Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 2, Problem 32ESP
An interesting procedure has been applied for assessing the chromosomal balance of potential secondary oocytes for use in human in vitro fertilization. Using fluorescence in situ hybridization (FISH), Kuliev and Verlinsky (2004) were able to identify individual chromosomes in first polar bodies and thereby infer the chromosomal makeup of “sister” oocytes. Assume that when examining a first polar body you saw that it had one copy (dyad) of each chromosome but two dyads of chromosome 21. What would you expect to be the chromosomal 21 complement in the secondary oocyte? What consequences are likely in the resulting zygote, if the secondary oocyte was fertilized?
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Kuliev and Verlinsky (2004) state that there was a relatively high number of separation errors at meiosis I. In these cases the centromere underwent a premature division, occurring at meiosis I rather than meiosis II. Regarding chromosome 21, what would you expect to be the chromosome 21 complement in the secondary oocyte in which you saw a single chromatid (monad) for chromosome 21 in the first polar body? If this secondary oocyte was involved in fertilization, what would be the expected consequences?
Because of the relatively high frequency of meiotic errors that lead to developmental abnormalities in humans, many research efforts have focused on identifying correlations between error frequency and chromosome morphology and behavior. Tease et al. (2002) studied human fetal oocytes of chromosomes 21, 18, and 13 using an immunocytological approach that allowed a direct estimate of the frequency and position of meiotic recombination. Below is a summary of information [modified from Tease et al. (2002)] that compares recombination frequency with the frequency of trisomy for chromosomes 21, 18, and 13. (Note: You may want to read appropriate portions of Chapter 8 for descriptions of these trisomic conditions.) Trisomic Mean Recombination Frequency Live-born Frequency Chromosome 21 1.23 1/700 Chromosome 18 2.36 1/3000–1/8000 Chromosome 13 2.50 1/5000–1/19,000 (a) What conclusions can be drawn from these data in terms of recombination and nondisjunction frequencies? How might…
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The chromosomes drawn below are found in a human skin cell. The lines represent the chromosomes themselves. The letters indicate the location of a particular version of a gene.
a) T/F An individual’s offspring are more likely to inherit a D1 and R together than a D1 and T9 together.
b) T/F About 50% of the egg cells from this individual will have both T9 and BA5.
c) T/F Because r is recessive, the frequency at which the r version of the gene occurs in a population is lower than the frequency at which the R version occurs in the population.
d) T/F If the protein produced by the BA gene is involved in the formation of ovaries, then the BA gene is likely located on a sex chromosome.
e) T/F Because the genes shown here are found in skin cells, these genes code for proteins required for skin cell function.
Chapter 2 Solutions
Concepts of Genetics (12th Edition)
Ch. 2 - With the initial appearance of the feature we call...Ch. 2 - An organism has a diploid number of 16 in a...Ch. 2 - Examine Figure 2.12, which shows oogenesis in...Ch. 2 - Over a period of two years, a man in his early 20s...Ch. 2 - Over a period of two years, a man in his early 20s...Ch. 2 - Over a period of two years, a man in his early 20s...Ch. 2 - In this chapter, we focused on how chromosomes are...Ch. 2 - Review the Chapter Concepts list on page 14. All...Ch. 2 - What role do the following cellular components...Ch. 2 - Discuss the concepts of homologous chromosomes,...
Ch. 2 - If two chromosomes of a species are the same...Ch. 2 - Describe the events that characterize each stage...Ch. 2 - How are chromosomes named on the basis of their...Ch. 2 - Contrast telophase in plant and animal mitosis.Ch. 2 - Describe the phases of the cell cycle and the...Ch. 2 - Define and discuss these terms: (a) synapsis, (b)...Ch. 2 - Contrast the genetic content and the origin of...Ch. 2 - Given the end results of the two types of...Ch. 2 - Contrast spermatogenesis and oogenesis. What is...Ch. 2 - Explain why meiosis leads to significant genetic...Ch. 2 - A diploid cell contains three pairs of homologous...Ch. 2 - Considering Problem 15, predict the number of...Ch. 2 - During oogenesis in an animal species with a...Ch. 2 - What is the probability that, in an organism with...Ch. 2 - The nuclear DNA content of a single sperm cell in...Ch. 2 - Describe the role of meiosis in the life cycle of...Ch. 2 - Contrast the chromatin fiber with the mitotic...Ch. 2 - Describe the folded-fiber model of the mitotic...Ch. 2 - Prob. 23PDQCh. 2 - If one follows 50 primary oocytes in an animal...Ch. 2 - In mitosis, what chromatid combination(s) will be...Ch. 2 - During meiosis I, assuming no crossing over, what...Ch. 2 - Are there any possible combinations present during...Ch. 2 - Draw all possible combinations of chromatids...Ch. 2 - Assume that during meiosis I none of the C...Ch. 2 - Assume that each gamete resulting from Problem 29...Ch. 2 - A species of cereal rye (Secale cereale) has a...Ch. 2 - An interesting procedure has been applied for...Ch. 2 - Assume that you were examining a first polar body...Ch. 2 - Kuliev and Verlinsky (2004) state that there was a...
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- A diploid species has 3 pairs of chromosomes in its somatic cells. In males, the first pair is large submetacentric[1]; the second is medium acrocentric[2], and the third is small telocentric[3]. In females, the first two pairs are like those of the males while the third is large metacentric[4][5], with satellite4 Illustrate the karyograms (drawing/picture of the chromosome) of the following: A triploid cell in females tetrasomic cell in males tetraploid cell in females [1] submetacentric --centrosome is just above the middle of the chromosome [2] acrocentric --centrosome is much higher location than submetacentric so that the “p” arm of the chromosome is much shorter than the q arm [3] telocentric --the centromere is at the end of the chromosome [4] metacentric --centrosome is in the middle of the chromosome; thus the “p-arm” and the “q-arm” or both arms of the chromosome are equal in length [5] satellite-a constriction in an arm of a chromosome, aside…arrow_forwardA species of cereal rye (Secale cereale) has a chromosome number of 14, while a species of Canadian wild rye (Elymus canadensis) has a chromosome number of 28. Sterile hybrids can be produced by crossing Secale with Elymus. (a) What would be the expected chromosome number in the somatic cells of the hybrids? (b) Given that none of the chromosomes pair at meiosis I in the sterile hybrid (Hang and Franckowlak, 1984), speculate on the anaphase I separation patterns of these chromosomes.arrow_forwardAlleles of genes A and B were analyzed in Neurospora according to the cross shown below. Ordered tetrads are summarized in each horizontal row with the number of tetrads in each category listed alongside. A) Analyze the data to determine the recombination frequency (RF) between A and B, along with any additional information that is available from this data. B) Draw a map of the chromosome or chromosomes with appropriate map distances. C) Use the Perkins formula to reanalyze any relationship between A and B.arrow_forward
- Select the best process listed below for each of the following statements. Note that you need to know that the sequence of amino acids directly reflects the sequence of genes that coded for their placement. Homologous sets of genes ABCDEF and aBCdEF are located on - vnonhomologous chromosomes. Crossing over between them is suppressed because their locations are the result of this. Homologous sets of genes ABCDEF and AEDCBF are located on homologous chromosomes. A small region of a protein from three species is sequenced and found to be as follows: species X is alanine, glycine, glycine, threonine, alanine v species Y is alanine, glycine, threonine, alanine species Z is alanine, glycine, glycine, threonine, alanine The difference in the amino acid sequence of species Y is most likely due to this. a. Inversion b. Deletion A small region of a protein from three species is sequenced and found to be as follows: c. Gene duplication d. Translocation e. Insertion species X is alanine, glycine,…arrow_forwardn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forwardIn rice, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of rice plants (i.e. the stamen) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile rice plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Give the result(s) of the cross and explain the phenotype of the offspring.arrow_forward
- In 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_forwardIn corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm of each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspringarrow_forwardSketch a series of diagrams showing each of the following, making sure to end each series with haploid cells:(a) How a pair of alleles for a single locus segregate in meiosis(b) How the alleles of two unlinked loci assort independently in meiosis(c) How the alleles of two linked loci undergo genetic recombinationarrow_forward
- The mutations called bobbed in Drosophila result from variable reductions (deletions) in the number of amplified genes coding for rRNA. Researchers trying to maintain bobbed stocks have often documented their tendency to revert to wild type in successive generations. Propose a mechanism based on meiotic recombination which could account for this reversion phenomenon. Why would wild-type flies become more prevalent in Drosophila cultures?arrow_forwardConsider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forwardMap distances were determined for four differentgenes (MAT, HIS4, THR4, and LEU2) on chromosome III of the yeast Saccharomyces cerevisiae:HIS4 ↔ MAT 37 cMTHR4 ↔ LEU2 35 cMLEU2 ↔ HIS4 23 cMMAT ↔ LEU2 16 cMMAT ↔ THR4 20 cMWhat is the order of genes on the chromosome?arrow_forward
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