A small population of Ashkenazi Jewish settlers made a home for themselves in ru

Question

A small population of Ashkenazi Jewish settlers made a home for themselves in rural upstate New York. Since the group’s settlement after WWII, no migration of marrying from the outside has occurred, as is their custom. 2 of the initial 200 settlers were carriers for the recessive Tay-Sachs allele. The population has remained almost unchanged over the last 65 years other than growing to the size of 500 people.

What was the frequency of q in the initial 1950 population?

If the frequency of the allele has been steady, how many people of the current population would be expected to be carriers? What assumptions are you making to answer these questions?

What is the likelihood a random couple from this group would have an affected child?

Despite Tay-Sachs being lethal at a very young age, the allele persists in the population. How is this possible?

Explanation / Answer

Answer to Part 1 of 5 questions:

Frequency of q in the initial 1950 population:

If A represents the dominant allele (P) and the recessive allele (q), then the following holds for the initial 1950 population of 200 Ashkenazi Jewish settlers:

Genotype

# of Individuals

Genotypic frequencies

AA

198

AA = 198/200 = 99%

Aa

2

Aa = 2/200 = 1%

aa

0

aa = 0/200 = 0%

Total

200

The 1950 population of 200 Ashkenazi Jewish settlers is said to have only 2 carriers of the recessive Tay -Sachs allele, thus in out of the 200 initial settlers only 2 individuals have the Aa allele (since carriers possess the recessive allele and are heterozygous).

The entire population consists of 400 (200 X 2) alleles. To determine the allelic frequencies, we simply count the number of dominant or recessive alleles and divide by the total number of alleles. So, the allelic frequency for the recessive allele (i.e. by convention, q) will be:

f(q) = [(2 x 0) + 2]/400 = 0.005.

Frequency of q in the initial 1950 population is 0.005.

P.S.: As per the answering policy, since the number of questions is more than 4, have answered only the first full question.

Genotype

# of Individuals

Genotypic frequencies

AA

198

AA = 198/200 = 99%

Aa

2

Aa = 2/200 = 1%

aa

0

aa = 0/200 = 0%

Total

200

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