Question

Based on their family histories, a couple knows that they are each heterozygous for alkaptonuria. The alkaptonuria phenotype is autosomal recessive to the normal phenotype. The couple plan to have seven children.

A. in a single birth what is the probability of the couple having an affected child? 25 percent

B. In a single birth, what is the probability of the couple having an unaffected child? 25 percent

C. what is the probability that three children will be alkaptonuric and the other four normal?

D. what is the probability that AT LEAST ONE of the children will be alkaptonuric?

NEED C and D answered

Answer 1

c) 3 affected and 4 normal

probability of affected = 1/4 (aa phenotype)

probabilty of normal = 3/4 (AA, Aa and Aa phenotype)

the probability of 3 affected and 4 normal children = 1/4*1/4*1/4 + 3/4*3/4*3/4*3/4

= 1/64 +81/256

= 85/256

= 33.2% = 33%

D) Atleast one is affected

probability of all seven being unaffected = 3/4*3/4*3/4*3/4*3/4*3/4*3/4 = 2187/16384

probability of one or more being affected = 1- 2187/16384 = (16384-2187)/16384 = 14197/16384 = 86.65%

= 87%

Answer is 87%

to check, all probabilities must be equal to 1

= (2187/16384)+(14197/16384)=1