Solutions For An Introduction to Genetic Analysis Chapter 3 Problem 56P

b) If the 4 genes assort independently and the couple have 4 children the probability of all of them being brachydactylous would be:

c) For none of the 4 children to be brachydactylous the probability would be:

d) For all the children to be tasters of PTC, the probability would be:

e) For all the children to be nontasters of PTC, the probability would be:

f) For all the children to be brachydactylous PTC tasters, the probability would be:

g) For none of the children to be brachydactylous PTC tasters, the probability would be 1 – chance of a brachydactylous taster.

h) For at least one of the children to be a brachydactylous taster, the probability would be 1 – none of the children being a brachydactylous taster.