Human Heredity: Principles and Issues (MindTap Course List)
11th Edition
ISBN: 9781305251052
Author: Michael Cummings
Publisher: Cengage Learning
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Chapter 6, Problem 15QP
A woman gives birth to monozygotic twins. One boy has a normal genotype (46, XY), but the other boy has trisomy 13 (47, +13). What events—and in what sequence—led to this situation?
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Chapter 6 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
Ch. 6 - Genetics in Practice case studies are...Ch. 6 - Genetics in Practice case studies are...Ch. 6 - Analyzing Karyotypes 1. Originally, karyotypic...Ch. 6 - Given the karyotype shown at right, is this a male...Ch. 6 - A colleague e-mails you saying that she has...Ch. 6 - What are the two most commonly used methods of...Ch. 6 - Prob. 5QPCh. 6 - Discuss the following sets of terms: a. trisomy...Ch. 6 - What chromosomal abnormality can result from...Ch. 6 - Tetraploidy may result from: a. lack of...
Ch. 6 - A cytology student believes he has identified an...Ch. 6 - An individual is found to have some tetraploid...Ch. 6 - A spermatogonial cell undergoes mitosis before...Ch. 6 - A teratogen is an agent that produces nongenetic...Ch. 6 - As a physician, you deliver a baby with protruding...Ch. 6 - Variations in Chromosome NumberAneuploidy Describe...Ch. 6 - A woman gives birth to monozygotic twins. One boy...Ch. 6 - Assume that a meiotic-nondisjunction event causes...Ch. 6 - Prob. 17QPCh. 6 - What is the genetic basis and phenotype for each...Ch. 6 - The majority of nondisjunction events leading to...Ch. 6 - Prob. 20QPCh. 6 - If all the nondisjunction events leading to Turner...Ch. 6 - Identify the type of chromosomal aberration...Ch. 6 - Describe the chromosomal alterations and phenotype...Ch. 6 - A geneticist discovers that a girl with Down...Ch. 6 - Albinism is caused by an autosomal recessive...Ch. 6 - Fragile-X syndrome causes the most common form of...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Human sex chromosomes are XX for females and XY for males. a. With respect to an X-linked gene, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele? c. If a female is heterozygous for an X-linked allele, how many different types of gametes can she produce with respect to this allele?arrow_forwardAssume that a meiotic-nondisjunction event causes trisomy 8 in a newborn. If two of the three copies of chromosome 8 are absolutely identical, at what point during meiosis did the nondisjunction event take place?arrow_forwardCould an individual with blood type O (genotype ii) be a legitimate child of parents in which one parent had blood type A and the other parent had blood type B?arrow_forward
- A WOMAN IS HETEROZYGOUS FOR TWO HARMFUL RECESSIVE ALLELES IN DIFFERENT CHROMOSOMES, ONE FOR PHENYLKETONURIA (PKU) AND THE OTHER FOR CYSTIC FIBROSIS (CF). SHE MARRIES AN UNAFFECTED MAN WHO IS A CARRIER FOR NEITHER DISEASE. IF SHE HAS A DAUGHTER, WHAT IS THE PROBABILITY THAT THE CHILD WILL CARRY NEITHER OF THE RECESSIVE ALLELES? EXACTLY ONE? BOTH?arrow_forwardTable 8.1 shows that Turner syndrome occurs when an individual inherits one X chromosome but lacks a second sex chromosome. Can Turner syndrome be due to nondisjunction during oogenesis, spermatogenesis, or both? If a phenotypically normal couple has a color-blind child (due to a recessive X-linked allele) with Turner syndrome, did nondisjunction occur duringoogenesis or spermatogenesis in this child’s parents? Explain your answer.arrow_forwardA human female with Turner syndrome (45,X) alsoexpresses the X-linked trait hemophilia, as did her father.Which of her parents underwent nondisjunction during meiosis,giving rise to the gamete responsible for the syndrome?arrow_forward
- A homozygous mother has retinoblastoma, a dominant sex-linked disorder carried on the X chromosome. The father does not have this disorder. What is the chance they will have a child with this disorder? 1) 0% 2) 25% 3) 50% 4) 75% 5) 100% 身arrow_forwardHumans have two different sex chromosomes, X and Y. Figure 4 below shows the inheritance of sex in humans. Figure 4 Mother Father XX XX Y XY XY a) Copy the diagram into your answer booklet and draw a circle on the part of Figure 4 that shows an egg cell. b) What is the genotype of the male offspring? c) A man and a woman have two sons. The woman is pregnant with a third child. What is the chance that this child will also be a boy?arrow_forwardA patient has two parents with Huntington's disease. They may not have inherited this autosomal dominant disorder due to: 1) increased DNA repeats (CAG) during spermatogenesis O2) incomplete penetrance 3) inheriting one recessive allelearrow_forward
- 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 gynandromorph, 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. Question is also in the picture.arrow_forwardOne form of the bleeding disorder known as von Willebrand disease is an autosomal recessive disease. A man who is a carrier marries a woman who is also a carrier of the disease. (a) What percentage of their children are likely to have a disease phenotype? (b) What percentage of their children are likely to have a normal phenotype? (c) What percentage of their children are likely to be carriers of the disease?arrow_forwardHemophilia is an X-linked recessive disease. A hemophilic man marries a woman who is not a carrier of the disease. (a) Draw a Punnett square showing the genotypes of their children. (b) What are the chances that their daughters will be carriers of the disease? (c) What percentage of their children are likely to have the disease?arrow_forward
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