Essentials of Genetics (9th Edition) - Standalone book
9th Edition
ISBN: 9780134047799
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Videos
Textbook Question
Chapter 7, Problem 1PDQ
HOW DO WE KNOW?
In this chapter, we focused on linkage, chromosomal mapping, and many associated phenomena. In the process, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions?
(a) How was it established experimentally that the frequency of recombination (crossing over) between two genes is related to the distance between them along the chromosome?
(b) How do we know that specific genes are linked on a single chromosome, in contrast to being located on separate chromosomes?
(c) How do we know that crossing over results from a physical exchange between chromatids?
(d) How do we know that sister chromatids undergo recombination during mitosis?
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long
bristles. Mutant strains have been isolated with either curled wings or short bristles.
The genes representing these two mutant traits are located on separate
chromosomes. Carefully examine the data from the five crosses below. (a) For each
mutation, determine whether it is dominant or recessive. In each case, identify
which crosses support your answer; and (b) define gene symbols and determine the
genotypes of the parents for each cross.
Cross
1. straight, short X
straight, short
2. straight, long X
straight, long
3. curled, long X
straight, short
4. straight, short X
straight, short
5. curled, short X
straight, short
straight
wings,
long
bristles
30
120
40
40
20
Number of Progeny
straight
curled
wings, wings,
short
long
bristles
bristles
90
10
0
40
120
60
40
40
0
20
curled
wings,
short
bristles
30
0
40
0
60
Alleles of genes A and B were analyzed in Neurospora according to the cross shown below. Ordered tetrads are summarized in each horizontal row with the number of tetrads in each category listed alongside.
A) Analyze the data to determine the recombination frequency (RF) between A and B, along with any additional information that is available from this data.
B) Draw a map of the chromosome or chromosomes with appropriate map distances.
C) Use the Perkins formula to reanalyze any relationship between A and B.
Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL.
A. What would be the physical phenotype of these flies?
B. If PCR was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PCR products?
C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker?
D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?
Chapter 7 Solutions
Essentials of Genetics (9th Edition) - Standalone book
Ch. 7 -
CASE STUDY | Links to autism
As parents of an...Ch. 7 -
CASE STUDY | Links to autism
As parents of an...Ch. 7 - CASE STUDY | Links to autism As parents of an...Ch. 7 -
CASE STUDY | Links to autism
As parents of an...Ch. 7 -
HOW DO WE KNOW?
1. In this chapter, we focused on...Ch. 7 -
CONCEPT QUESTION
2. Review the Chapter Concepts...Ch. 7 - Describe the cytological observation that suggests...Ch. 7 - Why does more crossing over occur between two...Ch. 7 - Why is a 50 percent recovery of single-crossover...Ch. 7 - Why are double-crossover events expected less...
Ch. 7 - What is the proposed basis for positive...Ch. 7 - What three essential criteria must be met in order...Ch. 7 - The genes dumpy wings (dp), clot eyes (cl), and...Ch. 7 - Colored aleurone in the kernels of corn is due to...Ch. 7 - In the cross shown here, involving two linked...Ch. 7 - In a series of two-point map crosses involving...Ch. 7 -
13. Two different female Drosophila were...Ch. 7 -
14. In Drosophila, a cross was made between...Ch. 7 -
15. A cross in Drosophila involved the recessive,...Ch. 7 -
16. Drosophila melanogaster has one pair of sex...Ch. 7 -
17. Drosophila females homozygous for the third...Ch. 7 - In Drosophila, the two mutations Stubble bristles...Ch. 7 -
19. A female of genotype
produces 100 meiotic...Ch. 7 - In a plant, fruit color is either red or yellow,...Ch. 7 - In Drosophila, Dichaete (D) is a mutation on...Ch. 7 - An organism of the genotype AaBbCc was testcrossed...Ch. 7 - Based on our discussion of the potential...Ch. 7 - Prob. 24PDQCh. 7 - DNA markers have greatly enhanced the mapping of...Ch. 7 - Are sister chromatid exchanges effective in...
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
- The wild-type (normal) fruit fly, Drosophila melanogaster, has straight wings and long bristles. Mutant strains have been isolated that have either curled wings or short bristles. The genes representing these two mutant traits are located on separate chromosomes. Carefully examine the data from the following five crosses shown below (running across both columns). (a) Identify each mutation as either dominant or recessive. In each case, indicate which crosses support your answer. (b) Assign gene symbols and, for each cross, determine the genotypes of the parents.arrow_forwardYou are studying four linked genes located on chromosome 2 in the fruit fly Drosophila melanogaster: adp (which contributes to obesity), b (which contributes to body color), pr (which contributes to metabolism), and vg (which contributes to wing formation). A series of crosses between pair-wise combinations of these mutations yielded the following recombination frequencies between the indicated loci: pr and adp 9% adp and b 6% pr and b 3% vg and b 10% vg and pr 7% What is the genetic distance in map units (cM) between the adp and vg loci? 16 9 6 4arrow_forwardPlease label the tetrad type in the table as PD (parental ditype), NPD (non parental ditype) or T (tetratype) and answer the following questions a) Are the genes linked? Please explain SPECIFICALLY how you can distinguish between linked and unlinked genes in this instance. b) If the two genes are linked, calculate the % recombination between ser and thr. Show the formula used, as well as all of your calculations. c) Draw a single map illustrating the arrangement of the two genes on the chromosome with respect to each other and to the centromere of the chromosome. Make sure to map ALL three distancesarrow_forward
- Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forwardIn an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n = 7) moves slowly but eventually forms seven bands, which represent DNA fractions that are of different sizes and hence have moved at different speeds. These bands are presumed to be the seven chromosomes. How would you show which band corresponds to which chromosome?arrow_forwardIs )Given the experimental set up below, and knowing the genes for stalked eyes (st) and black bodies (b) are on the same chromosome and recessive to wild type (un-stalked eyes and light body), answer the questions below the diagram. 1. st st F1 F2 A = 489 B = 56 C= 497 D= 49 A) ) How many recombinant animals are in the F2 generation? B) woH) How far apart, (as both a % recombination and in map units) are the stalk eyed and black bodied genes on this chromosome?arrow_forward
- Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual): 563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…arrow_forwardBaker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3? After crossing the F1 generation of the cross between mutant strains 1…arrow_forwardBaker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other? B. What phenomenon is occurring in the cross between mutant strains 1 and 3?arrow_forward
- Ensure answers are clearly labelled a) & b). a) Consider the following pedigree. The solid symbols represent affected individuals. Which of the following is / are possible genotypes for II-2 with respect to this disease? Please type 1 - 4, and indicate yes or no only for each. II III 1. xAxa 2. xaxa 3. Aa 4. AA2 b) Genes A, B and C are on the same chromosome linked in cis (coupling) conformation. A'is 16 cM from B, and B is 22 cM from C. The distance between A and C is 38 cM. The coefficient of coincidence is 0.55 for a trihybrid test cross. How many individuals with the genotype AabbCc do you expect to see among the offspring of the cross if 1000 offspring are obtained? Please show your calculations and round your answer off to the nearest whole number.arrow_forwardIf you wanted to make a mouse model for any of the following human genetic conditions (a–d), indicate which of thefollowing types of mice (i–vi) would be useful to your studies. If more than one answer applies, state which type ofmouse would most successfully mimic the human disease:(i) transgenic mouse overexpressing a normal mouse protein; (ii) transgenic mouse expressing normal amounts of amutant human protein; (iii) transgenic mouse expressing adominant negative form of a protein; (iv) a knockout mouse;(v) a conditional knockout mouse; and (vi) a knockin mousein which the normal allele is replaced with a mutant allelethat is at least partially functional. In all cases, the transgeneor the gene that is knocked out or knocked in is a form of thegene responsible for the disease in question.a. Marfan syndrome (a dominant disease caused byhaploinsufficiency for the FBN1 gene);b. A dominantly inherited autoinflammatory diseasecaused by a hypermorphic missense mutation in thegene PLCG2;c.…arrow_forwardAssume you are mapping genes A, B, C, and D in Drosophila. You know that these genes are linked on the same chromosome, and you determine the recombination frequencies between each pair of genes to be as follows: A–B, 8%; A– C, 28%; A–D, 25%; B–C, 20%; B–D, 33%. (a) Describe how you determined the recombination frequency for each pair of genes. (b) Draw a chromosome map based on your data.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Human Biology (MindTap Course List)BiologyISBN:9781305112100Author:Cecie Starr, Beverly McMillanPublisher:Cengage Learning
Human Biology (MindTap Course List)
Biology
ISBN:9781305112100
Author:Cecie Starr, Beverly McMillan
Publisher:Cengage Learning
What are Mutations and what are the different types of Mutations?; Author: Science ABC;https://www.youtube.com/watch?v=I16YlE8qTBU;License: Standard youtube license