20. In a family of seven children, what is the probability of obtaining the foll
ID: 274889 • Letter: 2
Question
20. In a family of seven children, what is the probability of obtaining the following numbers of boys and girls? (Section 3.2) a. All boys b. All children of the same sex c. Six girls and one boy d. Four boys and three girls e. Four girls and three boys 21. Phenylketonuria (PKU) is a disease that results from a recessive allele.Two normal parents produce a child with PKU. (Section 3.2) a. What is the probability that a sperm from the father will contain the PKU allele? b. What is the probability that an egg from the mother will contain the PKU allele? c. What is the probability that their next child will have PKU? d. What is the probability that their next child will be heterozygous for the PKU gene?Explanation / Answer
??I am answering only the questions you have asked:
20)
Let the probability of the child being a boy be 'b' and that of it being a girl be 'g'
We know that b=g= 1/2
Now, using a binomial expansion we can say that
(g + b)7= g7+ 7g6b + 21g5b2+ 35g4b3+ 35g3b4+ 21g2b5+ 7gb6+ b7
c) This probability is given by the term 7g6b = 7*0.5^6*0.5 = 7/128
d) This is given by the term 35g3b4 = 25*0.5^3*0.5^4 = 35/128
e) This probability is given by the term 35g4b3 = 35* 0.5^4 * 0.5^3 = 35/128
21)
Since the child has PKU but both parents are normal, the genotype of the parents must be pP ( heterozygous expression).
Since the alleles separate into two during gamete formation, each gamete of the father or mother will have either p allele or P allele.
Therefore,
a) probability of having the p allele( lower case PKU allele/ recessive allele)= 1/2
b)probability of having the p allele( lower case PKU allele/ recessive allele) =1/2
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