Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 5, Problem 49P
A diploid strain of yeast has a wild-type phenotype but the following genotype:
a, b, c, d, and e all represent recessive alleles that yield a visible phenotype, and leth represents a recessive lethal mutation. All genes are on the same chromosome, and a is very tightly linked to the centromere (indicated by a small circle). Which of the following
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Chapter 5 Solutions
Genetics: From Genes to Genomes
Ch. 5 - Choose the phrase from the right column that best...Ch. 5 - a. A Drosophila male from a true-breeding stock...Ch. 5 - Prob. 3PCh. 5 - The Punnett square in Fig. 5.4 shows how Mendels...Ch. 5 - In mice, the dominant allele Gs of the X-linked...Ch. 5 - In Drosophila, males from a true-breeding stock...Ch. 5 - If the a and b loci are 20 m.u. apart in humans...Ch. 5 - CCDD and ccdd individuals were crossed to each...Ch. 5 - In mice, the autosomal locus coding for the...Ch. 5 - In a particular human family, John and his mother...
Ch. 5 - Albino rabbits lacking pigment are homozygous for...Ch. 5 - In corn, the allele A allows the deposition of...Ch. 5 - If the a and b loci are 40 cM apart and an AA BB...Ch. 5 - Write the number of different kinds of phenotypes,...Ch. 5 - A DNA variant has been found linked to a rare...Ch. 5 - Figure 5.7a shows chromosomes during prophase of...Ch. 5 - Figure 5.7b shows bivalents in mouse primary...Ch. 5 - Cinnabar eyes cn and reduced bristles rd are...Ch. 5 - In Drosophila, the autosomal recessive dp allele...Ch. 5 - From a series of two-point crosses, the following...Ch. 5 - Map distances were determined for four different...Ch. 5 - In the tubular flowers of foxgloves, wild-type...Ch. 5 - In Drosophila, the recessive allele mb of one gene...Ch. 5 - A snapdragon with pink petals, black anthers, and...Ch. 5 - In Drosophila, three autosomal genes have the...Ch. 5 - Drosophila females heterozygous for each of three...Ch. 5 - Male Drosophila expressing the autosomal recessive...Ch. 5 - a. In Drosophila, crosses between F1 heterozygotes...Ch. 5 - A true-breeding strain of Virginia tobacco has...Ch. 5 - Prob. 30PCh. 5 - The following list of four Drosophila mutations...Ch. 5 - Do the data that Mendel obtained fit his...Ch. 5 - Two genes control color in corn snakes as follows:...Ch. 5 - A mouse from a true-breeding population with...Ch. 5 - Neurospora of genotype a c are crossed with...Ch. 5 - A cross was performed between one haploid strain...Ch. 5 - Prob. 40PCh. 5 - Prob. 41PCh. 5 - Indicate the percentage of tetrads that would have...Ch. 5 - Prob. 43PCh. 5 - This problem leads you through the derivation of a...Ch. 5 - a. In ordered tetrad analysis, what is the maximum...Ch. 5 - Prob. 46PCh. 5 - A single yeast cell placed on a solid agar will...Ch. 5 - Figure 5.29 shows mitotic recombination leading to...Ch. 5 - A diploid strain of yeast has a wild-type...Ch. 5 - In Drosophila, the yellow y gene is near the...Ch. 5 - Neurofibromas are tumors of the skin that can...Ch. 5 - Two important methods for understanding the...
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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
- In yeast, LYS5, ADE1, and URA2 are genes required to synthesize lysine, adenine and uracil, respectively. A MATa lys5 ADE1 URA2 haploid was mated to a MATA LYS5 ade1 ura2 haploid to make a triply heterozygous diploid. This diploid was put through meiosis, 100 tetrads were dissected and the spore colonies transferred to medium lacking either lysine, adenine or uracil to determine which spores were prototrophic or auxotrophic for those nutrients. The following results were obtained: LYS5-ADE1 ADE1-URA2 LYS5-URA2 PD NPD Ι 20 22 58 18 20 62 50 8 42 Based on this information, select the statements below that are TRUE. Select 4 correct answer(s) A) The LYS5 and URA2 genes are linked. B) The genotypes of the four spores in the ADE1-URA2 tetratypes is ade1 ura2 ade1 URA2 ADE1 ura2 ADE1 URA2 C) The map distance between ADE1 and URA2 is 51 CM. D) ADE1 and URA2 are likely located on the same chromosome. E) The map distance between LYS5 and URA2 is 29 CM. ☐ F) LYS5 and ADE1 independently assort…arrow_forwarda. Manually, using a pencil, draw a cell in anaphase II from an organism in which 2n = 2 and each chromosome is metacentric. b. Given that each G1 nucleus from this organism contains 16 picograms of DNA, how many picograms of chromosomal DNA would you expect in the cell shown here?arrow_forwardIn a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally dimerizes to catalyzes a biochemical reaction necessary for the production of a dark pigment. Ed represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E*/Ed, and why? O wild type (normal production of the dark pigment), as E is haplosufficient mutant (no pigment production), as no dimers will form in the heterozygous mutant (no pigment production), as the mutant allele Eg is dominant O wild type (normal production of the dark pigment), as dimers of wild-type and mutant protein E will be formed in the heterozygousarrow_forward
- Mitotic nondisjunction can occur, particularly in tumor cells. Assume a single nondisjunction event involving the X chromosome occurs in a cell undergoing mitosis in a human female. First, consider the daughter cell with the smaller number of chromosomes: How many total chromosomes will this cell have? How many X chromosomes will this cell have? Next, consider the daughter cell with the larger number of chromosomes: How many total chromosomes will this cell have? How many X chromosomes will this cell have?arrow_forwardHumans and many other complex organisms are diploid, possessing two sets of genes, one inherited from the mother and one from the father. However, a number of eukaryotic organisms spend most of their life cycles in a haploid state. Many of these eukaryotes, such as Neurospora and yeast, still undergo meiosis and sexual reproduction, but most of the cells that make up the organism are haploid. Considering that haploid organisms are fully capable of sexual reproduction and generating genetic variation, why are most complex eukaryotes diploid? In other words, what might be the evolutionary advantage of existing in a diploid state instead of a haploid state? And why might a few organisms, such as Neurospora and yeast, exist as haploids?arrow_forwardDuring a practical, a group of biology students are tasked to study the cell cycle of an insect. This insect has the karyotype 2n=6, with genotype Aa Bb Ee HH. The students must draw a G2 cell about to enter meiosis, using red lines for maternal chromosomes, blue lines for paternal chromosomes and black dots to represent centromeres. The drawing provided by the students is shown below. a) Identify three mistakes the students made when producing this diagram. In your answer only refer to elements already present in the figure. (3) b) This fly is used in a testcross. What is the probability of producing a fly with genotype Aa bb ee Hh?arrow_forward
- To identify genes controlling the cell cycle in budding yeast, a genetic screen was carried out. In this screen, haploid yeast cells were exposed to a DNA damaging agent to introduce random mutations in the genome. By culturing cells at an elevated temperature (e.g. 37 degrees), where many mutated genes lose their function, scientists identified yeast mutants that showed growth defects and arrest at specific stages of the cell cycle (e.g. in mitosis with large buds). In this screen, mutants of the cyclin-dependent kinase were identified, but not mutants of cyclins. Explain the reason for this outcome.arrow_forwardA cross between yeast strains ab x AB produces the following unordered tetrads: a. Label each group of tetrads as parental ditypes (PD), non-parental ditypes (NPD), or tetratypes (T). AB AB ab ab 43 6 22 b. Are the genes linked or on separate chromosomes? Why? aB aB Ab Ab Ab AB ab aB c. If the genes are linked determine the distance between them.arrow_forwardIn the plant, Haplopappus gracile there is one long pair and one short pair of chromosomes. In the diagrams below, anaphase of individual cells in meiosis or mitosis in a plant that is heterozygous for the genes, A and B (which are on separate chromosomes) are shown. The lines represent chromosomes or chromatids and the points of the "V" are the centromeres. For each case, indicate if the cell represents meiosis I, meiosis II, mitosis, or impossible situation. Provide a brief (one sentence) reason for your decision. 1. A A a В a B 9. A a B A a 2. а В В a A a B 9. B. a A 3.arrow_forward
- 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_forwardIn a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…arrow_forwardThe image below shows a somatic cell from an organism for which 2n = 4 (the cell from the image below). Which of the cells beneath it (cells A, B, C or D) could be a gamete from this organism? Explain why you chose the cell you did discussing the changes in chromosome number nd DNA content of a cell throughout the cell cycle. You should also explain if any of the other images could represent a cell from this organism in stages of meiosis or mitosis or other stages of the cell cycle. X Xx C "₁ XX B XX X Xarrow_forward
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Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY