Question

Sickle-cell disease (SCD) is a type of blood disorder caused by a mutation in the HBB gene. The mutation results in the production of abnormal red blood cells; these sickle-shaped cells are unable to efficiently transport oxygen to body tissues. Individuals must inherit two abnormal copies of HBB in order to have SCD. Heterozygous individuals, with one normal copy of HBB and one abnormal copy, are said to have sickle-cell trait (SCT); they generally do not experience problems with their red blood cells, and in fact are less susceptible to malaria. Thus, the mutated HBB allele is particularly common among people from parts of the world where malaria is or was common, such as sub-Saharan Africa. In the United States, SCD occurs among about 1 out of 365 African-American births (i.e. this is the proportion of individuals who have 2 abnormal copies of HBB).

a. The abnormal copy of HBB is written as HBB -; the normal copy of HBB is written as HBB+. Suppose that both members of a couple have SCT (genotype HBB+ HBB-). What is the probability that their first two children have SCD (HBB - HBB -) and the next three do not (either genotype HBB+ HBBor genotype HBB+ HBB+)?

b. Suppose that both members of a couple have SCT (genotype HBB+ HBB-), and that they have five children. What is the probability that exactly two of their children have SCD?

c. Find the expected frequencies of each possible genotype, for African-Americans living

Answer 1

Let us define,

A_i=the ith born child has SCT (i=1(1)5)

A_i^c=the ith born child does not have SCT (i=1(1)5)

So we've to find :

i.P[1st 2 child have SCT] = P[A₁.A₂.A₃^c.A₄^c.A₅^c]

= P[A₁].P[A₂].P[A₃^c].P[A₄^c].P[A₅^c]

= (1/365)*(1/365)*(1-(1/365))*(1-(1/365))*(1-(1/365))

ii.

iii.Since we are not given the total number of population we will consider that 1 out of 365 african-american have 2 negative HBBs then the probability of an African-American not having 2 negative HBBs is 364/365.

Now if we consider that the chances of getting 1 negative HBB and 1 positive HBB and 2 positive HBBs are equally likely, then the probability of getting 1 negative HBB and 1 positive HBB is (1/2*364/365)=182/365 and the probability of getting 2 positive HBBs is also 182/365.

So, the frequencies of :

#2 negative HBBs is 1 out of 365 African-Americans

#1 negative HBB and 1 positive HBB is 182 out of 365 African-Americans

#2 positive HBBs is also 182 out of 365 African-Americans.