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Genetics Problem Help Please

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

  1. moto_za

    moto_za Member 10+ Year Member

    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
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  3. Jumb0

    Jumb0 5+ Year Member

    Aug 24, 2012
    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
    Hadi7183 and Czarcasm like this.
  4. moto_za

    moto_za Member 10+ Year Member

    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

    Jumb0 5+ Year Member

    Aug 24, 2012

    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. Czarcasm

    Czarcasm Hakuna matata, no worries. 2+ Year Member

    Jun 22, 2013
    Crypts of Lieberk├╝hn
    Excellent explanation!!

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