Biology: The Dynamic Science (MindTap Course List)
4th Edition
ISBN: 9781305389892
Author: Peter J. Russell, Paul E. Hertz, Beverly McMillan
Publisher: Cengage Learning
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Chapter 13, Problem 13TYK
Summary Introduction
To review:
The construction of a linkage map to determine the order of six linked genes, which include a, b, c, d, e, and f, to determine the map units between them.
Introduction:
A linkage map (also called the genetic map) is a chromosomal map that indicates the genes of species and variously related markers. The recombination frequency is necessary to construct a linkage map. In the condition when two or more genes are linked together, the chances of crossover increases, producing recombination. A linkage map is used to determine the position of two genes on the same chromosome. It is not used to identify the physical distance between the genes.
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Proteins A,B,C, and D in the diagram are encoded by different genes and interact
with each other. Imagine that a mutation in the gene for protein A changes one of the
charged amino acid in the red circle area from positive to negative charge (blue
arrow). this mutation results in a mutant phenotype. Assume a mutation in the gene
for protein B occurs and the double mutants have a phenotype that is almost wild
type. How would you best describe the mutation in gene B?
Protein C
Proten A
Protein B
Protein B
Wild type
Mutation in the gene for protein A
In this gene order question, would the order be a/b -- kf -- nm since genes with shorter map unit distances are closely linked?
In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele t) was crossed with a strain that isauxotrophic for methionine (mutant allele m). Linear asci were isolated and classified into the following groups: a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right).
Chapter 13 Solutions
Biology: The Dynamic Science (MindTap Course List)
Ch. 13.1 - You want to determine whether genes a and b are...Ch. 13.2 - You have a true-breeding strain of...Ch. 13.3 - What mechanisms are responsible for: (a)...Ch. 13.4 - A man has Simpson syndrome, an addiction to a...Ch. 13.4 - Prob. 2SBCh. 13.5 - Prob. 1SBCh. 13 - In humans, redgreen color blindness is an X-linked...Ch. 13 - The following pedigree shows the pattern of...Ch. 13 - Individuals affected by a condition known as...Ch. 13 - A number of genes carried on the same chromosome...
Ch. 13 - Prob. 5TYKCh. 13 - Discuss Concepts Can a linkage map be made for a...Ch. 13 - In Drosophila, two genes, one for body color and...Ch. 13 - Another gene in Drosophila determines wing length....Ch. 13 - Prob. 9TYKCh. 13 - You conduct a cross in Drosophila that produces...Ch. 13 - Discuss Concepts Crossing-over does not occur...Ch. 13 - Prob. 12TYKCh. 13 - Prob. 13TYKCh. 13 - Prob. 14TYKCh. 13 - Prob. 1ITDCh. 13 - Prob. 2ITDCh. 13 - Prob. 3ITDCh. 13 - Prob. 4ITD
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- Draw all possible linkage maps for these three genes.arrow_forwardA pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and Barrow_forwardConsider two maize plants:a. Genotype C/c m ; Ac/Ac+, where cm is an unstable allele caused by a Ds insertionb. Genotype C/c m, where cm is an unstable allele caused by Ac insertionWhat phenotypes would be produced and in what proportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a is crossed with the plant in part b? Assume that Ac and c are unlinked, that the chromosome-breakage frequency is negligible, and that mutant c /C is Ac+.arrow_forward
- Construct a map for the genes d,e,f. Assume that: d and e = 3%; e and f = 5%. Give 2arrangements of the genes/maps.arrow_forwardIn a three-point mapping experiment for the genes a-w-ec, the following percentages of events are observed: NCO events: 57%; SCO events between a and w: 18%; SCO events between w and ec: 22%; DCO events: 2% Assuming that w is in the middle, what is the map distance between w and ec? What is the map distance between a and w? Draw the map.arrow_forwardThe HbβS(sickle-cell) allele of the human β-globingene changes the sixth amino acid in the β-globinchain from glutamic acid to valine. In HbβC, the sixthamino acid in β-globin is changed from glutamic acidto lysine. What would be the order of these two mutations within the map of the β-globin gene?arrow_forward
- Considering Figure 2-13, if you had a homozygous double mutant m3/m3 m5/m5, would you expect it to be mutant in phenotype? (Note: This line would have two mutant sites in the same coding sequence.)arrow_forwardYou have the following DNA coding sequence of a wild-type allele: 5’-ATG TTC CAG CTA GAT GAT ATG CTG GTA ATT GGG GAA CGC GCG CGG TAA-3’ For each of the following mutations: A. State whether the mutation is missense, nonsense, frameshift, or silent. B. Write the codon change that occurs for the missense, nonsense, and silent mutations (ex. GAA -- GAT). C. For frameshift mutations, write out the entire mutant sequence with each codon clearly indicated (if the frameshift creates a new stop codon, end the sequence at the new stop). Using the wild type DNA sequence above as a guide : Write the amino acid sequence of the mutants. Mutant 1: transition at nucleotide 23 Mutant 2: T --> G transversion at nucleotide 29 Mutant 3: an insertion of “A” after nucleotide 14 Mutant 4: transition at nucleotide 7 Mutant 5: An insertion of GG after nucleotide 40 Mutant 6: transition at nucleotide 15 Mutant 7: a deletion of nucleotide 25arrow_forwardA panel of cell lines was created by human-mouse somatic-cell hybridization. Each cell line was examined for the presence of human chromosomes and for the production of three enzymes. The following results were obtained: Human chromosomes Enzyme 1 2 3 Cell line 4 8 9 12 15 16 17 22 X A D - + + + + On the basis of these results, give the chromosomal locations of the genes encoding enzyme 1, enzyme 2, and enzyme 3.arrow_forward
- Calculate the map distances among the genes.arrow_forwardA yeast strain with a mutant spo11- allele has been isolated. The mutant allele is nonfunctional; it makes no spo11 protein. What do you suppose is the phenotype of this mutant strain?arrow_forwardColchicine is a chemical mutagen that inhibits the spindle formation and prevents anaphase, which retains the cell’s single restitution nucleus (doubled chromosome number). Suppose that an onion (2n=16) is subjected to three consecutive rounds of colchicine treatment, what will be the resulting chromosome number of the treated onion?arrow_forward
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