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Answer Key to Problem Set 4

by: AllieStarks

Answer Key to Problem Set 4 2200

GPA 3.76

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These are ALL of the correct answers to problem set #4 in General Genetics. I've taken this class and passed with an "A" last semester and saved all of the answer keys to the problem sets.
General Genetics
David Braun
Class Notes
Genetics, answers, Answer Key
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This 14 page Class Notes was uploaded by AllieStarks on Saturday January 2, 2016. The Class Notes belongs to 2200 at University of Missouri - Columbia taught by David Braun in Fall 2015. Since its upload, it has received 57 views. For similar materials see General Genetics in Biological Sciences at University of Missouri - Columbia.


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Date Created: 01/02/16
BioSci 2200 General Genetics Problem Set 1 Answer Key Introduction and Mitosis/ Meiosis Introduction - Fields of Genetics To answer the following question, review the three traditional subdivisions of genetics from Chapter 1, the lecture notes and the syllabus, then read the story about albinism in the Hopi tribe of Native Americans on pg 1-2 of the book. Additional Q1 1. Which aspects of the story about Albinism in the Hopis illustrates each of the three major fields of genetics. Answer: a) Transmission or Mendelian Genetics Determining that albinisim is inherited as an autosomal recessive trait. b) Molecular Genetics Determining that albinism is due to a lack of production of melanin likely caused by a mutation in the OCA2 gene on chromosome 15 c) Population Genetics Determining that the frequency of albinism in the Hopis is one hundred times more frequent than other populations and what forces may have contributed to the increased frequency (positive selection, small population size). Mitosis and Meiosis Once you have convinced yourself that you know what occurs at each stage of mitosis and meiosis, test your understanding by solving the following problems from Chapter 2. To compare and contrast mitosis and meiosis, it may be helpful to review Fig. 2.18 and Connecting concepts pg 33. 16. List the similarities and differences between mitosis and meiosis. Which differences do you think are most important and why? Mitosis Meiosis A single cell division produces two Two cell divisions usually result in four progeny genetically identical progeny cells. cells that are not genetically identical. Chromosome number of progeny cells and Daughter cells are haploid and have half the the original cell remain the same. chromosomal complement of the original diploid cell as a result of the separation of homologous pairs during anaphase I. Daughter cells and the original cell are Crossing over in prophase I and separation of genetically identical. No separation of homologous pairs during anaphase I produce homologous chromosomes or crossing over daughter cells that are genetically different from takes place. each other and from the original cell. 1 Homologous chromosomes do not synapse. Synapsis of homologous chromosomes takes place during prophase I. In metaphase, individual chromosomes line In metaphase I, homologous pairs of chromosomes up on the metaphase plate. line up on the metaphase plate. Individual chromosomes line up in metaphase II. In anaphase, sister chromatids separate. In anaphase I, homologous chromosomes separate. Separation of sister chromatids takes place in anaphase II. The most important difference is that mitosis produces cells genetically identical to each other and to the original cell, resulting in the orderly passage of information from one cell to its progeny. In contrast, by producing progeny that do not contain pairs of homologous chromosomes, meiosis results in the reduction of chromosome number from the original cell. Meiosis also allows for genetic variation through crossing over and the random assortment of homologous chromosomes. 28. All of the following cells, shown in various stages of mitosis and meiosis, come from the same rare species of plant. What is the diploid number of chromosomes in this plant? Give the names of each stage of mitosis or meiosis shown. AAanapaapIease Answer: To determine the diploid chromosome number in this plant, the number of centromeres present within a cell that contains homologous pairs of chromosomes must be determined. Remember, each chromosome possesses a single centromere. The location and presence of a centromere are determined by the attachment of the spindle fibers to the chromosome, which occurs at the centromere in the above diagram. Only the cell in stage (a) clearly has homologous pairs of chromosomes. So the diploid chromosome number for cells of this species of plant is six. a. This cell is undergoing anaphase I of meiosis I, as indicated by the separation of the homologous pairs of chromosomes. b. In this cell, sister chromatids have separated, resulting in a doubling of the chromosome number within the cell from six to 12. Based on the number of chromosomes, the separation of sister chromatids in this cell must be occurring during anaphase of mitosis. c. Again, sister chromatids are being separated, but the number of chromosomes present in the cell is only six. This indicates that no homologs are present within the cell, so in this cell the separation of sister chromatids is occurring in anaphase II of meiosis II. 2 Additional Q2 2. The cells in the following figure were all taken from the same individual (a mammal). Identify the cell division events occurring in each cell and explain your reasoning. What is the diploid chromosome number? Answer: To reconcile how all the cells illustrated could come from one individual, consider that the cells shown could come from either somatic or germ-line cells. Cell (a) shows three pairs of previously synapsed homologs disjoining and must therefore illustrate anaphase I of meiosis. That three pairs of chromosomes are present indicates that the organism has 2N = 6 chromosomes, so that N = 3. Cell (b) shows the disjoining of chromatids. Since the organism has 2N = 6 chromosomes, and the daughter cells that will form as a result of this cell division will have 6 chromosomes, (b) must illustrate part of mitosis, specifically anaphase. Cell (c) also shows the disjoining of chromatids. Since the daughter cells will receive three chromosomes, this must be anaphase II of meiosis. Cell (d) shows the pairing of homologs and therefore illustrates metaphase I of meiosis. Because the individual has three pairs of identically appearing chromosomes, there are two identical sex (= X) chromosomes, indicating that the animal is female. Additional Q3 3. DNA synthesis occurs prior to mitosis in the cell cycle. During mitosis, the 2 daughter cells each receive half of the DNA in the parent cell. If the parent has 2N chromosomes prior to mitosis (but after DNA synthesis) and a daughter cell receives half of the DNA, why is the chromosome number of the daughter 2N? Shouldn’t each daughter receive N chromosomes? 3 Comparing and contrasting mitosis and meiosis, continued To attempt the following problems it might be helpful to review Fig. 2.12, and Connecting concepts on pg 27. 26. A cell in G1 of interphase has 12 chromosomes. How many chromosomes and DNA molecules will be found per cell when this original cell progresses to the following stages? Answer: The number of chromosomes and DNA molecules depends on the stage of the cell cycle. Each chromosome contains only one centromere, but after the completion of S phase, and prior to anaphase of mitosis or anaphase II of meiosis, each chromosome will consist of two DNA molecules. a. G2 of interphase. G2 of interphase occurs after S phase, when the DNA molecules are replicated. Each chromosome now consists of two DNA molecules. So a cell in G2 will contain 12 chromosomes and 24 DNA molecules. b. Metaphase I of meiosis. Neither homologous chromosomes nor sister chromatids have separated by metaphase I of meiosis. Therefore, the chromosome number is 12, and the number of DNA molecules is 24. c. Prophase of mitosis. This cell will contain 12 chromosomes and 24 DNA molecules. d. Anaphase I of meiosis. During anaphase I of meiosis, homologous chromosomes separate and begin moving to opposite ends of the cell. However, sister chromatids will not separate until anaphase II of meiosis. The number of chromosomes is still 12, and the number of DNA molecules is 24. e. Anaphase II of meiosis. Homologous chromosomes were separated and migrated to different daughter cells at the completion of meiosis I (6 chromosomes, 12 DNA molecules). However, in anaphase II of meiosis, sister chromatids separate, resulting in a temporary doubling of the 4 chromosome number in the now haploid daughter cell (compare to f). The number of chromosomes and the number of DNA molecules present will both be 12. f. Prophase II of meiosis. The daughter cells in prophase II of meiosis are haploid. The haploid cells will contain six chromosomes and 12 DNA molecules. g. After cytokinesis following mitosis. After cytokinesis following mitosis the daughter cells will enter G1. Each cell will contain 12 chromosomes and 12 DNA molecules. h. After cytokinesis following meiosis II. After cytokinesis following meiosis II, the haploid daughter cells will contain six chromosomes and six DNA molecules. 29. The amount of DNA per cell of a particular species is measured in cells found at various stages of meiosis, and the following amounts are obtained (3.7picogram, 7.3pg, 14.6pg). Match the amounts of DNA with the corresponding stages of the cell cycle. You may use more than one stage for each amount of DNA. Stage of meiosis Amount of DNA per cell a. G1 a. 7.3 pg b. Prophase I b. 14.6 pg c. G2 c. 14.6 pg d. Following telophase II and cytokinesis d. 3.7 pg e. Anaphase I e. 14.6 pg f. Metaphase II f. 7.3 pg Answer: The amount of DNA in the cell will be doubled after the completion of S phase in the cell cycle and prior to cytokinesis in either mitosis or meiosis I. At the completion of cytokinesis following meiosis II, the amount of DNA will be halved. a. G1 occurs prior to S phase and the doubling of the amount of DNA and prior to the completion of the meiosis II and cytokinesis, which will result in a haploid cell containing one-half the amount of DNA that was contained in the cell in G . 1 b. During prophase I of meiosis, the amount of DNA in the cell is two times the amount in G 1 The homologus chromosomes are still located within a single cell, and there are two sister chromatids per chromosome. c. G2takes place directly after the completion of S phase, so the amount of DNA is two times the amount prior to the S phase. d. Following cytokinesis associated with meiosis II, each daughter cell will contain only one-half the amount of DNA of a mother cell found in G 1 of interphase. By the completion of cytokinesis associated with meiosis II, both homologous pairs of chromosomes and sister chromatids have been separated into different daughter cells. Therefore, each daughter cell will contain only one-half the amount of DNA of the original cell in G . 1 e. During anaphase I of meiosis, the amount of DNA in the cell is two times the amount in G 1. The homologus chromosomes are still located within a single cell, and there are two sister chromatids per chromosome. f. Metaphase II takes place after the cytokinesis associated with meiosis I and results in the daughter cells receiving only one-half the DNA found in their mother cell. In metaphase II of 5 meiosis, the amount of DNA in each cell is the same as G be1ause each chromosome still consists of two DNA molecules (two sister chromatids per chromosome). 31. A cell in prophase II of meiosis contains 12 chromosomes. How many chromosomes would be present in a cell from the same organism if it were in prophase of mitosis? Prophase I of meiosis? Answer: A cell in prophase II of meiosis will contain the haploid number of chromosomes. For this organism, 12 chromosomes represent the haploid chromosome number of a cell, or one complete set of chromosomes. A cell from the same organism that is undergoing prophase of mitosis would contain a diploid number of chromosomes, or two complete sets of chromosomes, which means that homologous pairs of chromosomes are present. So a cell in this stage should contain 24 chromosomes. Homologous pairs of chromosomes have not been separated by prophase I of meiosis. During this stage, a cell of this organism will contain 24 chromosomes. 32. A cell has eight chromosomes in G1 of interphase (ie: 2N=8). Draw a picture of this cell with its chromosomes at the following stages. Indicate how many DNA molecules are present at each stage. Prophase I of meiosis 8 chromosomes 8 chromosomes 8 DNA molecules 16 DNA molecules 6 ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯Answer Key▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯A▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯a▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯Wx▯▯▯▯▯wx▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯AA▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯A/A▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯a▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯a/a▯▯▯▯▯a▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯i ▯▯▯▯i ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯L_▯▯▯▯▯▯▯▯▯ll▯▯▯ ▯▯▯▯▯▯▯▯ Hint:▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ Answers online. Try choosing an incorrect answer to see the tutorials at the Biology Project website. ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ Answer: The principle of segregation, or Mendel’s first law, states that an organism possesses two alleles for any one particular trait and that these alleles separate during the formation of gametes. In other words, one allele goes into each gamete. The principle of segregation is important because it explains how the genotypic ratios in the haploid gametes are produced.▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ Answer: The concept of dominance states that when two different alleles are present in a genotype, only the dominant allele is expressed in the phenotype. ▯ ▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ Answer: According to the principle of independent assortment, genes for different characteristics that are at different loci segregate independently of one another. The principle of segregation indicates that the two alleles at a locus separate; the principle of independent assortment indicates that the separation of alleles at one locus is independent of the separation of alleles at other loci. ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ Answer: In anaphase I of meiosis, homologous chromosomes separate. This explains the principle of segregation (Mendel’s 1 Law). Also in anaphase I, each pair of homologous chromosomes segregate independently of all other pairs of homologous chromosomes (Mendel’s nd 2 Law). The assortment is dependent on how the homologs line up during metaphase I. This assortment of homologs explains how genes located on different pairs of chromosomes will separate independently of one another. ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯R▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯r▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯ ▯ Answer a. The cross of a homozygous cucumber plant that produces orange fruit (RR) with a homozygous cucumber plant that produces cream fruit (rr) will result in an F1 generation heterozygous for the orange fruit phenotype. R R P × rr F1 Rr orange fruit Intercrossing the F1 will produce F2 that are expected to show a 3:1 orange-to cream-fruit phenotypic ratio. F1 Rr × Rr F2 1 RR orange fruit, 2 Rr orange fruit, 1 rr cream fruit b. The backcross of the F1 orange offspring (Rr) with homozygous orange parent (RR) will produce progeny that all have the orange fruit phenotype. However, one-half of the progeny will ▯▯ ▯ be expected to be homozygous for orange fruit and one-half of the progeny will be expected to be heterozygous for orange fruit. ) 1 F ( r R × RR (orange parent) ½ RR orange fruit, ½ Rr orange fruit c. The backcross of the F1 offspring (Rr) with the cream parent (rr) is also a testcross. The product of this testcross should produce progeny, one-half of which are heterozygous for orange fruit and one-half of which are homozygous for cream fruit. ) 1 F ( r R × rr (cream parent) ▯ ½ Rr orange fruit, ½ rr cream fruit ▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯ ▯ ▯ ▯ ▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯I ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯i ▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ Answer: a. Male with blood type A × Female with blood type B B B Because the female parentA As bloA Btype B, she must have the genotype i i . The male parent could be either I I or I i . However, as some of the offspring are kittens with blood B type B, the male parent must have contributed an i allele to these kittens. Therefore, the A B male must have the genotype of I i . b. Male with blood type B × Female with blood type B ▯▯ ▯ B Because blood type B is caused by the recessive allele i , both parents must be homozygous B B B for the recessive allele or i i . Each contributes only the i allele to the offspring. c. Male with blood type B × Female with blood type A B B Again, the male with type B blood must be i i . A female with type A blood could have either the I I or I i genotypes. Because all of her kittens have type A blood, this suggests that she A A A A is homozygous for the for I allele (I I ) and contributes only the I allele to her offspring. It is possible that she is heterozygous for type A blood, but if so it is unlikely that chance alone would have produced eight kittens with blood type A. ▯ d. Male with blood type A × Female with blood type A Because kittens with blood type A and blood type B are found in the offspring, both parents A B must be heterozygous for blood type A, or I i . With both parents being heterozygous, the offspring would be expected to occur in a 3:1 ratio of blood type A to blood type B, which is close to the observed ratio. ▯ e. Male with blood type A × Female with blood type A Only kittens with bloA Atype A are produced, which suggests that each parent is homoA Aous for blood type A (I I ), or that one parent is homozygous for blood type A (I I ), and the A B other parent is heterozygous for blood type A (I i ). The data from the offspring will not allow us to determine the precise genotype of either parent. f. Male with blood type A × Female with blood type B B B On the basis of her phenotype, the female will be i i . In the offspring, one kitten with blood B type B is produced. This kitten would require that both parents contribute an ▯ to produce its A B genotype. Therefore, the male parent’s genotype is ▯ ▯ . From this cross, the number of kittens with blood type B would be expected to be similar to the number of kittens with blood type A. However, due to the small number of offspring produced, random chance could have resulted in more kittens with blood type A than kittens with blood type B.▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯B▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯b▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯S▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯s▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯ ▯ Answer: a. P: homozygous bitter fruit, yellow spots × homozygous sweet fruit and no spots BB SS x bb ss F1: All progeny have bitter fruit and yellow spots Bb Ss The F1 are intercrossed to produce the F2: Bb Ss × Bb Ss The F2 phenotypic ratios are as follows: 9/16 bitter fruit and yellow spots, 3/16 bitter fruit and no spots, 3/16 sweet fruit and yellow spots, 1/16 sweet fruit and no spots b. The backcross of an F1 plant (Bb Ss) with the bitter, yellow-spotted parent (BB SS) will produce all bitter, yellow-spotted offspring. c. The backcross of a F1 plant (Bb Ss) with the sweet, nonspotted parent (bb ss) will produce the following phenotypic proportions in the offspring: ¼ bitter fruit and yellow spots, ¼ bitter fruit and no spots, ¼ sweet fruit and yellow spots, ¼ sweet fruit and no spots ▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯P▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯p▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯S▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯s▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯ ▯ ▯ ▯▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ ▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯▯ Answer: First use the symbols for the alleles specifying each trait and the information about which allele is dominant or recessive to make initial assignments of possible genotypes. For example, if a plant is white, it must have a dominant W allele, but as it may be either WW or Ww, it would be initially noted as W–. If a plant is yellow, it has to be ww. Then, by considering just one pair of allelic traits at a time, and recalling the Mendelian progeny ratios you have seen [a 1:1 ratio follows from a testcross (Aa x aa), an all-to-none ratio follows if at least one parent is homozygous dominant (AA x A– or aa), and a 3:1 ratio follows from a monohybrid cross (Aa x Aa)], you can ascertain whether a parental W– plant is Ww or WW. ▯ ▯▯ ▯


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