Question 1

Fill in the blank. Another name for a genetic screen is a ______ screen.

Accepted Answer

mutant

Question 2

Tryptophan is an essential amino acid that Neurospora crassa needs in order to be able to grow. The tryptophan biosynthetic pathway has three main steps by which the amino acid tryptophan is produced from chorismate. In an experiment, three N. crassa strains that were suspected of having defects in tryptophan synthesis were cultured on minimal medium (without tryptophan) and on minimum medium (plus tryptophan), with the following results. Which strain is defective in the tryptophan biosynthetic pathway?
$$\begin{array} { | l | c | c | } 
\hline & \text { Minimal medium } & \begin{array} { c } 
\text { Minimal medium } \
+ \text { tryptophan }
\end{array} \
\hline \text { Strain 1 } & \text { no growth } & \text { growth } \
\hline \text { Strain 2 } & \text { no growth } & \text { no growth } \
\hline \text { Strain 3 } & \text { growth } & \text { growth } \
\hline
\end{array}$$

Accepted Answer

Strain 1 is defective in the tryptophan biosynthetic pathway because it cannot grow on minimal medium without tryptophan, but it can grow on minimal medium with tryptophan. This indicates that Strain 1 is unable to synthesize tryptophan on its own and must obtain it from the medium.

Question 3

Arginine is synthesized in a pathway involving several enzymes and biosynthetic intermediates. Five Neurospora crassa strains were cultured on minimal media containing various additives, and the results are shown. Match each N. crassa strain with its description.

$$\begin{array} { | c | c | c | c | c | } 
\hline & \text { glutamate } & \text { ornithine } & \text { citrulline } & \text { arginine } \
\hline \text { Strain A } & \text { growth } & \text { growth } & \text { growth } & \text { growth } \
\hline \text { Strain B } & \text { no growth } & \text { no growth } & \text { growth } & \text { growth } \
\hline \text { Strain C } & \text { no growth } & \text { no growth } & \text { no growth } & \text { no growth } \
\hline \text { Strain D } & \text { no growth } & \text { no growth } & \text { no growth } & \text { growth } \
\hline \text { Strain E } & \text { no growth } & \text { growth } & \text { growth } & \text { growth } \
\hline
\end{array}$$

Accepted Answer

Answer: Answers are shown in the following table. Strain C is defective in a different biosynthetic pathway, because supplementation with arginine does not rescue the growth phenotype. Note that strain A is also technically not defective in the arginine pathway, but shows normal growth so this is wild-type with respect to the conditions under consideration here. The remaining strains can grow if they are provided with a supplement that occurs after the enzyme defect in the pathway. Thus, growth in strain E with ornithine, citrulline, and arginine but not with glutamate indicates that the defect is in enzyme 1.
 $$\begin{array} { | c | c | c | c | c | c | } 
\hline & \text { glutamate } & \text { ornithine } & \text { citrulline } & \text { arginine } & \
\hline \text { Strain A } & \text { growth } & \text { growth } & \text { growth } & \text { growth } & \text { Wild-type } \
\hline \text { Strain B } & \text { no growth } & \text { no growth } & \text { growth } & \text { growth } & \text { defective in enzyme 2 } \
\hline \text { Strain C } & \text { no growth } & \text { no growth } & \text { no growth } & \text { no growth } & \begin{array} { c } 
\text { not defective in the } \
\text { arginine biosynthesis } \
\text { pathway }
\end{array} \
\hline \text { Strain D } & \text { no growth } & \text { no growth } & \text { no growth } & \text { growth } & \text { delective in enzyme 3 } \
\hline \text { Strain E } & \text { no growth } & \text { growth } & \text { growth } & \text { growth } & \text { defective in enzyme 1 } \
\hline
\end{array}$$

Question 4

Forward genetics involves disruption of a known gene to see what phenotypes will be produced. Reverse genetics involves introducing random mutations and examining the resultant phenotypes to discover the causative gene.

Accepted Answer

In forward genetics, unbiased mutagenesis of a genome produces a range of mutants with different phenotypes. The mutant with the phenotype of interest is then examined to discover the responsible gene. In reverse genetics, a known gene is disrupted to determine its phenotype.

Question 5

Which of the following features of genetic screens was not key to the success of Beadle and Tatum's early Neurospora crassa experiments with nutritional auxotrophs?

Accepted Answer

Targeted insertional mutagenesis was not a key feature of Beadle and Tatum's experiments. In fact, they used X-rays to induce random mutations in the Neurospora crassa genome. The other three choices (easily-screened phenotype, efficient screening procedure, and appropriate model organism) were essential to the success of their experiments.

Question 6

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity.
-IPTG

Accepted Answer

IPTG binds to the lac repressor, preventing it from binding to the operator region of the lac operon. This allows for constitutive expression of lac genes regardless of lactose presence.

Question 7

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity.
-allolactose

Accepted Answer

Allolactose derepresses the lac operon when lactose is present in the cell, by binding to the lac repressor and preventing it from binding to the operator.

Answer: A
 The lac repressor binds to the operator region of the lac operon in the absence of lactose, preventing RNA polymerase from binding and transcribing the genes involved in lactose metabolism. Allolactose binds to the repressor, causing it to release from the operator and allowing transcription to occur.

Answer: D
 Beta-galactosidase is the enzyme involved in breaking down lactose into glucose and galactose. It is encoded by the lacZ gene in the lac operon.

Answer: C
 Beta-galactosidase catalyzes the hydrolysis of lactose into glucose and galactose, but it also cleaves a phosphate group from the substrate ONPG (ortho-nitrophenyl beta-D-galactopyranoside), producing a blue color. This activity is frequently used as a reporter gene to assay for the activity of the lac operon.

Question 8

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity.
-X-gal

Accepted Answer

X-gal is a substrate that is cleaved by the beta-galactosidase enzyme, which is produced when the lac operon is turned on. This results in a blue color, allowing researchers to easily visualize when the lac operon is active.

Answer: A and B are incorrect and D is not a molecule.

Question 9

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity.
-β-galactosidase

Accepted Answer

β-galactosidase breaks down lactose into its components, glucose and galactose.

Question 10

Fill in the blank. In the lac operon, cAMP-CAP is produced and binds to the CAP binding site when ____ is low

Accepted Answer

The answer of Fill in the blank. In the lac...

Question 11

Which of the following is true of dominant loss-of-function mutations?

Accepted Answer

Dominant loss-of-function mutations produce a mutant phenotype even if a wild-type copy of the gene is present. This is because the mutant protein interferes with the function of the wild-type protein. The mutant allele is present in either the homozygous or heterozygous state.

Question 12

During investigation of the lac operon, Jacob and Monod found that mutations in the same gene did not always lead to the same phenotypic outcome. For example, some mutations in the lacI gene led to constitutive lactose breakdown, whereas other mutations led to an inability to ferment lactose under any conditions. Which of the following statements are true about lacI^s mutations? Please select all that apply.

Accepted Answer

Normally, lacI encodes the lac repressor protein. LacI binding at the operator represses the lac operon. Allolactose binding to LacI releases LacI from the operator and the lac operon is de-repressed. However, lacI^smutants encode a 'super-repressor' protein that cannot bind allolactose and is therefore always bound to DNA. This means that the lac operon is always repressed, even in the presence of lactose, and so lacI^s mutants cannot utilize lactose. The operator sequence is not changed in lacI^s mutants, though it would be possible for a mutation in the operator sequence to have a similar effect as a lacI^s mutation. Introduction of a functional operon on a plasmid (i.e., complementation in trans) is not effective at rescuing this phenotype. This is because the LacI^s protein never releases from the operator protein, so a 'good' copy of LacI cannot compensate for the 'bad' copy by replacing it at the operator.

Question 13

Nüsslein-Volhard and Wieschaus used Drosophila melanogaster as a model organism in genetic screens to examine developmental biology. Why was Drosophila a good model organism for this study?
Please select all that apply.

Accepted Answer

Drosophila was a well-established model system about which much was known. This included well-established embryology that allowed the segmentation pattern of embyros to act as the screening phenotype. Drosophila has a short generation time, meaning that many embryos can be generated and studied efficiently. Balancer chromosomes are available in Drosophila, which facilitates the maintenance of lethal mutations in a heterozygote. Haploids in Drosophila cannot survive, so this is not a feature of this model system. In addition, although experiments in Drosophila that use eye color as a marker are common, this was not a factor in the developmental screens here.

Question 14

A diagram of the mutagenesis and crossing strategy used by Nüsslein-Volhard and Wieschaus to generate and maintain segmentation mutants is shown. Male flies were exposed to EMS (a mutagen) and mated to a non-mutagenized female. How did the researchers ensure that the offspring of Cross B (i.e., the offspring seen in generation C in the diagram) were heterozygotes that carried a mutagenized chromosome? Please select all that apply.

Accepted Answer

CyO homozygotes die, so offspring in generation C with curly wings must be heterozygotes. This means that the offspring could have either the mutagenized chromosome (with markers cn, bw, sp here) or the other parental chromosome (with markers b, pr, can, sca, DTS-91, here). To ensure that the only surviving flies were those carrying a mutagenized chromosome, the researchers used a temperature sensitive allele on the non-mutagenized chromosome (DTS-91 here). The researchers raised the temperature, and flies with the DTS-91 allele died. This means that the only surviving flies were heterozygous flies with CyO and the mutagenized chromosome. Although it is true that homozygous flies with a severe segmentation mutation would likely die, at this stage the researchers did not know which of their mutations were lethal or which were segmentation mutants - in fact, the majority of mutants would not have an aberrant segmentation phenotype. In any case, no offspring of cross B would be homozygous for the mutagenized chromosome. Here, male flies were mutagenized, but this did not have an impact on selection of heterozygotes in generation

Question 15

Match the missing words in the following phrase. In the Nüsslein-Volhard and Wieschaus experiment, embryos lacking in a contiguous region across segments had mutations in [A] genes. Mutants lacking every other segment had defects in [B] genes. Mutants that had alterations in specific regions of the embryo were defective in [C] genes. Finally, mutants that exchanged the identity of a region for that of another region were defective in [D] genes.

Accepted Answer

The answer of Match the missing words in the following...

Question 16

Fill in the blank. In genetic screens, the chance of discovering a novel gene decreases as the screen proceeds. When the chance of finding a new gene is very low and the effort expended to discover a new gene is not justifiable, the screen has been _________.

Accepted Answer

The answer of Fill in the blank. In genetic screens,...

Question 17

Which of the following is an advantage of a genome-wide mutagenesis screen compared to a random mutagenesis screen?

Accepted Answer

In a genome-wide mutagenesis screen, each gene can be targeted, allowing for a comprehensive analysis of gene function. This is not possible in a random mutagenesis screen, where there is no targeted approach. Option A is incorrect because both types of screens can be unbiased if designed properly. Option C is incorrect because phenotypic analysis is still needed in a genome-wide screen to determine the effects of mutations on gene function. Option D is incorrect because random mutagenesis does not specifically identify important regulatory regions.

Question 18

Hartwell and colleagues used the yeast Saccharomyces cerevisiae to examine the cell cycle. However, most cell cycle mutants are lethal. How did the researchers get around this problem in their experiments?

Accepted Answer

The researchers used conditional mutants, which are only affected by the mutation under specific conditions or in the presence of specific molecules. This allowed them to study the effects of the mutant gene without it being lethal to the cells.

Question 19

Fill in the blank. Temperature-sensitive mutations exhibit the mutation when grown at the restrictive temperature. They exhibit a wild-type phenotype when grown at a _____ temperature.

Accepted Answer

The answer of Fill in the blank. Temperature-sensitive mutations exhibit...

Question 20

Match the missing words in the following phrase. Experiments showed that cell cycle mutants stalled at specific points in the cycle. Mutants could be classified in relation to recognizable steps called [A] events. Cells with a particular mutation stalled at [B] point in the cell cycle, despite the individual cells starting the experiment at [C] point in the cycle. This indicated that the gene product was critical at a specific point in the cycle but was not always needed throughout the cycle. These critical points were termed the [D] point for each gene.

Accepted Answer

The answer of Match the missing words in the following...

Quiz 15: Genetic Analysis of Cellular Processes

Fill in the blank. Another name for a genetic screen is a ______ screen.

Forward genetics involves disruption of a known gene to see what phenotypes will be produced. Reverse genetics involves introducing random mutations and examining the resultant phenotypes to discover the causative gene.

Which of the following features of genetic screens was not key to the success of Beadle and Tatum's early Neurospora crassa experiments with nutritional auxotrophs?

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity. -IPTG

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity. -allolactose

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity. -X-gal

Components and analogs of the lac operon are used by researchers as experimental tools. Match the molecule with its activity. -β-galactosidase