1.E: Fundamental Properties of Genes (Exercises)
- Page ID
- 6918
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Question 1.5. Calculating recombination frequencies
Corn kernels can be colored or white, determined by the alleles C(colored, which is dominant) or c(white, which is recessive) of the coloredgene. Likewise, alleles of the shrunkengene determine whether the kernels are nonshrunken (Sh, dominant) or shrunken (sh, recessive). The geneticist Hutchison crossed a homozygous colored shrunken strain (CC shsh) to a homozygous white nonshrunken strain (cc ShSh) and obtained the heterozygous colored nonshrunken F1. The F1 was backcrossed to a homozygous recessive white shrunken strain (cc shsh). Four phenotypes were observed in the F2 progeny, in the numbers shown below.
Phenotype Number of plants
colored shrunken 21,379
white nonshrunken 21,096
colored nonshrunken 638
white shrunken 672
a) What are the predicted frequencies of these phenotypes if the coloredand shrunkengenes are not linked?
b) Are these genes linked, and if so, what is the recombination frequency between them?
Question 1.6. Constructing a linkage map:
Consider three genes, A, B and C, that are located on the same chromosome. The arrangement of the three genes can be determined by a series of three crosses, each following two of the genes (referred to as two-factor crosses). In each cross, a parental strain that is homozygous for the dominant alleles of the two genes (e.g. AB/AB) is crossed with a strain that is homozygous for the recessive alleles of the two genes (e.g. ab/ab), to yield an F1 that is heterozygous for both of the genes (e.g. AB/ab). In this notation, the slash (/) separates the alleles of genes on one chromosome from those on the homologous chromosome. The F1 (AB/ab) contains one chromosome from each parent. It is then backcrossed to a strain that is homozygous for the recessive alleles (ab/ab) so that the fates of the parental chromosomes can be easily followed. Let's say the resulting progeny in the F2 (second) generation showed the parental phenotypes (AB and ab) 70% of the time. That is, 70% of the progeny showed only the dominant characters (AB) or only the recessive characters (ab), which reflect the haploid genotypes AB/aband ab/ab, respectively, in the F2 progeny. The remaining 30% of the progeny showed recombinant phenotypes (Aband aB) reflecting the genotypes Ab/aband aB/abin the F2 progeny. Similar crosses using F1's from parental AC/ACand ac/acbackcrossed to a homozygous recessive strain (ac/ac) generated recombinant phenotypes Acand aCin 10% of the progeny. And finally, crosses using F1's from parental BC/BCand bc/bcbackcrossed to a homozygous recessive strain (bc/bc) generated recombinant phenotypes Bcand bCin 25% of the progeny.
a. What accounts for the appearance of the recombinant phenotypes in the F2 progeny?
b. Which genes are closer to each other and which ones are further away?
c. What is a linkage map that is consistent with the data given?
Question 1.7
Why are the distances in the previous problem not exactly additive, e.g. why is the distance between the outside markers (A and B) not 35 map units (or 35% recombination)? There are several possible explanations, and this problem explores the effects of multiple crossovers. The basic idea is that the further apart two genes are, the more likely that recombination can occur multiple times between them. Of course, two (or any even number of) crossover events between two genes will restore the parental arrangement, whereas three (or any odd number of) crossover events will give a recombinant arrangement, thereby effectively decreasing the observed number of recombinants in the progeny of a cross.
For the case examined in the previous problem, with genes in the order A___C_______B, let the term abrefer to the frequency of recombination between genes Aand B, and likewise let acrefer to the frequency of recombination between genes Aand C, and cbrefer to the frequency of recombination between genes Cand B.
a) What is the probability that when recombination occurs in the interval between Aand C, an independent recombination event also occurs in the interval between Cand B?
b) What is the probability that when recombination occurs in the interval between Cand B, an independent recombination event also occurs in the interval between Aand C?
c) The two probabilities, or frequencies, in a and b above will effectively lower the actual recombination between the outside markers Aand Bto that observed in the experiment. What is an equation that expresses this relationship, and does it fit the data in problem 3?
d. What is the better estimate for the distance between genes Aand Bin the previous problem?
Question 1.8 Complementation and recombination in microbes.
The State College Bar Association has commissioned you to study an organism, Alcophila latrobus, which thrives on Rolling Rock beer and is ruining the local shipments. You find three mutants that have lost the ability to grow on Rolling Rock (RR).
a) Recombination between the mutants can restore the ability to grow on RR. From the following recombination frequencies, construct a linkage map for mutations 1, 2, and 3.
Recombination between Frequency
1- and 2- 0.100
1- and 3- 0.099
2- and 3- 0.001
b) The following diploid constructions were tested for their ability to grow on RR. What do these data tell you about mutations 1, 2, and 3?
Grow on RR?
1) 1- 2+ / 1+ 2- yes
2) 1- 3+ / 1+ 3- yes
3) 2- 3+ / 2+ 3- no
Question 1.9 Using recombination frequencies and complementation to deduce maps and pathways in phage.
A set of four mutant phage that were unable to grow in a particular bacterial host (lets call it restrictive) were isolated; however, both mutant and wild type phage will grow in another, permissive host. To get information about the genes required for growth on the restrictive host, this host was co-infected with pairs of mutant phage, and the number of phage obtained after infection was measured. The top number for each co-infection gives the total number of phage released (grown on the permissive host) and the bottom number gives the number of wild-type recombinant phage (grown on the restrictive host). The wild-type parental phage gives 1010 phage after infecting either host. The limit of detection is 102 phage.
Phenotypes of phage, problem 1.9:
Assays after co-infection with mutant phage:
Results of assays, problem 1.9:
Number of phage
mutant 1 mutant 2 mutant 3 mutant 4
mutant 1 total <102
recombinants <102
mutant 2 total 1010 <102
recombinants 5x106 <102
mutant 3 total 1010 1010 <102
recombinants 107 5x106 <102
mutant 4 total 105 1010 1010 <102
recombinants 105 5x106 107 <102
a) Which mutants are in the same complementation group? What is the minimum number of genes in the pathway for growth on the restrictive host?
b) Which mutations have the shortest distance between them?
c) Which mutations have the greatest distance between them?
d) Draw a map of the genes in the pathway required for growth on the restrictive host. Show the positions of the genes, the positions of the mutations and the relative distances between them.
Question 1.10
One of the classic experiments in bacterial genetics is the fluctuation analysisof Luria and Delbrück (1943, Mutations of bacteria from virus sensitivity to virus resistance, Genetics 28: 491-511). These authors wanted to determine whether mutations arose spontaneouslywhile bacteria grew in culture, or if the mutations were inducedby the conditions used to select for them. They knew that bacteria resistant to phage infection could be isolated from infected cultures. When a bacterial culture is infected with a lytic phage, initially it “clears” because virtually all the cells are lysed, but after several hours phage-resistant bacteria will start to grow.
Luria and Delbrück realized that the two hypothesis for the source of the mutations could be distinguished by a quantitative analysis of the number of the phage-resistant bacteria found in many infected cultures. The experimental approach is outlined in the figure below. Many cultures of bacteria are grown, then infected with a dose of phage T1 that is sufficient to kill all the cells, except those that have acquired resistance. These resistant bacteria grow into colonies on plates and can be counted.
a. What are the predictions for the distribution of the number of resistant bacteria in the two models? Assume that on average, about 1 in 107 bacteria are resistant to infection by phage T1.
b. What do results like those in the figure and table tell you about which model is correct?
Figure for question 1.10.
The actual results from Luria and Delbrück are summarized in the following table. They examined 87 cultures, each with 0.2 ml of bacteria, for phage resistant colonies.
Number of resistant bacteria | Number of cultures |
0 | 29 |
1 | 17 |
2 | 4 |
3 | 3 |
4 | 3 |
5 | 2 |
6-10 | 5 |
11-20 | 6 |
21-50 | 7 |
51-100 | 5 |
101-200 | 2 |
201-500 | 4 |
501-1000 | 0 |
Contributors and Attributions
Ross C. Hardison, T. Ming Chu Professor of Biochemistry and Molecular Biology (The Pennsylvania State University)