Essentials of Genetics - Masteringgenetic
9th Edition
ISBN: 9780134143699
Author: KLUG
Publisher: PEARSON
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Chapter 6, Problem 24PDQ
In a human genetic study, a family with five
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An individual is heterozygous for a reciprocal translocation, with the following chromosomes:
A • B C D E F
A • B C V W X
R ST • U D E F
R ST • U V W X
Q. Draw the products of alternate, adjacent-1, and adjacent-2 segregations.
In a human genetic study, a family with five phenotypicallynormal children was investigated. Two children were “homozygous”for a Robertsonian translocation between chromosomes19 and 20 (they contained two identical copies of the fusedchromosome). They have only 44 chromosomes but a completegenetic complement. Three of the children were “heterozygous”for the translocation and contained 45 chromosomes,with one translocated chromosome plus a normal copy of bothchromosomes 19 and 20. Two other pregnancies resulted instillbirths. It was later discovered that the parents were firstcousins. Based on this information, determine the chromosomecompositions of the parents. What led to the stillbirths? Whywas the discovery that the parents were first cousins a key pieceof information in understanding the genetics of this family?
An individual is heterozygous for a reciprocal translocation, with the following chromosomes:
A • B C D E F
A • B C V W X
R ST • U D E F
R ST • U V W X
Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.
Chapter 6 Solutions
Essentials of Genetics - Masteringgenetic
Ch. 6 - CASE STUDY| Fish tales Aquatic vegetation...Ch. 6 - CASE STUDY |Fish tales Aquatic vegetation...Ch. 6 - CASE STUDY |Fish tales
Aquatic vegetation...Ch. 6 - HOW DO WE KNOW? In this chapter, we focused on...Ch. 6 -
CONCEPT QUESTION
2. Review the Chapter Concepts...Ch. 6 -
3. Define these pairs of terms, and distinguish...Ch. 6 -
4. For a species with a diploid number of 18,...Ch. 6 - What explanation has been proposed to explain why...Ch. 6 - Contrast the fertility of an allotetraploid with...Ch. 6 -
7. Why do human monosomies most often fail to...
Ch. 6 -
8. Describe the origin of cultivated American...Ch. 6 - Predict how the synaptic configurations of...Ch. 6 - Inversions are said to “suppress crossing over.”...Ch. 6 -
11. Predict the genetic composition of gametes...Ch. 6 - Human adult hemoglobin is a tetramer containing...Ch. 6 -
13. The primrose, Primula kewensis, has 36...Ch. 6 - Certain varieties of chrysanthemums contain 18,...Ch. 6 - Drosophila may be monosomic for chromosome 4, yet...Ch. 6 - Mendelian ratios are modified in crosses involving...Ch. 6 -
17. Having correctly established the F2 ratio in...Ch. 6 -
18. In a cross between two varieties of corn,...Ch. 6 -
19. A couple planning their family are aware that...Ch. 6 -
20. A woman who sought genetic counseling is...Ch. 6 - The woman in Problem 20 has had two miscarriages....Ch. 6 -
22. In a recent cytogenetic study on 1021 cases...Ch. 6 -
23. A boy with Klinefelter syndrome (47,XXY) is...Ch. 6 - In a human genetic study, a family with five...Ch. 6 - A 3-year-old child exhibited some early indication...Ch. 6 - A normal female is discovered with 45 chromosomes,...
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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
- Two phenotypically normal parents produce a phenotypically abnormal child in which chromosome 5 is missing part of its long arm but has a piece of chromosome 7 attached to it. The child also has one normal copy of chromosome 5 and two normal copies of chromosome 7. With regard to chromosomes 5 and 7, what do you think are the chromosomal compositions of the parents? Would it most likely be reciprocal translocation? It wouldn't be simple translocation because then the child would have the entirety of one chromosome and only some of the other, but in this case, there is only partial chromosome 5 and chromosome 7?arrow_forwardAn individual has the following reciprocal translocation:What would be the outcome of alternate segregation and of adjacent-1segregation?arrow_forwardIn the following schematic drawing of a Holliday junction,one chromatid is shown in red, and the homologous chromatidis shown in blue. The red chromatid carries a dominant allelelabeled A and a recessive allele labeled b, whereas the blue chromatidcarries a recessive allele labeled a and a dominant allelelabeled B. Where would the DNA strands have to be cut to produce recombinantchromosomes? Would they be cut at sites 1 and 3, or at sites2 and 4? What would be the genotypes of the two recombinantchromosomes?arrow_forward
- A chromosome of genotype C D recombines with a homolog of genotype c d during meiosis when Spo11 produces a double-strand break between genes C and D. If anticrossover helicase disentangles the invading strand, the likely outcome would be A) two C D and two c d gametes. B) two C d and two c D gametes. C) one C D, one C d, one c D, and one c d gamete. D) four C D gametes. E) four c d gametesarrow_forwardThree autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progenyarrow_forwardGenes R, H and F are located on the same chromosome. On one chromosome R, h and f alleles are attached, while on the homologue contains r, H and F. Which of the following would be an example of a recombination event?arrow_forward
- A pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and Barrow_forwardA cell has two pairs of submetacentric chromosomes, which we will call chromosomes Ia, Ib, IIa, and IIb (chromosomes Ia and Ib are homologs, and chromosomes IIa and IIb are homologs). Allele M is located on the long arm of chromosome Ia, and allele m is located at the same position on chromosome Ib. Allele P is located on the short arm of chromosome Ia, and allele p is located at the same position on chromosome Ib. Allele R is located on chromosome IIa and allele r is located at the same position on chromosome IIb.a. Draw these chromosomes, identifying genes M, m, P, p, R, and r, as they might appear in metaphase I of meiosis. Assume that there is no crossing over.b. Taking into consideration the random separation of chromosomes in anaphase I, draw the chromosomes (with genes identified) present in all possible types of gametes that might result from this cell’s undergoing meiosis. Assume that there is no crossing over.arrow_forwardPeople with Down syndrome have an extra copy of chromosome 21, for a total of 47 chromosomes. However, in a few cases of Down syndrome, 46 chromosomes are present. This total includes two normal-looking chromosomes 21, one normal chromosome 14, and a longer-than-normal chromosome 14. Interpret this observation. How can these individuals have 46 chromosomes?arrow_forward
- A normal chromosome A B C D * E F G H I J K has mutated toA B C F E * D G H I J K J K.This is an example of a _________________ chromosomal mutation. ( * denotes the centromere) Group of answer choices Pericentric inversion of DEF and tandem duplication of JK. Pericentric inversion of DEF and displaced duplication of JK Paracentic inversion of DEF and displaced duplication of JK. Paracentric inversion of DEF and tandem duplication of JK.arrow_forwardIn a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed: pr–adp 29% pr–vg 13 pr–c 21 pr–b 6 adp–b 35 adp–c 8 adp–vg 16 vg–b 19 vg–c 8 c–b 27 (a) Given that the adp gene is near the end of chromosome II (locus 83), construct a map of these genes. (b) In another set of experiments, a sixth gene, d, was tested against b and pr: d–b 17% d–pr 23% Predict the results of two-point mapping between d and c, d and vg, and d and adp.arrow_forwardWhen a female melanotic fly is crossed with a normal male, the progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses between melanotic females and normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inhertiacne of the melanotic mutation (Hint: The cross produces twice as many female progeny as male progeny)arrow_forward
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