Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Chapter 16, Problem 10P
| a. | Assume that two transcription factors are required for expression of the blue pigmentation genes in pansies. (Without the pigment, the flowers are white.) What |
| b. | Now assume that either transcription factor is sufficient to get blue color. What phenotypic ratios would you expect from crossing the same two heterozygous strains? |
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A. Deletion of the SOX9 gene leads to sex reversal resulting in a person with karyotype 46XY being phenotypically Explain the genetic basis for this.
B. Describe what would happen to the phenotype of a male with a mutation in the gene encoding SF1? Explain your answer.
In wheat, aleurone cells form a thin layer of the seed coat that is critical to early gene expression in plant development. The color of this layer of cells is controlled by two alleles of a gene [colored aleurone (R) is dominant to colorless (r)]. A second gene is known to control the color of leaf tips [green leaf tip (G) is dominant to yellow (g)]. Two plants, each heterozygous for both characteristics, are test crossed to homozygous recessives, and their progeny are combined to produce the following totals:
colored green 102
colored yellow 98
colorless green 103
colorless yellow 97
a) Use chi-square analysis to test these data for an independent assortment of the two characteristics (table provided). Please show work, how your expected values are calculated, and explain what your results indicate about the data.
b) You decide to be cautious in your analysis, and decide to analyze the progeny from each of the crosses individually (instead of adding them together as shown above).…
The following genotypes of two independently assorting autosomal genes determine coat color in rats:A-B- (gray); A-bb (yellow); aaB-(black); aabb (cream)A third gene pair on a separate autosome determines whetherany color will be produced. The CC and Cc genotypes allow coloraccording to the expression of the A and B alleles. However,the cc genotype results in albino rats regardless of the A and Balleles present. Determine the F1 phenotypic ratio of the followingcrosses: (a)AAbbCC * aaBBcc; (b) AaBBcc * AABbcc;(c) AaBbCc * AaBbcc.
Chapter 16 Solutions
Genetics: From Genes to Genomes, 5th edition
Ch. 16 - For each of the terms in the left column, choose...Ch. 16 - Does each of the following types of gene...Ch. 16 - List five events other than transcription...Ch. 16 - Which eukaryotic RNA polymerase RNA pol I, pol II,...Ch. 16 - You have synthesized an enhancerless GFP reporter...Ch. 16 - Prob. 6PCh. 16 - Yeast genes have cis-acting elements upstream of...Ch. 16 - A single UASG regulates the expression of three...Ch. 16 - Prob. 9PCh. 16 - a. Assume that two transcription factors are...
Ch. 16 - a. You want to create a genetic construct that...Ch. 16 - In Problem 12, you identified a genomic region...Ch. 16 - Prob. 13PCh. 16 - Prob. 14PCh. 16 - Genes in both prokaryotes and eukaryotes are...Ch. 16 - Prob. 16PCh. 16 - Prader-Willi syndrome is caused by a mutation in...Ch. 16 - The human IGF-2R gene is autosomal and maternally...Ch. 16 - Follow the expression of a paternally imprinted...Ch. 16 - Reciprocal crosses were performed using two inbred...Ch. 16 - Interestingly, imprinting can be tissue-specific....Ch. 16 - Prob. 22PCh. 16 - a. How can a single eukaryotic gene give rise to...Ch. 16 - A hunchback gene, a gene necessary for proper...Ch. 16 - You know that the mRNA and protein produced by a...Ch. 16 - You are studying a transgenic mouse strain that...Ch. 16 - Prob. 27PCh. 16 - Scientists have exploited the siRNA pathway to...Ch. 16 - Researchers know that Fru-M controls male sexual...Ch. 16 - The Drosophila gene Sex lethal Sxl is deserving of...Ch. 16 - Prob. 31P
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- Imagine a scenario in which prenatal testing of a human female fetus indicates that the baby will have a normal XX karyotype but is heterozygous for a mutation that inactivates the Xist promoter. Allele “Xr” represents the mutated version of the Xist promoter, and “XR” represents the normal version of the Xist promoter. How will this mutation affect the process of X inactivation?A. "X inactivation will still be random with both mutant and normal X chromosomes being randomly inactivated" B. "The chromosome with the mutant Xist promoter will always be active" C. "The chromosome with the mutant Xist promoter will always be inactivated" D. "The normal X chromosome (no mutation in Xist) will always be active" E. "The normal X chromosome (no mutation in Xist) will always be inactivated"arrow_forwardExplain how (a) the absence of class B gene expression produces the flower structures seen in class B mutants (see Figure 22.15c) and (b) the absence of class C gene expression produces the structures seen in class C mutants (see Figure 22.15d).arrow_forwardWhen the organization of adjacent genes with highly similar sequences is in an inverted orientation, this can reduce the expression of other genes that have similar sequence and are located on other chromosomes. Explain the mechanism of how this generally occurs.arrow_forward
- The Drosophila gene Sex lethal (Sxl) is deserving of itsname. Certain alleles have no effect on XY animals butcause XX animals to die early in development. Other alleles have no effect on XX animals but cause XY animals to die early in development. Thus, some Sxl allelesare lethal to females, while others are lethal to males.a. Would you expect a null mutation in Sxl to causelethality in males or in females? b. Why do Sxl alleles of either type cause lethality ina specific sex?The gene transformer (tra) gets its name from sexualtransformation, as some tra alleles can change XXanimals into morphological males, while other traalleles can change XY animals into morphologicalfemales.c. Which of these sex transformations would becaused by null alleles of tra and which would becaused by constitutively active alleles of tra?d. In contrast with Sxl, null tra mutations do notcause lethality either in XX or in XY animals.However, the Sxl protein regulates the productionof the Tra protein. Why…arrow_forwardIn rats, the following genotypes of two independently assorting autosomal genes determine coat color: A-B- (gray) A-bb (yellow) aaB- (black) aabb (cream) A third gene pair on a separate autosome determines whether or not any color will be produced. The CC and Cc genotypes allow color according to the expression of the A and B alleles. However, the cc genotype results in albino rats regardless of the A and B alleles present. Determine the F1 phenotypic ratio of the following crosses: (a) AAbbCC * aaBBcc (b) AaBBCC * AABbcc (c) AaBbCc * AaBbcc (d) AaBBCc * AaBBCc (e) AABbCc * AABbccarrow_forwardUse the dropdowns to indicate the pattern of methylation and gene expression expected for a maternally-imprinted gene in the somatic cell of a male individual. The Somatic cell would show [Select] male's male parent, and [Select] male's female parent. As a result, [Select] of the allele inherited from the of the allele inherited from the could be expressed. The pattern of methylation would be [Select] imprinted gene in a somatic cell of a female individual. for the same maternally-arrow_forward
- Coat color of dogs depend upon the action of at least two genes. At one locus a dominant epistatic inhibitor of coat color pigment (I_) prevents the expression of color alleles at another independently assorting locus, producing white coat color. When the recessive condition exists in the inhibitor locus (ii), the alleles of the hypostatic locus may be expressed, iiB_ producing black and iibb producing brown. When dihybrid white dogs are mated together, determine the: (a) the type of gene interaction (b) the phenotypic proportions expected in the progeny (c) the chance of choosing, from among the white progeny, a genotype that is homozygous at both lociarrow_forward. In the plant Arabidopsis thaliana, a geneticist is interested in the development of trichomes (small projections). A large screen turns up two mutant plants (A and B) that have no trichomes, and these mutants seem to be potentially useful in studying trichome development. (If they were determined by single-gene mutations, then finding the normal and abnormal functions of these genes would be instructive.) Each plant is crossed with wild type; in both cases, the next generation (F1) had normal trichomes. When F1 plants were selfed, the resulting F2’s were as follows: F2 from mutant A: 602 normal; 198 no trichomes F2 from mutant B: 267 normal; 93 no trichomesa. What do these results show? Include proposed genotypes of all plants in your answer. b. Under your explanation to part a, is it possible to confidently predict the F1 from crossing the original mutant A with the original mutant B?arrow_forwardWhite eyes in Drosophila melanogaster result from an X-linked recessive mutation. Occasionally, white-eyed mutants give rise to offspring that possess white eyes with small red spots. The number, distribution, and size of the red spots are variable. Explain how a transposable element could be responsible for this spotting phenomenon.arrow_forward
- In drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? Show COMPLETE cross for both cases. If m-m- females produce useless eggs, then how are m-m- produced?arrow_forwardIn drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. From the offspring, backcross the recessive female with the paternal strain. 1. What are the genotypes and phenotypes of the F2s? 2. If m-m- females produce useless eggs, then how are m-m- produced?arrow_forwardIn drosophila, a recessive mutation (m-) of a maternal effect gene results in an abnormal phenotype wherein homozygous (m-m-) females produce eggs that cannot support embryonic development. Homozygous (m-m-) males, however, can still produce viable sperm. (A) Using m+ to denote a normal gene, determine the genotypes and phenotypes of the F1s produce by a cross between a heterozygous female and a recessive male. (B) From the offspring, backcross the recessive female with the paternal strain. What are the genotypes and phenotypes of the F2s? (C) If m-m- females produce useless eggs, then how are m-m- produced?arrow_forward
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