Of 31 children born from father-daughter matings, 8 died as infants, 7 exhibited

Question

Of 31 children born from father-daughter matings, 8 died as infants, 7 exhibited birth defects, and 16 appeared normal. A) From this information calculate how many recessive deleterious alleles, on average, are present in the human genome? Hint: If the answer were 1, then a daughter would have a 50% chance of carrying a deleterious allele from the father, and these two recessive alleles would combine with a frequency of 25% so 1/8 of the offspring would exhibit a deleterious mutation. B) How would undetected fatalities that occur in utero affect your result?

Explanation / Answer

(a)

Of 31 children 16 appeared normal whereas 8 died as infants and 7 exhibited birth defects

So , out of 31 children 15 have deleterious mutation .

So, 15/31 of the offspring would exhibit a deleterious mutation i.e 1/(31/15) of of the offspring would exhibit a deleterious mutation

Now 1/8 of offspring exhibit deleterious mutation with frequency 25%

So, 1/31/15 of offspring exhibit deleterious mutation for 15/31 = 48.4%

then a daughter would have a 97% chance of carrying a deleterious allele from the father

So,

1.94 recessive deleterious alleles, on average, are present in the human genome.

(b) undetected fatalities in utero if used in the result will lead to increase the number of offspring that would exhibit a deleterious mutation so recessive deleterious alleles will increase in number

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