something. If you don't mind showing the calculations for how you came about an answer that would be greatly appreciated.

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1 Answer

up vote 2 down vote accepted

Question 1

Okay, so I'll go through my own process for you here step by step, moving down the tree. Here's an annotated version of your diagram with my own thoughts (it's been a while since I've done this but should hopefully be accurate):

Annotated pedigree chart

Generation 2: As you realised, the trait is autosomal recessive so the female (II:2) has the genotype aa. The male is considered wild type unless informed otherwise, giving him a genotype of AA. I think you had worked all this out, but they are shown in red on the diagram.

Generation 3: Using the two genotypes of the parents from II (red), we know that all the progeny in generation III are carriers - i.e. Aa genotypes. This is indicated for III:6 in orangey-brown. I think you also worked this out successfully. However in your original chart, III:7 is noted as Aa. III:7 is from outside of the affected family and would therefore again be considered to have be wild type and therefore instead have the genotype AA - shown in purple.

Generation 4: In order to work out the potential genotypes of IV:1 we have to do a cross between the two parents III:6 and III:7 - which works out as follows:

        |          Male          |
        |     A            a     |
        |------------------------|
F       |                        |
e     A |    AA            Aa    |
m       |                        |
a       |                        |
l     A |    AA            Aa    |
e       |                        |

As you can see, there is a 50% chance of IV:1 being healthy AA and a 50% chance of being a carrier Aa - shown in green. As IV:2 is affected, we know his genotype is aa (pink).

Generation V: Now for answer to your first question. If IV:1 had the AA genotype then the child could never be affected, reducing the odds to 50% immediately. Of this remaining 50%:

        |          Male          |
        |     a            a     |
        |------------------------|
F       |                        |
e     A |    Aa            Aa    |
m       |                        |
a       |                        |
l     a |    aa            aa    |
e       |                        |

50% of the offspring would be affected.

Therefore the probability of V:I being affected is $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$


Question 2

The second question seems very odd to me. If IV:2 was unaffected in the literal sense of the word they would be wild-type AA. This would make the probability of them having an affected child as 0.

As this isn't one of your options, I assume they mean that IV:2 is a carrier (Aa). We are this time therefore crossing either AA or Aa (IV:1) with Aa (IV:2). Again, if IV:1 is heterozygous dominant (AA) none of her children will be affected. This means the odds are again reduced by a half. In the case that she is heterozygous Aa then the cross becomes:

        |          Male          |
        |     A            a     |
        |------------------------|
F       |                        |
e     A |    AA            Aa    |
m       |                        |
a       |                        |
l     a |    Aa            aa    |
e       |                        |

Giving 25% healthy, 50% carrier and 25% affected.

As you're looking for affected children: $\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$

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Thanks so much! –  user4410 Sep 5 '13 at 22:58
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