Of 31 children born of father–daughter matings, 6 died in infancy, 12 were very

Question

Of 31 children born of father–daughter matings, 6 died in infancy, 12 were very abnormal and died in childhood, and 13 were normal. From this information, calculate roughly how many recessive lethal genes we have, on average, in our human genomes. For example, if the answer is 1, then a daughter would stand a 50 percent chance of carrying the gene, and the probability of the union’s producing a lethal combination would be 1/2 × 1/4 = 1/8. (So obviously 1 is not the answer.) Consider also the possibility of undetected fatalities in utero in such matings. How would they affect your result?

Explanation / Answer

The probability of not getting a recessive lethal genotype for one gene is 1 – 1/8 = 7/8. If there are n lethal genes, the probability of not being homozygous for any of them is (7/8) n= 13/31. Solving for n, an average of 6.5 recessive lethals is predicted. If the actual percentage of “normal” children is less owing to missed in utero fatalities, the average number of recessive lethals would be higher.

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