Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 6, Problem 33P
From a cross between e f g and e f g strains of Neurospora, recombination between these linked genes resulted in a few octads containing the following ordered set of spores:
e+ f+ g+
e+ f+ g+
e+ f- g+
e+ f- g+
e- f- g-
e- f- g-
e- f- g-
e- f- g-
| a. Where was recombination initiated? |
| b. Did crossing-over occur between genes e and g? Explain. |
| c. Why do you end up with 2 f+:6 f− but 4 e+: 4 e− and 4g+: 4g−? |
| d. Could you characterize these unusual octads as MI or MII for any of the three genes involved? Explain. |
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For each genotype in the table below, determine whether or
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ma
Chromosome
F' lac (plasmid)
- Inducer
+ Inducer
----
---
---
I*O*Z
I*O°Z+
I*O*Z*
I* = wildtype repressor
|- - no functional repressor produced
O* - wildtype operator
OC = operator mutation prevents repressor from binding
%3D
An Hfrstrain that is a *b*c*d* e*f* g *h* is mated with an F strain that is a b e d
e f gh. The mating is interrupted at 5 minutes interval, and the genotypes of the F
recombinants are determined. The results obtained are tabulated in Table 2. Draw the map of
the Hfrchromosome and indicate the position of the origin of transfer, the direction of the
transfer and the minutes between genes.
Table 2:Entry time of Hfr chromosome into recipient cell.
Time
a
d
e
f
h
5
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
85
+ +
+ + + + +
+
+ + +
+ + +
+ + + +
+ + +
+ + +
+ + + +
+ + + + +
An Hfr strain that is hisE + and pheA + was conjugated to a strain that is hisE − and pheA −. The conjugation was interrupted at different times, and the percentage of recombinants for each gene was determined by streaking on media that lacked either histidine or phenylalanine. The following results were obtained:
A. Determine the map distance (in minutes) between these twogenes.B. In a previous experiment, it was found that hisE is 4 minutesaway from pabB and that PheA is 17 minutes from pabB. Drawa genetic map showing the locations of all three genes.
Chapter 6 Solutions
Genetics: From Genes to Genomes
Ch. 6 - Griffith, in his 1928 experiments, demonstrated...Ch. 6 - Griffith, in his 1928 experiments, demonstrated...Ch. 6 - During bacterial transformation, DNA that enters a...Ch. 6 - Nitrogen and carbon are more abundant in proteins...Ch. 6 - If 30 of the bases in human DNA are A, a what...Ch. 6 - Which of the following statements are true about...Ch. 6 - Imagine you have three test tubes containing...Ch. 6 - What information about the structure of DNA was...Ch. 6 - A portion of one DNA strand of the human gene...Ch. 6 - When a double-stranded DNA molecule is exposed to...
Ch. 6 - A particular virus with DNA as its genetic...Ch. 6 - The underlying structure of DNA is very simple,...Ch. 6 - Prob. 13PCh. 6 - Bacterial transformation and bacteriophage...Ch. 6 - The CAP protein is shown bound to DNA in Fig....Ch. 6 - In Meselson and Stahls density shift experiments...Ch. 6 - When Meselson and Stahl grew E. coli in 15N medium...Ch. 6 - If you expose human tissue culture cells for...Ch. 6 - Draw a replication bubble with both replication...Ch. 6 - a. Do any strands of nucleic acid exist in nature...Ch. 6 - As Fig. 6.21 shows, DNA polymerase cleaves the...Ch. 6 - The bases of one of the strands of DNA in a region...Ch. 6 - Replicating structures in DNA can be observed in...Ch. 6 - Indicate the role of each of the following in DNA...Ch. 6 - Draw a diagram of replication that is occurring at...Ch. 6 - Figure 6.18 depicts Watson and Cricks initial...Ch. 6 - Researchers have discovered that during...Ch. 6 - A DNA synthesizer is a machine that uses automated...Ch. 6 - Bacterial cells were coinfected with two types of...Ch. 6 - A yeast strain with a mutant spo11- allele has...Ch. 6 - Imagine that you have done a cross between two...Ch. 6 - The Neurospora octad shown came from a cross...Ch. 6 - From a cross between e f g and e f g strains of...Ch. 6 - In Step 6 of Fig. 6.27, the resolvase enzyme...Ch. 6 - Figure 6.31shows four potential outcomes of...Ch. 6 - Each of the substrates for site-specific...Ch. 6 - Prob. 37PCh. 6 - Suppose that you could inject a wild-type mouse...Ch. 6 - C31 is a type of bacteriophage that infects...Ch. 6 - Cre is a recombinase enzyme encoded by a gene in...Ch. 6 - Like Cre/loxP recombination, site-specific...
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- An Hfr strain that is leuA+ and thiL+ was mixed with a strain thatis leuA− and thiL−. In the data points shown in the following graph,the conjugation was interrupted at different time points, and thepercentage of recombinants for each gene was determined bystreaking on a medium that lacked either leucine or thiamine.What is the map distance (in minutes) between these two genes?arrow_forwardWhen the interrupted mating technique was used withfive different strains of Hfr bacteria, the following orders ofgene entry and recombination were observed. On the basisof these data, draw a map of the bacterial chromosome.Do the data support the concept of circularity?HfrStrain Order1 T C H R O2 H R O M B3 M O R H C4 M B A K T5 C T K A Barrow_forwardTwo strains of S. cerevisae (yeast) are crossed. One has the genotype A B and the other a b. Which statement correctly describes the tetrads that can be produced by one reciprocal exchange between the A and B genes without gene conversion in the dihybrid? A) Recombination occurs at the four-strand stage to produce a tetrad with A B, A B, a B, and a b chromosomes. B) Recombination occurs at the four-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes. C) Recombination occurs at the two-strand stage to produce a tetrad with A B, A b, a B, and a b chromosomes. D) Recombination occurs at the two-strand stage to produce a tetrad with A b, A b, a B, and a B chromosomes.arrow_forward
- A 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_forwardA particular Hfr strain normally transmits the pro+marker as the last one in conjugation. In a cross of thisstrain with an F− strain, some pro+ recombinants are recovered early in the mating process. When these pro+cells are mixed with F−cells, the majority of the F− cellsare converted into pro+ cells that also carry the F factor.Explain these results.arrow_forwardA series of Hfr strains that have genotype m+ n+ o+ p+ q+ r+ are mixed with an F− strain that has genotype m− n− o− p− q− r−. Conjugation is interrupted at regular intervals and the order of the appearance of genes from the Hfr strain is determined in the recipient cells. The order of gene transfer for each Hfr strain is What is the order of genes on the circular bacterial chromosome? For each Hfr strain, give the location of the F factor in the chromosome and its polarity. Hfr5 m+ q+ p+ n+ r+ o+ Hfr4 n+ r+ o+ m+ q+ p+ Hfr1 o+ m+ q+ p+ n+ r+ Hfr9 q+ m+ o+ r+ n+ p+arrow_forward
- An Hfr strain that is leuA+ and thiL+ was mixed with a strain thatis leuA− and thiL−. In the data points shown in the following graph,the conjugation was interrupted at different time points, and thepercentage of recombinants for each gene was determined bystreaking on a medium that lacked either leucine or thiamine.Make a calculation.arrow_forwardYou've crossed two strains: (Hfr) pro+ lac+ arg+ with (F-) pro- lac- arg- . Interrupted mating studies indicated that arg+ enters the cell last, so you want to select for arg+ recombinants in order to map the genes by recombination. Which type of plates will you use for your initial selection? Containing proline, lactose, and arginine. Containing proline, glucose, and arginine. Containing arginine and glucose. Containing lactose and proline. Containing glucose and proline. Minimal media with nothing added.arrow_forwardThe following recombinants are recovered when conjugation occurs between an a*d*g+ donor and an adg recipient. at d+ g+ = 84% a d g+ = 6% at d g+ = 10% a dt g+ = less than 1% What is the map distance between the a and d genes? 10 map units 74 map units less than 1 map unit 84 map units 6 map unitsarrow_forward
- The 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_forwardBelow are two pictures of perithecium on microscope slides that have been squished under cover slips so that the asci inside have spread out. In the asci, we can clearly see the tan ascospores (light colored) and wildtype ascospores (dark colored). We will perform our tetrad analysis by counting how many of the asci are non-recombinant and how many of the acsi are recombinant. We can tell which are recombinant and non-recombinant by observing the patterns of the ascospores in the asci. See the patterns of ascospores for non- recombinant and recombinant asci in the table below. Ascospore pattern of non-recombinant asci OOOO OOOOO You will count the total number of non-recombinant and recombinant asci in the two pictures below. You will only count the asci that have a star next to them because these asci can be clearly identified as non-recombinant or recombinant. Select the correct number below after you are done. Total number of non-recombinant asci [Select] Ascospore pattern of…arrow_forwardDNA obtalned from the indicated donor strains is used to TRANSFORM the indicated recipient strains. The resulting progeny are plated on minimal medium so that only wild- type recombinants are scored. The number of wild-types for each cross is given in the chart below. What is the order of the genes? Donor Recipient wid type colonies a-b- c+ a+ b+ c- 273 a- b+ c- a+ b-c+ 462 a- b+ C+ a+ b-c- 2 a b- C+ a- b+ c+ O 4-b-c a-o-b b-a-c cannot be determined Question 18 of 26arrow_forward
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genetic recombination strategies of bacteria CONJUGATION, TRANSDUCTION AND TRANSFORMATION; Author: Scientist Cindy;https://www.youtube.com/watch?v=_Va8FZJEl9A;License: Standard youtube license