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
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 14, Problem 15PDQ
The synthesis of flower pigments is known to be dependent on enzymatically controlled biosynthetic pathways. For the crosses shown here, postulate the role of mutant genes and their products in producing the observed
- (a) P1: white strain A × white strain B
F1: all purple
F2: 9/16 purple: 7/16 white
- (b) P1: white × pink
F1: all purple
F2: 9/16 purple: 3/16 pink: 4/16 white
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
In sweet peas, the given enzymatic pathway result to pigment formation in the flowers:
A dihybrid plant is crossed to a white one which is heterozygous at the “C” locus (provided in the image).
What is the genotype of the dihybrid plant?
What is the genotype of the white plant?
What kinds of flowers, colored or white, are to be expected from the cross above? Include the ratio.
Schizosaccharomyces pombe, also known as "fission yeast," is a powerful model organism in molecular
and cell biology. While performing a genetic screen, you discover an auxotrophic S. pombe strain that
is unable to synthesize one or more vitamins. The following table represents the key experiments you
performed during your genetic screen. Fill in the table with the outcome of each experiment for your
mutant strain (using + for growth and - for no growth).
Medium
Rich media
Minimal media
Minimal media +
all vitamins
Minimal media +
all amino acids
Growth
Wild-type
+
+
+
+
Mutant
+
+
+
>
>
>
For the following genotype, indicate if beta-galactosidase will be made in the presence of lactose and absence of lactose.
P₁+1+PL+OcZ/
P₁1 P₁O+Z+
O O O
beta-galactosidase will be made in both the presence and absence of lactose (+; +)
beta-galactosidase will be made in the presence but not in the absence of lactose (+;-)
beta-galactosidase will not be made in the presence but will be made in the absence of lactose (-; +)
beta-galactosidase will not be made in either the presence or absence of lactose (-;-)
Chapter 14 Solutions
Concepts of Genetics (12th Edition)
Ch. 14 - Prob. 1NSTCh. 14 - A series of mutations in the bacterium Salmonella...Ch. 14 - HbS results from the substitution of valine for...Ch. 14 - Given that a faulty ribosomal protein is the...Ch. 14 - A couple with a child affected with DBA undergoes...Ch. 14 - Prob. 3CSCh. 14 - HOW DO WE KNOW? In this chapter, we focused on the...Ch. 14 - CONCEPT QUESTION Review the Chapter Concepts list...Ch. 14 - Contrast the roles of tRNA and mRNA during...Ch. 14 - Francis Crick proposed the adaptor hypothesis for...
Ch. 14 - During translation, what molecule bears the codon?...Ch. 14 - The chain of eukaryotic hemoglobin is composed of...Ch. 14 - Assuming that each nucleotide in an mRNA is 0.34...Ch. 14 - Summarize the steps involved in charging tRNAs...Ch. 14 - To carry out its role, each transfer RNA requires...Ch. 14 - What are isoaccepting tRNAs? Assuming that there...Ch. 14 - When a codon in an mRNA with the sequence 5-UAA-3...Ch. 14 - Discuss the potential difficulties of designing a...Ch. 14 - Prob. 13PDQCh. 14 - Prob. 14PDQCh. 14 - The synthesis of flower pigments is known to be...Ch. 14 - The study of biochemical mutants in organisms such...Ch. 14 - Explain why the one-gene: one-enzyme concept is...Ch. 14 - Why is an alteration of electrophoretic mobility...Ch. 14 - Prob. 19PDQCh. 14 - Prob. 20PDQCh. 14 - Prob. 21PDQCh. 14 - Prob. 22PDQCh. 14 - Several amino acid substitutions in the and ...Ch. 14 - Define and compare the four levels of protein...Ch. 14 - What are the two common types of protein secondary...Ch. 14 - How do covalent disulfide bonds, hydrogen bonds...Ch. 14 - Prob. 27PDQCh. 14 - List three different types of posttranslational...Ch. 14 - Prob. 29PDQCh. 14 - How does an enzyme function? Why are enzymes...Ch. 14 - Prob. 31PDQCh. 14 - Three independently assorting genes (A, B, and C)...Ch. 14 - How would the results vary in cross (a) of Problem...Ch. 14 - Deep in a previously unexplored South American...Ch. 14 - Many antibiotics are effective as drugs to fight...Ch. 14 - The flow of genetic information from DNA to...Ch. 14 - Prob. 37ESP
Knowledge Booster
Learn more about
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
- Time mapping is performed in a cross involving the genes his,leu, mal, and xyl. The recipient cells are auxotrophic for all fourgenes. After 25 minutes, mating is interrupted, with the resultsin recipient cells shown below. Diagram the positions of thesegenes relative to the origin (O) of the F factor and to one another.(a) 90% are xyl+(b) 80% are mal+(c) 20% are his+(d) None are leu+arrow_forwardTwo morphotypes of the newly discovered plant in Mt. Banahaw de Lucban were tested for linkage of three genes such as the presence of tendril (t+), dense trichomes (d+), and the presence of secretory cells (sc+). The loci for the mutant genes have been mapped and are separated by the following map distances: t and d = 20CM; d and sc = 12CM. The interference between these genes is 0.4. The first morphotype is located at the lower elevation and is characterized by the absence of tendril, sparse trichome, and absence of secretory cells. On the other hand, the second morphotype is located at the higher elevation and is characterized by the presence of tendril, dense trichome, and presence of secretory cells. Further genetic analysis showed that the morphotypes were true breeding. The two morphotypes were intercrossed and the resulting F1 is testcrossed with the first morphotype. The cross resulted in 1800 progeny. Give the genotypes, phenotypes, and the expected numbers of phenotypes in…arrow_forwardSome sweet-pea plants have purple flowers and others have white flowers. A homozygous variety of sweet pea that has purple flowers is crossed with a homozygous variety that has white flowers. All the F1have purple flowers. When these F1 self-fertilize, the F2 appear in a ratio of 916 purple to 716 white. a.Draw a hypothetical biochemical pathway to explain the production of purple and white flowers in sweet peas.arrow_forward
- Time mapping is performed in a cross involving the genes his, leu, mal, and xyl. The recipient cells were auxotrophic for all four genes. After 25 minutes, mating was interrupted with the following results in recipient cells. Diagram the positions of these genes relative to the origin (O) of the F factor and to one another. (a) 90% were xyl+ (b) 80% were mal+ (c) 20% were his+ (d) none were leu+arrow_forwardThe following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.arrow_forwardThree haploid fungal mutants that require compound W for growth were isolated. Each mutant contains a recessive allele in a single gene. Three compounds (A, B and C) in the biosynthetic pathway to W are known, but their order in the pathway is unknown. Each compound is tested for its ability to support the growth of each of the three mutants. Phenotypes of all of the three mutants are shown in the following table (“+" indicates growth, "-" indicates no growth). A C W Mutant 1 Mutant 2 Mutant 3 What would be the phenotype of a haploid mutant that contains both mutant alleles in mutant 2 and 3? Phenotype refers to growth or absence of growth on compounds A, B, C and WN. O Like mutant 1 O Like mutant 2 Like mutant 3 O Like wild typearrow_forward
- You are studying a plant with tissue comprising both green and white sectors. You wish to decide whether this phenomenon is due (1) to a chloroplast mutation of the type considered in this chapter or (2) to a dominant nuclear mutation that inhibits chlorophyll production and is present only in certain tissue layers of the plant as a mosaic. Outline the experimental approach that you would use to resolve this problem.arrow_forwardA complex biochemical pathway is shown below, along with the alleles that either promote or inhibit each step of the pathway leading to a phenotype. Gene A has alleles A and a, B has alleles B and b, and so forth. Genes B and C are duplicate dominant epistatic lethal as heterozygotes (i.e. Bb Cc are lethal). Genes D and E are duplicate dominant epistatic (i.e. dd eg = desired phenotype). If I were to cross AA Bb cc Dd Ee with aa BB Cc Dd e, (i) (ii) What proportion of all offspring don't show the phenotype? What proportion of offspring survive? Gene A Gene B B Gene D a Gene C Gene Earrow_forwardIn roses, the synthesis of red pigment is produced by two steps in a pathway. gene O magenta intermediate - gene P colorless intermediate- red pigment What would the phenotype be of a plant homozygous for a null mutation of gene P? What would the phenotype be of a plant homozygous for a null mutation of gene Q? What would the phenotype be of a plant homozygous for null mutations of genes P and Q? magenta red Match a genotype to each strain. colorless Strain P locus Q locus homozygous null mutation of gene P homozygous null mutation of gene Q homozygous null mutations of genes P and Q Answer Bank plp PIP What F2 ratio is expected from crossing a plant that is homozygous for a null mutation of gene P with a plant that is homozygous for a null mutation of gene Q? Assume independent assortment. 9 colorless : 4 magenta : 3 red 9 red : 4 colorless : 3 magenta O 9 red : 4 magenta : 3 colorlessarrow_forward
- Multiple crosses were made between true-breeding lines of black and yellow Labrador retrievers. All the F1 progeny were yellow. When these progeny were intercrossed, they produced an F2 consisting of 121 yellow, 9 black and 30 chocolate. What epistatic ratio and what kind of epistasis is approximated in the F2? Propose a biochemical pathway for coat color in Labrador retrievers based on the type of epistasis. Correlate each genotype with the phenotype that would occur in your pathway. Also show the frequency of each genotype. A-B- A-bb aaB- aabbarrow_forwardIn a species of tree, seed color is determined by four independently assorting genes: A, B, C, and D. The recessive alleles of each of these genes (a, b, C, and d) produce abnormal enzymes that cannot catalyze a reaction in the biosynthetic pathway for seed pigment. This pathway is diagrammed as follows: A White precursor Yellow ----Orange----- Red --- Blue When both red and blue pigments are present, the seeds are purple. Trees with the genotypes Aa Bb Cc Dd and Aa Bb Cc dd were crossed. (a) What color are the seeds in these two parental genotypes? (b) What proportion of the offspring from the cross will have white seeds? (c) Determine the relative proportions of red, white, and blue offspring from the cross.arrow_forwardA yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies.These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic,25% adenine-requiring auxotrophsa. Explain the difference between these two types ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Anatomy & Physiology (11th Edition)BiologyISBN:9780134580999Author:Elaine N. Marieb, Katja N. HoehnPublisher:PEARSON
Biology 2eBiologyISBN:9781947172517Author:Matthew Douglas, Jung Choi, Mary Ann ClarkPublisher:OpenStax
Anatomy & PhysiologyBiologyISBN:9781259398629Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa StouterPublisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)BiologyISBN:9780815344322Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter WalterPublisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & PhysiologyBiologyISBN:9781260159363Author:Martin, Terry R., Prentice-craver, CynthiaPublisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)BiologyISBN:9781260231700Author:Sylvia S. Mader, Michael WindelspechtPublisher:McGraw Hill Education
Human Anatomy & Physiology (11th Edition)
Biology
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:PEARSON
Biology 2e
Biology
ISBN:9781947172517
Author:Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher:OpenStax
Anatomy & Physiology
Biology
ISBN:9781259398629
Author:McKinley, Michael P., O'loughlin, Valerie Dean, Bidle, Theresa Stouter
Publisher:Mcgraw Hill Education,
Molecular Biology of the Cell (Sixth Edition)
Biology
ISBN:9780815344322
Author:Bruce Alberts, Alexander D. Johnson, Julian Lewis, David Morgan, Martin Raff, Keith Roberts, Peter Walter
Publisher:W. W. Norton & Company
Laboratory Manual For Human Anatomy & Physiology
Biology
ISBN:9781260159363
Author:Martin, Terry R., Prentice-craver, Cynthia
Publisher:McGraw-Hill Publishing Co.
Inquiry Into Life (16th Edition)
Biology
ISBN:9781260231700
Author:Sylvia S. Mader, Michael Windelspecht
Publisher:McGraw Hill Education
Mitochondrial mutations; Author: Useful Genetics;https://www.youtube.com/watch?v=GvgXe-3RJeU;License: CC-BY