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Chapter 15: Q15.2-2CC (page 300)

Neither Tim nor Rhoda had Duchenne muscular dystrophy, but their firstborn son does. What is the probability that a second child will have the disease? What is the probability if the second child is a boy? A girl?

Short Answer

Expert verified

The probability that the second child will have Duchenne muscular dystrophy is one-fourth (1/4) because they have a half-chance each of receiving Y chromosome from father and being a son and receives defective allele from mother.

The probability that the second child will be diseased if he is a boy is half (1/2) because they have a half-chance of receiving defective X-chromosome from mother.

The probability that the second child will be diseased if it is a girl is zero chance because they receive just either X chromosome.

Step by step solution

01

Duchenne muscular dystrophy

Duchenne muscular dystrophy is a recessive disorder. It is an X-linked disease that results in the degeneration of muscles.The condition occurs due to the defect caused by protein dystrophin.

Here, the parents are not diseased, but their first child is diseased. This means the parents must be carriers for the disease and transmit their defective alleles to the first child.

02

Probability for the second child with Duchenne muscular dystrophy

A son (XY) receives an X chromosome from his mother and a Y chromosome from his father. Girls (XX) receive one X chromosome from their mother and the other from their father.

The probability that the second child will also be diseased is 录. This is because the second child has 陆 chance that he receives the Y chromosome from the father or X-chromosome from the mother and 陆 chance that he will receive the defective X-chromosome from his mother.

Thus, the second child has a 录 (1/2 from Y chromosome x 陆 from defective X chromosome) chance of having Duchenne muscular dystrophy.

03

Probability for a second child to be diseased if it is a boy

Son receives an X chromosome from his mother, and Duchenne muscular dystrophy is an X-linked disease.The son receiving an X-chromosome will have 陆 chances of being diseased because it can either receive a diseased X-chromosome or a normal X-chromosome from his mother.

Thus, the probability for a second child to be diseased if it is a boy is 陆.

04

Explanation for probability for a second child to be diseased if it a girl

Girls possess two X-chromosomes, andDuchenne muscular dystrophy is an X-linked recessive disease.This means the presence of both diseased X-chromosomes is a must for the disease to occur in girls.

The presence of one diseased X-chromosome makes the girl a carrier for the disease. Thus, the probability for the second child to be diseased if it is a girl is zero. However, the girl has 陆 chance to be the carrier for the disease if she receives the diseases X-chromosome.

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Most popular questions from this chapter

The ABO blood type locus has been mapped on chromosome 9. A father with type AB blood and a mother who has type O blood have a child with trisomy nine and type A blood. Using this information, can you tell in which parent the non-disjunction occurred? Explain your answer. (See Figures 14.11 and 15.13).

The\({\chi ^2}\)value means nothing on its own- it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are consistent with the hypothesis, and thus the hypothesis should be rejected, A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the\({\chi ^2}\)value is 0.05or considered statistically significant, the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically significant: the observed data are consistent with the hypothesis.

To find the probability, locate your\({\chi ^2}\)value in the\({\chi ^2}\)Distribution table in Appendix F. The 鈥渄egree of freedom鈥� (pdf) of your data set is the number of categories (here,4 phenotypes), minus 1, so df=3.

(a). Determines which values on the df =3 line of the table your calculated\({\chi ^2}\)value lies between.

(b). The column headings for these values show the probability range for your\({\chi ^2}\)number. Based on whether there is non-significant (p\( \le \)0.05) or significant (p>0.05) difference between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

A wild-type fruit fly (heterozygous for the gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for the body color and eye color? Using information for problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

Butterflies have an X-Y sex determination system that is different from that of flies or humans. Female butterflies may be either XY or X0, while butterflies with two or more X chromosomes are males. This photograph shows a tiger swallowtail gynandromorphy, which is half male (left side) and half female (right side). Given that the first division of the zygote divides the embryo into the future right and left halves of the butterfly, propose a hypothesis that explains how nondisjunction during the first mitosis might have produced this unusual looking butterfly.

Review the description of meiosis (see Figure 13.8) and Mendel鈥檚 laws of segregation and independent assortment (see Concept 14.1). What is the physical basis for each of Mendel鈥檚 laws?
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