Discussion in 'MCAT Study Question Q&A' started by moto_za, Sep 28, 2014.

1. ### moto_zaMember 10+ Year Member

1,574
42
Jan 16, 2006
Hi All,

I am having trouble figuring out the best way to solve the following problems (1-3) that are related to each other. Any help would be appreciated. I think once you figure out the first one then then rest are easy? Thanks!

The pregnant woman represented in the pedigree shown (see image) had a brother who died of
sickle cell disease. For the purpose of this problem, assume that the population birth frequency
of individuals affected with sickle cell disease is 1/400. What is the mother's probability of being
a carrier?

A. 1/4
B. 1/2
C. 2/3
D. 1/20

In the sickle cell pedigree in the previous question (see image), what is the child's probability of
having sickle cell disease?
A. 1/16
B. 1/20
C. 1/60
D. 1/120

In the sickle cell pedigree in the previous question (see image), what is the father's probability of
being a carrier?
A. 2/3
B. 1/10
C. 1/20
D. 1/40

3. ### Jumb0 5+ Year Member

239
106
Aug 24, 2012
USA
There are two possibilities for the woman's parents. They could have been Rr and Rr or they could have been Rr and rr. They could not have both been homozygous because then their kids would have been all the same. So, if the parents were both heterozygotes, then the chance that their daughter would be born a carrier is 2/3, since she was nor rr, but there is a 1/3 chance that is RR. If they were a heterozygote and a homozygote recessive, then the chance that she would be a carrier would be 1, since she is not rr and the only other combination is the carrier Rr . We don't know which is the case, but we can figure out the probability using Hardy Weinberg.

p^2 + 2pq + q^2 =1
p + q = 1

q^2 = 1/400
Therefore, q = 1/20
Therefore, p = 1- (1/20) = 19/20

Therefore, chance of being a carrier (2pq) = 2*(1/20)*(19/20) = .095 = 38/400

The chance of being a double recessive is given = 1/400

In either case, one of her parents was definitely a heterozygote. We just don't know whether the other parent was doubly recessive or a heterozygote: Was the parent the 1 in 400 or the 38 in 400?

So, the chance of one of hers parent being a homozygote recessive (1/400) AND her being born a carrier (1) = (1/400) * (1)
And the chance of that parent being a heterozygote (38/400) AND her being born a carrier (2/3) = (38/400) * (2/3)

Since these events are mutually exclusive, the probability that she was born a carrier from either situation is the sum of the probabilities:

[ (1/400) * (1) ] + [ (38/400) * (2/3) ] = 2/3

So she has a 2/3 chance of being a carrier.

For the child to have sickle cell disease. The mom would need to be a carrier (2/3). The dad (who is not diseased) would need to be a carrier (38/400), and their child would need to be born homozygous recessive (1/4):

2/3 * 38/400 * 1/4 = 1/60

The kid has a 1 in 60 chance of being diseased.

The dad's chance of being a carrier we just said was 38/400 or ~ 1/20.

DISCLAIMER = The numbers don't work out to be exact. They are off by very small fractions. I guess this was because they intend for you to round i.e. They want you to realize that 38/400 is approximately 1/20.

Last edited: Sep 28, 2014
4. ### moto_zaMember 10+ Year Member

1,574
42
Jan 16, 2006
Thanks so much! I really appreciate it.

Can I ask you one last genetics questions if you have time:

This pedigree illustrates a family of Northern European descent with lamellar ichthyosis, an
autosomal recessive condition in which there is a mutation in transglutaminase-1, a protein
required for normal formation of the stratum corneum. The probability that individual III-3 is a
carrier for this disease is closest to which of the following?

A. 1/4
B. 1/3
C. 1/2
D. 2/3

5. ### Jumb0 5+ Year Member

239
106
Aug 24, 2012
USA

Ok, so the parental generation must have been heterozygotes since neither of them were affected and they had affected children.
The father of Individual III-3 is not affected, so he could be either RR or Rr. The chance of RR = 1/3. The chance of Rr = 2/3.
We are not explicitly told that the disease is rare, but it is recessive and we are not given any information about its prevalence in the population, so we can assume it is rare.
Therefore, we can assume the mother of III-3 (not shown) was RR.

Therefore, in order for III-3 to be a carrier: His dad must have been a carrier (2/3) and he must have passed on this allele (1/2) = 2/3 * 1/2 = 2/6 = 1/3

There is a 1/3 chance that III-3 is a carrier.

Czarcasm likes this.
6. ### CzarcasmHakuna matata, no worries. 2+ Year Member

966
405
Jun 22, 2013
Crypts of Lieberkühn
Excellent explanation!!