Question

In: Biology

Of 31 children born of father-daughter matings, 6 died in infancy, 12 were abnormal and died...

Of 31 children born of father-daughter matings, 6 died in infancy, 12 were abnormal and died in childhood, and 13 were normal. From this information calculate roughly how many recessive lethal genes we have, on average, in the human genome. (Hint: if the answer were 1, then the daughter would stand a 50% chance of carrying the lethal allele, and the probability of the union’s producing a lethal combination would be 1/2 * 1/4 = 1/8. So, 1 is not the answer.) Discuss the possibility of undetected fatalities in utero in such matings. How would they affect your result?

Expert Solution

The number of infants that inherited the recessive mutation are 18 (6+12) out of 31. Hence, 18/31= 0.5806.

The number of normal children is 13 out of 31. This is equal to a value of 0.41935 or 42%.

The probability of not obtaining a recessive mutation for one gene is 1-1/8= 7/8.

Hence, for n number of recessive lethal genes, probability of not being homozygous for any of them is (7/8)^n= 13/31

To predict n:

If n=1; (7/8)^1= 7/8=0.875

If n=2; (7/8)^2= 0.765

If n=3; (7/8)^3= 0.66992

If n= 4; (7.8)^4= 0.586

If n= 5; (7/8)^5= 0.5129

If n= 6, (7/8)^6= 0.4487

If n=7; (7/8)^7= 0.392

The fraction of normal children obtained is 13/31 or 0.41935.

Hence, n should be between 6 and 7.

If n= 6.5; then (7/8)^n= 0.41980

Hence, the average number of recessive lethal are predicted to be approximately 6.5.

When there are in utero fatalities, the actual percentages of normal children obtained are less.

For example, it is 9 normal children out of 31 are obtained. Then this ratio is 0.290, which is 29%.

Then (7/8)^n= 0.290.

n is approximately 9.25 in this scenario.

Hence, if there are undetected fatalities in such matings, the percentages of normal children are less. In such cases, the average recessive lethals will increase or be higher.

gladiator answered 1 year ago

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