Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 7, Problem 6EQ
In your laboratory, you have an
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In your laboratory, you have an F− strain of E. coli that is resistantto streptomycin and is unable to metabolize lactose, but it can metabolizeglucose. Therefore, this strain can grow on a medium thatcontains glucose and streptomycin, but it cannot grow on a mediumcontaining lactose. A researcher has sent you two E. colistrains in two separate tubes. One strain, let’s call it strain A, hasan F′ factor (an F prime factor) that carries the genes that are requiredfor lactose metabolism. On its chromosome, it also has thegenes that are required for glucose metabolism. However, it is sensitiveto streptomycin. This strain can grow on a medium containinglactose or glucose, but it cannot grow if streptomycin is addedto the medium. The second strain, let’s call it strain B, is an F−strain. On its chromosome, it has the genes that are required forlactose and glucose metabolism. Strain B is also sensitive to streptomycin.Unfortunately, when strains A and B were sent to you, thelabels had fallen…
In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?
BACTERIA STRAIN A IS AUXOTROPHIC FOR METHIONINE AND STRAIN B IS AUXOTROPHIC FOR LEUCINE.
A. WILL THESE STRAIN GROW ON MINIMAL MEDIUM (A MEDIUM THAT DOES NOT CONTAIN EITHER METHIONINE OR LEUCINE)?
B. WHAT TYPE OF GENETIC RECOMBINATION MUST OCCUR TO RECOVER A PROTOTROPH?
Chapter 7 Solutions
Genetics: Analysis and Principles
Ch. 7.1 - 1. A form of genetic transfer that involves the...Ch. 7.2 - 1. A bacterial cell with an F factor conjugates...Ch. 7.2 - 2. Which of the following is a type of plasmid?...Ch. 7.3 - 1. With regard to conjugation, a key difference...Ch. 7.3 - 2. In mapping experiments, ______ strains are...Ch. 7.4 - Prob. 1COMQCh. 7.4 - Cotransduction may be used to map bacterial genes...Ch. 7.5 - Prob. 1COMQCh. 7.5 - Prob. 2COMQCh. 7.6 - 1. Which of the following is an example of...
Ch. 7 - 1. The terms conjugation, transduction, and...Ch. 7 - Prob. 2CONQCh. 7 - If you mix together an equal number of F+ and F...Ch. 7 - What is the difference between an F+ and an Hfr...Ch. 7 - 5. What is the role of the origin of transfer...Ch. 7 - 6. What is the role of sex pili during...Ch. 7 - Prob. 7CONQCh. 7 - Prob. 8CONQCh. 7 - Prob. 9CONQCh. 7 - 10. What is cotransduction? What determines the...Ch. 7 - Prob. 11CONQCh. 7 - Prob. 12CONQCh. 7 - Describe the steps that occur during bacterial...Ch. 7 - Prob. 14CONQCh. 7 - Prob. 15CONQCh. 7 - Antibiotics such as tetracycline, streptomycin,...Ch. 7 - Prob. 1EQCh. 7 - 2. In the experiment of Figure 7.1, Lederberg and...Ch. 7 - Explain how a U-tube apparatus can distinguish...Ch. 7 - Prob. 4EQCh. 7 - 5. In a conjugation experiment, what is meant by...Ch. 7 - In your laboratory, you have an F strain of E....Ch. 7 - 7. As mentioned in question 2 of More Genetic...Ch. 7 - An Hfr strain that is hisE+ and pheA+ was mixed...Ch. 7 - Acridine orange is a chemical that inhibits the...Ch. 7 - Prob. 10EQCh. 7 - Prob. 11EQCh. 7 - Lets suppose a new strain of P1 phage has been...Ch. 7 - If two bacterial genes are 0.6 minute apart on the...Ch. 7 - 14. In a cotransduction experiment involving P1,...Ch. 7 - Prob. 15EQCh. 7 - Prob. 16EQCh. 7 - 1. Discuss the advantages of the genetic analysis...Ch. 7 - Prob. 2QSDC
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- So if you see growth on a histidine-deficient medium in a yeast 1 hybrid and you assume that it is a false positive. What could have caused this? The yeast cells are normal and this wasn't caused by other molecules.arrow_forwardIn a transformation experiment, donor DNA was obtained from aprototroph bacterial strain (a+b+c+), and the recipient was a tripleauxotroph (a-b-c-). What general conclusions can you draw aboutthe linkage relationships among the three genes from the followingtransformant classes that were recovered? a+ b- c- 180 a- b+ c- 150 a+ b+ c- 210 a- b- c+ 179 a+ b- c+ 2 a- b+ c+ 1 a+ b+ c+ 3arrow_forwardType S Streptococcus pneumoniae bacterium is lethal and will kill its host. If heat inactivated the S strain dies and becomes nonlethal. Type R Streptococcus pneumoniae is a nonvirulent strain of bacteria. What would occur if one were to inject both the R strain and heat-killed S strains into a host organism such as the mouse? The R strain would be transformed into the virulent S strain and kill the host. Neither the S nor the R strain would change. The R strain would be transformed into the virulent S strain and not affect the host. The S strain would be transformed into the nonvirulent R strain and not affect the host The S strain would be transformed into the nonvirulent R strain and kill the host.arrow_forward
- An F' merozygote is formed as a result of the following events: 1) the recipient cell receives genes which used to be a part of an Hfr bacterium chromosome 2) an F' bacterium excuses an F factor from another bacterium chromosome 3) an F factor that was imprecisely excised from Hfr chromosome was transferred to an F- bacterium 4)an F' bacterium conjugate with an F- bacterium 5) an F' bacterium incorporates an F factor into its chromosome 6) an F- bacterium incorporates an F factor into its chromosomearrow_forwardMatch the following Conjugation terms with the most appropriate description in the image: F- recombinant F- strain F+ strain HFR strainarrow_forwardIn Hershey-Chase experiment, bacteriophages protein coats were tagged with radioactive isotope S-32. These phages were used to infect E. coli cells and the cells were further centrifuged to form pellets. Why was the radioactivity level of S-32 found greater outside the cells compared to the E. coli cell pellets? Explain briefly. If the experiment is repeated in the same manner but this time the phage protein coats are labelled with isotope X and the phage DNA with isotope Y, which isotope’s radioactivity will be found in greater amounts in the E. coli cell pellets after centrifugation? Explain briefly.arrow_forward
- Recombinant protein production by a genetically modified Escherichia coli strain is proportional to cell growth. Ammonia is used as a nitrogen source for aerobic glucose respiration. The recombinant protein has the general formula CH1,55O0,31N0,25, while that of the cellular biomass is CH1,77O0,49N0,24. The biomass yield from glucose equals 0.50 g/g, while the recombinant protein yield from glucose corresponds to 20% of the cell yield from substrate.a) How much ammonia is required? What is the oxygen demand? (b) If the biomass yield remains the same, what are the ammonia and oxygen requirements for a wild-type strain of E. coli, with cell biomass of the same elemental composition, but unable to synthesize the recombinant protein? (c) On an industrial scale, cultivation takes place in a continuous fermenter at 28°C and the desired recombinant protein production rate is 7 g/h. Since the viscosity of the culture broth is considerable, the energy input due to agitation cannot be neglected.…arrow_forwardWhy is the observed mutation rate of E. coli 10–8 to 10–10 per base pair replicated, even though the error rates of Pol I and Pol III are 10–6 to 10–7 per base pair replicated?arrow_forwardProduction of a recombinant protein by E. coli is proportional to cell growth. Ammonia is used as a source of N and glucose as a source of C, under aerobic conditions. The recombinant protein has a general formula CH1.55N0.31O0.25 and the cell CH1.77N0.4900.24. The yield of biomass from glucose is 0.48 gcel/gglic, and the yield of recombinant protein from glucose is about 20% of that for biomass.a) How much ammonia is needed to produce 50 g of cells producing the recombinant protein?b) What is the oxygen demand in this process?c) For the cultivation of a wild E. coli not producing the recombinant protein, how different would the ammonia and oxygen demand be if the biomass yield remained at 0.48 gcel/gglyc ?arrow_forward
- In the name E. coli 4X5B, which part is called the serotype, serovar, or strain? E coli 4X5Barrow_forwardA bacterial transformation is performed with a donor strain that is resistant to four drugs, A, B, C, and D, and a recipient strain that is sensitive to all four drugs. The resulting recipient cell population is divided and plated onmedia containing various combinations of the drugs. The following table shows the results.a. One of the genes is distant from the other three, which appear to be closely linked. Which is the distant gene? b. What is the likely order of the three closely linked genesarrow_forwardIn five Hfr strains, each of which was used to build a time-of-transfer map, the genes entered the recipient cells as follows: Strain 1: S L A C T F Strain 2: N P F T C A Strain 3: T F P N U Y Strain 4: S H Y U N P Strain 5: U N P F T C Which of the following represents a correct gene map of these results? N P F T S L A C H U Y S L A C T F P N H Y U C T F P N U Y H S L A T C A L S P N U Y H F U N P C A L S F T H Yarrow_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