genetics unit III

Why are most recombinant human proteins produced in animal or plant hosts instead of bacterial host cells? A. Bacteria might not process all the eukaryotic proteins that might be useful to process the protein molecule. B. Bacterial cells might be harmful for the researchers. C. Bacteria do not process any eukaryotic proteins. D. Only transgenic eukaryotes are able to process eukaryotic proteins.

A. Bacteria might not process all the eukaryotic proteins that might be useful to process the protein molecule.

Fluorescence in situ hybridization is another way to visualize the presence of a nucleotide sequence. Which of the following is an advantage of FISH over Northern and Southern blots? A. Blotting is not used in FISH. B. Fluorophores are not used in Northern or Southern blots. C. There are no advantages to FISH over Northern and Southern blots. D. Probes are not needed in FISH.

A. Blotting is not used in FISH.

What is the main purpose of genome-wide association studies (GWAS)? A. GWAS involve scanning the genomes of thousands of unrelated individuals with a particular disease and comparing them with the genomes of individuals who do not have the disease. B. GWAS involve scanning the genomes of thousands of unrelated individuals with a particular mutation and comparing them with the genomes of individuals who do not have the mutation.

A. GWAS involve scanning the genomes of thousands of unrelated individuals with a particular disease and comparing them with the genomes of individuals who do not have the disease.

Why does radiotherapy often have significant side effects? A. In addition to killing cancer cells, radiotherapy can also kill normal cells too, causing side effects. B. It produces internal sources of radiation in the organism. C. A radiation dose that was too low to kill or slow the growth of cancer cells was used.

A. In addition to killing cancer cells, radiotherapy can also kill normal cells too, causing side effects.

Which of the following statements is true? A. The chromosomal theory of inheritance denotes that linked alleles will never be separated. B. A centromere and its surrounding genes constitute a linkage group. C. Linkage without crossing over leads to all recombinant chromosomes. D. If two gene loci are on nonhomologous chromosomes, genes at these loci are expected to assort independently. E. The linkage ratio is best seen in genes on nonhomologous chromosomes.

D. If two gene loci are on nonhomologous chromosomes, genes at these loci are expected to assort independently.

How can the order of three linked genes (A, B, and C) on the same chromosome be determined? A. Gene order cannot be determined by looking at the results of a cross. B. Look for single-crossover phenotypes involving the wild-type and mutant alleles of genes A, B, and C. C. Look for parental phenotypes. D. Look for double-crossover phenotypes involving the wild-type and mutant alleles of genes A, B, and C.

D. Look for double-crossover phenotypes involving the wild-type and mutant alleles of genes A, B, and C.

Knowing that tumors release free DNA into certain surrounding body fluids through necrosis and apoptosis [Kloten et al. (2013). Breast Cancer Res. 15(1):R4] choose an experimental protocol for using human blood as a biomarker for cancer and as a method for monitoring the progression of cancer in an individual. A. The products of the apoptosis should be looked for in blood. B. The protein content of blood should be checked. C. Extra inorganic components should be looked for in blood. D. Novel DNA should be looked for in blood.

D. Novel DNA should be looked for in blood.

T/F: Mutations in the genome of a fetus can be detected using a blood sample from the mother.

True (Three to six percent of the DNA circulating in a mother's bloodstream belongs to her baby. Some of this DNA is cfDNA, which when subjected to haplotype analysis can be distinguished from maternal DNA.)

Explain the three major steps in the polymerase chain reaction. "The first step is ______ of the target DNA molecule by exposure to _______. The second step is _____ of _____ to the target DNA. The third step is the ______ of the target DNA sequence by a(n) ______ DNA polymerase."

"The first step is denaturation of the target DNA molecule by exposure to high temperature. The second step is annealing of primers to the target DNA. The third step is the synthesis of the target DNA sequence by a(n) thermostable DNA polymerase."

When DNA breaks are detected in a cell, whether induced by Cas9 or otherwise, the endogenous double-strand break repair machinery of the cell attempts to repair the breaks. Double-stranded breaks in the genome may be repaired by either nonhomologous end-joining (NHEJ) or by homologous recombination called homology-directed repair (HDR). NHEJ simply involves the ligation of broken DNA fragments. HDR uses an undamaged homologous chromosome or sister chromatid as a template to repair a broken chromosome. Cells may use NHEJ or HDR, and the type of repair used has implications for gene editing. The two break repair mechanisms are different in terms of how error prone they are, and scientists can exploit these differences to achieve their genome editing goals. For instance, NHEJ is error prone and often results in small insertions or deletions (indels) at the repair site. Those indels can cause a shift of the coding sequence reading frame, often leading to non-functional proteins or to no protein being expressed in the modified cells. If the goal of genome editing is to disrupt a gene and create a nonfunctional (null) allele, scientists can use the Cas9 and sgRNA to make cuts, then let the endogenous repair mechanisms do their job without further action. While the HDR mechanism may repair Cas9-induced double-stranded breaks correctly in some cells, in many other cells, the error-prone NHEJ pathway will introduce indels and disrupt the gene. HDR is less error prone and better for making specific edits. The proportion of break repair completed by HDR (instead of NHEJ) increases if DNA breaks are only single stranded. Cas9 can be modified to make single strand breaks instead of double strand breaks if HDR repair is desired. While doing its repair, HDR can be "tricked" into using an artificial donor template (instead of the homologous chromosome) to make complex substitutions, deletions, or insertions, sometimes of relatively large stretches of DNA. The donor template is an experimentally introduced DNA molecule carrying a sequence with desired edits flanked by homology arms that match sequences adjacent to the genomic target. Through the HDR mechanism, the target sequence in the genome is replaced by the sequence on the donor template, resulting in complex modifications designed by the scientists. If no donor template is introduced by scientists, HDR will often repair the DNA breaks back to their original sequence. The precise nature of HDR repair stands in contrast to the random mutations caused by NHEJ.

(It is important to remember that while the Cas endonuclease cuts DNA, edits are made during the DNA repair stage. NHEJ and HDR are both DNA break repair mechanisms, but they can lead to different outcomes. Depending on the type of edit researchers are trying to induce, they can manipulate conditions in the cell to favor one mechanism over the other. Now that you understand the overall CRISPR-Cas9 process and how editing occurs during break repair, let's explore some of the limitations of moving the CRISPR-Cas9 technology from prokaryotes to eukaryotes.)

In a population of cattle, the following color distribution was noted: 36% red (RR), 48% roan (Rr), and 16% white (rr). This population is in a Hardy-Weinberg equilibrium. What will be the distribution of genotypes in the next generation if the Hardy-Weinberg assumptions are met? A. 36% red; 48% roan; 16% white B. 30% red; 60% roan; 10% white C. 30% red; 50% roan; 20% white D. 40% red; 40% roan; 20% white

A. 36% red; 48% roan; 16% white

Assume that in a Hardy-Weinberg population, 9% of the individuals are of the homozygous recessive phenotype. What percentage are homozygous dominant? A. 49% B. 91% C. 9% D. 51%

A. 49%

What is genetic equilibrium? A. A condition for a population in which the frequency of given alleles remains constant from generation to generation. B. A condition for a population in which the frequency of heterozygotes and homozygotes remains constant from generation to generation. C. A condition for a population in which the frequency of a given genotype remains constant from generation to generation. D. A condition for a population in which the frequency of X-linked genes remains constant from generation to generation.

A. A condition for a population in which the frequency of given alleles remains constant from generation to generation.

After double-stranded breaks are created by Cas9, repairs are attempted by the cell. If nonhomologous end-joining (NHEJ) is used to repair the double-stranded break, what is often the outcome? A. An insertion or deletion in the target gene, often disrupting gene function B. An edit with the desired version of the gene C. An insertion or deletion at a random place in the genome D. An insertion or deletion in the target gene, which doesn't affect gene function

A. An insertion or deletion in the target gene, often disrupting gene function (Nonhomologous end-joining (NHEJ) can repair double-stranded breaks in the genome, but it often includes small insertions or deletions called indels. If indels land in the coding sequence of the gene, they cause frameshift mutations. These frameshifts often create a nonfunctional version of the gene. Such mutations can be useful for research purposes.)

If a gene has three alleles in a population, their frequencies must add up to ________. A. 1.5 B. 1.0 C. 0 D. 3.0

B. 1.0

In a population of 10,000 individuals, where 3600 are MM, 1600 are Mm, and 4800 are mm, what are the frequencies of the M alleles and the m alleles? A. M = 0.56; m = 0.44 B. M = 0.44; m = 0.56 C. M = 0.70; m = 0.30 D. M = 0.30; m = 0.70

B. M = 0.44; m = 0.56

Identify two genetic mechanisms whereby proto-oncogenes can become overexpressed. Select the two mechanisms. A. mutations that result in an abnormal protein product B. mutations within gene-regulatory regions C. a gain-of-function alteration D. modification of proto-oncogenes products E. alterations in chromatin structure

B. mutations within gene-regulatory regions E. alterations in chromatin structure

Genome editing is based on ________. A. methylating specific DNA sequences with specific proteins B. targeting and manipulating specific DNA sequences with specific nucleases C. acetylating specific DNA sequences with specific nucleases D. transcribing specific DNA sequences with specific proteins

B. targeting and manipulating specific DNA sequences with specific nucleases

In a different population, suppose that 38.40% of people are blood type A, and 46.24% are blood type O. Assuming this population is in Hardy-Weinberg equilibrium, what percentage of the population is blood type B? What percentage is blood type AB? Enter your answers to two decimal places.

Blood Type B: 11.52% Blood Type AB: 3.84% (The key to solving the problem above is to calculate allele frequencies for IA , IB , and i, and to then use that information to calculate the percentage of people with the B phenotype (genotypes IBIB and IBi) and the percentage of people with the AB phenotype (genotype IAIB ).The question tells you that 46.24% of the population is blood type O. This is equal to r 2. The question also tells you that 38.40% of the population is blood type A. This is equal to p 2 + 2pr. Using this information, you can solve for both p and r, as shown below. Solve for r by taking the square root of r 2. The square root of 0.4624 is 0.68. Now that you know r = 0.68, you can solve for p using the equation p 2 + 2pr = 0.3840. Thus, p 2 + 1.36p = 0.3840. Rearrange the equation to be p 2 + 1.36p - 0.3840 = 0. Now you have a quadratic equation that you can solve. You may recall from high school math class that there is more than one way to solve a quadratic equation. One way is to use the quadratic formula: Recall that the formula above is used to solve ax 2 + bx + c = 0. Thus, plugging in the information from your quadratic equation, p 2 + 1.36p - 0.3840 = 0, gives you: p = 0.24 Now that you have identified p (0.24) and r (0.68), you can solve for the frequency of the IB allele (q) using the equation p + q + r = 1.)

The frequency of two alleles, Q and q, is 0.2 and 0.8, respectively. What is the frequency of Qq heterozygotes in this population? A. 1.0 B. 0.16 C. 0.32 D. 0.64

C. 0.32

Which E-value represents the highest probability of a BLAST result not being chance? A. 0.1 B. 1 C. 1 × 10-4 D. 0.01 E. 1 × 10-3

C. 1 × 10-4

Assume that a trait is caused by the homozygous state of a gene that is recessive and autosomal. Nine percent of the individuals in a given population express the phenotype caused by this gene. What percentage of the individuals would be heterozygous for the gene? Assume that the population is in Hardy-Weinberg equilibrium. A. 9% B. 91% C. 42% D. 58%

C. 42%

What is a probe in molecular biology? A. an instrument used to manipulate cells in culture B. a type of vector system C. a DNA or an RNA molecule used in hybridization reactions D. Probes are not used in molecular biology.

C. a DNA or an RNA molecule used in hybridization reactions (A probe is a labeled single strand of DNA or RNA used to locate its complementary sequence. Probes are commonly used in Southern and Northern blots, as well as in library screening.)

What does the variable 2pq represent? A. expected frequency of heritable traits in a population in the Hardy−−Weinberg equilibrium B. expected frequency of male and female individuals in a population in the Hardy−−Weinberg equilibrium C. expected frequency of heterozygotes in a population in the Hardy−−Weinberg equilibrium D. expected frequency of relevant individual alleles in a population in the Hardy−−Weinberg equilibrium

C. expected frequency of heterozygotes in a population in the Hardy−−Weinberg equilibrium

The primary benefit of CRISPR-Cas over TALENs and ZFNs is ________. A. the cost B. the specificity of the nuclease C. the ability of Cas9 to bind engineered sgRNA D. the speed of the process

C. the ability of Cas9 to bind engineered sgRNA

A certain form of albinism in humans is recessive and autosomal. Assume that 1% of the individuals in a given population are albino. Assuming that the population is in Hardy-Weinberg equilibrium, what percentage of the individuals in this population is expected to be heterozygous? A. 99% B. 9% C. 82% D. 18%

D. 18%

Which of the following is required to efficiently target DNA in vitro using CRISPR-Cas9? A. target RNA B. tRNA C. native crRNA D. Cas9

D. Cas9

dCas9 is a useful tool in the CRISPR-Cas toolkit because it can ________. A. fluoresce and help researchers locate target sequences B. deactivate the transcription of a gene C. bind DNA, but it cannot travel down the template strand D. bind DNA, but it cannot cut DNA

D. bind DNA, but it cannot cut DNA

One of the genes your research lab is interested in has a wild-type sequence that includes the sequence GATTCA. A previous graduate student used CRISPR-Cas9 to edit the gene so as to include the sequence GGTTCA instead. You have access to their old notes and reagents. Now, you've been tasked with creating another version of the gene with the sequence GCTTCA. Which component of the old experiment do you need to modify? A. Cas9 expression plasmid B. sgRNA expression plasmid C. 20-nucleotide targeting sequence D. donor template

D. donor template (If you already have the components to induce cuts in the right location, a new donor template with the newly desired sequence might be all you need. This donor template would induce homology-directed repair (HDR) to repair the double-strand break to GCTTCA. Your lab can then study the function of the gene with this new sequence.)

CRISPR-Cas functions in bacteria to ________. A. code for restriction enzymes B. edit genomes C. replicate viral DNA D. fight viruses

D. fight viruses

Which assumption pertains to a population in a Hardy-Weinberg equilibrium? A. a small population B. a selective advantage of one genotype C. a selectively mating population D. no migration, mutation or, genetic drift

D. no migration, mutation or, genetic drift

Fill in the Blanks: _________ are mutant forms of ______________ and normally function to promote cell division. _________________ normally function to inactivate or repress cell division.

Oncogenes are mutant forms of proto-oncogenes and normally function to promote cell division. Tumor-suppressor genes normally function to inactivate or repress cell division.

Below is an image of a typical cloning plasmid. Match each element in the plasmid with its most important function in cloning.

(A typical cloning plasmid includes these elements: - a multiple cloning site, which is a region that contains multiple restriction enzyme cut sites that are used to clone the DNA of interest - sequencing primer sites, which allow for the cloned DNA to be sequenced and identified - the origin of replication, which allows for the plasmid to be replicated in bacteria - the ampR gene, which allows for selection of cells that contain the plasmid - the lacZ gene, which allows for selection of cells that contain a cloned DNA fragment)

Bioinformatics programs can search DNA for specific sequences to help identify potential protein-coding regions. Label the sequences in the eukaryotic gene below that could be used by a computer program to identify a gene.

(Bioinformatics programs can use several consensus sequences found in most eukaryotic genes to identify the location of genes in a DNA sequence. Some sequences are critical in transcription; others are critical in translation. - Transcription regulatory sequences, such as enhancers or silencers, are found upstream of the promoter. - The promoter sequence is found upstream of the ATG start codon and is important for the initiation of transcription. - Start and stop codons signal the initiation and termination of translation. - Splice site sequences are found in introns and are important for splicing, an important step in mRNA processing that occurs before the mRNA is translated. - Transcription termination sequences are found downstream of the stop codon.)

You ligate a gene into the plasmid shown in Part A, transform the plasmid into bacteria, and plate the cells on three types of media: complete media, complete media + ampicillin, and complete media + ampicillin + X-gal.Complete the table by answering each question about the colonies of bacteria that grow on each type of media.

(The ampR gene produces a protein that breaks down ampicillin. Only cells that contain the plasmid with the ampR gene will be able to grow on plates that contain ampicillin.The lacZ gene encodes a protein that produces a blue color in the presence of X-gal. The multiple coding site (MCS) for the plasmid is located within the lacZ gene. Therefore, if a DNA fragment is ligated into the MCS, the lacZ gene will no longer function and will not produce the blue color. In the presence of X-gal, cells that contain a plasmid with a fragment inserted into the MCS will be white, whereas cells that contain a plasmid without a fragment inserted into the MCS will be blue.)

The CRISPR-Cas system functions as a form of adaptive immunity in prokaryotes to defend against pathogens. CRISPR stands for clustered regularly interspaced short palindromic repeats, which are encoded in the bacterial genome and are heritable. After a viral infection, bacteria can retain a "molecular memory" in the CRISPR locus of viral DNA fragments to help protect itself against future infections by the same virus. Transcription of the CRISPR locus in the bacterial genome produces crRNAs that are complementary to specific sequences of viral DNA. If a previously encountered virus infects a bacterium again, the crRNA and a noncoding RNA called a transactivating crRNA (tracrRNA) help guide an endonuclease, called Cas, to the target DNA in the viral genome. Upon reaching its target, domains on the Cas endonuclease act like molecular scissors and cut the viral DNA. After the discovery of the role of CRISPR-Cas in microbe immunity, scientists realized that the system could be reengineered to target any DNA sequence of interest and "edit" the genome of a living cell, including eukaryotic cells. By designing crRNAs that are complementary to sequences of interest in the target genome, scientists can essentially build customized scissors that cut in specific places, enabling more precise editing than was ever possible before. Furthermore, by fusing the crRNA and tracrRNA sequences into a single RNA molecule, called a single guide RNA (sgRNA), the machinery for cutting DNA is simplified to just two components: the sgRNA and the Cas endonuclease. Sometimes sgRNA is also simply called guideRNA, although this tutorial will use the term sgRNA. While different prokaryotes use several types of Cas endonucleases, scientists often use one particular version--Cas9. The Cas9 endonuclease is a single protein, unlike some others that require multi-protein complexes in order to cut. The following figure shows the sgRNA and Cas9 complex, after it has found the target DNA in the cell's genome. Researchers select genomic targets to edit, then engineer specific sequences in the sgRNA to be complementary to the target. Each genomic target must have an adjacent PAM (protospacer adjacent motif), a 5'-NGG-3' sequence, which is common in the genome. The sgRNA helps guide Cas9 to the target by binding to the target DNA via complementary base pairing; then Cas9 makes double-stranded DNA breaks at the target location. After the breaks are made, the breaks are repaired through endogenous mechanisms (the DNA repair mechanisms normally active in the cell, even for breaks not caused by Cas9). Depending on how the break is repaired, this technology not only allows the creation of random mutations at targeted locations in the genome, but also specific revisions to genomes. Use of this technology has wide implications in many fields, including basic research, genetically modified foods, and human health. Consequently, ethical considerations must also accompany use of this technology, due to the vast potential to edit eukaryotic genomes. When scientists use CRISPR-Cas9 to make specific cuts to a genome, they have a goal they're trying to achieve. Often, they either want to modify the protein that a particular gene encodes, or they want to block production of a particular protein. Depending on the goal, they customize the sgRNA they insert into the cell. For the CRISPR-Cas9 system to modify a eukaryotic genome, several steps need to happen. First, researchers must insert Cas9 and sgRNA into the desired cells. This can be accomplished through several techniques. For example, plasmid expression vectors encoding the components can be introduced into cells, or sgRNA and Cas9 protein can be directly injected into cells. Because eukaryotic cells have a nucleus, researchers add nuclear localization signals to Cas9, which help its protein translocate to the nucleus. When the sgRNA is also transcribed, it can guide Cas9 to cleave the target site in the cell's DNA. The resulting double-stranded DNA break can be repaired by one of two mechanisms, which we'll discuss in the next part. It is during this DNA break-repair step that the genome is edited, with new bases sometimes being inserted or deleted, resulting in a modification from the original sequence. After repair of the break is complete, future transcription and translation of the repaired sequence will include any modifications introduced during the repair process. The following figure shows the general steps that occur when a researcher uses the CRISPR-Cas9 system to modify a protein-encoding gene in a eukaryotic cell with the goal of modifying the protein product. Drag the descriptions of the steps to their appropriate locations on the figure.

(**at the double strand break the Cas9 protein cleaves the target sequence)

So far, you've learned how CRISPR-Cas9 technology is used to edit genomes, but it can serve other functions, too, with a few modifications. An important technological development was the creation of a "dead" version of Cas9, known as dCas9, which can still bind but cannot cut target DNA. This dCas9 allows researchers to target specific areas of the genome for other purposes besides making sequence changes. For example, transcription of any gene can now be activated by genetically fusing the activation domain of a transcriptional activator to dCas9 and designing an appropriate targeting sgRNA. Similarly, target genes can be turned off by attaching the repression domain of a transcriptional repressor to dCas9. The dCas9 technology can enable epigenetic regulation by altering the chromatin state of a particular region. If dCas9 is attached to the catalytic domain of an epigenetic modifier enzyme, such as a histone deacetylase or a DNA methyltransferase, epigenetic tags can be added or removed from sequences targeted by a sgRNA. This allows researchers increased flexibility in regulating gene expression. The dCas9 technology can also allow visualizations of where particular sequences are within a cell using fluorescent microscopy. This can be accomplished by designing an appropriate sgRNA and attaching fluorescent proteins, such as the green fluorescent protein (GFP), to dCas9. Fill in the Blanks 1. Researchers want to make an endonuclease that binds to target DNA, but does not cut the DNA. This will allow them to target specific areas of the genome for other modifications without making cuts. ____ 2. Researchers are studying a proposed oncogene in mice. They want to examine what happens when the gene is overexpressed. __________. 3. Researchers are studying a proposed tumor suppressor gene in fruit flies. They want to examine what happens when the gene has lower than normal expression, without inactivating the gene entirely. ___________ 4. Researchers are studying the three-dimensional organization of genomic structure. They want to visualize the location of specific genomic loci within a living cell. _________ 5. Researchers identify some previously uncharacterized genes that might play a role in diabetes. They want to make mouse mutants with null mutations in those genes and study what symptoms arise as a result. ________ 6. Researchers are studying an allele associated with a longer life span in humans, which varies from the normal variant by six base pairs. In order to study this allele in the laboratory, they want to modify the mouse homologue to have the same 6 base pair variation. _________

1. Researchers want to make an endonuclease that binds to target DNA, but does not cut the DNA. This will allow them to target specific areas of the genome for other modifications without making cuts. dCas9 2. Researchers are studying a proposed oncogene in mice. They want to examine what happens when the gene is overexpressed. dCas9 and activation domain. 3. Researchers are studying a proposed tumor suppressor gene in fruit flies. They want to examine what happens when the gene has lower than normal expression, without inactivating the gene entirely. dCas9 and repression domain 4. Researchers are studying the three-dimensional organization of genomic structure. They want to visualize the location of specific genomic loci within a living cell. dCas9 and fluorescent protein 5. Researchers identify some previously uncharacterized genes that might play a role in diabetes. They want to make mouse mutants with null mutations in those genes and study what symptoms arise as a result. Cas9 and NHEJ 6. Researchers are studying an allele associated with a longer life span in humans, which varies from the normal variant by six base pairs. In order to study this allele in the laboratory, they want to modify the mouse homologue to have the same 6 base pair variation. Cas9 and HDR (While the discovery of CRISPR-Cas function in prokaryotic immunity was ground-breaking in its own right, moving this technology to eukaryotic cells is revolutionizing the way scientists do basic genetic research. The simplicity of the two-part system -- Cas9 and sgRNA - has made applying this gene editing technology relatively inexpensive and fast. You have now seen how the creative modifications of CRISPR-Cas9 have moved its application beyond simple gene editing. The changes described here, many of which utilize dCas9, are likely only the first of many modifications that will be developed in the future. While there are still some technical hurdles, genome editing and related applications now seem limited primarily by the imagination of the researcher.)

The first step in annotating a gene is identifying the open reading frame. In bacteria, all genes that code for a protein have a start and stop codon with an open reading frame between them. A bioinformatics program can translate all the possible reading frames (series of 3 codons) in an attempt to find the longest open reading frame. How many possible open reading frames can a bioinformatics program translate from one DNA sequence?

6 (Each strand of DNA has six possible reading frames. A reading frame is a series of 3 bases (codons) that can be used to predict an open reading frame. There are three reading frames in the 5'- 3' direction and three reading frames in the 3'- 5' direction. You can also think of it as three possible frames in one direction on each strand of DNA.)

The CRISPR-Cas system of gene editing is based on what naturally occurring biological process? A. bacterial defense against viral infection B. viral defense against eukaryotic defenses C. viral defense against bacterial defenses D. eukaryotic defense against viral infection

A. bacterial defense against viral infection

In a three-point mapping experiment, which class do you expect to find the least? A. double crossover B. single crossover C. parental D. recombinant E. noncrossover

A. double crossover

Mutations in the ras gene family induce normally quiescent cells to proceed into the replication cycle. This converts the ras gene from a ________ gene to a ________ gene. A. oncogene; proto-oncogene B. mutant; oncogene C. proto-oncogene; oncogene D. pseudooncogene; proto-oncogene E. tumor suppressor; proto-oncogene

C. proto-oncogene; oncogene

DNA microarray analysis is excellent for ________. A. studying one gene's expression very closely B. studying all of a sample's unexpressed genes simultaneously C. studying all of a sample's expressed genes simultaneously D. studying all of a sample's genes simultaneously

C. studying all of a sample's expressed genes simultaneously

What was the first human protein to be produced by recombinant DNA technology? A. human clotting factor VIII B. interferons C. human growth hormone D. insulin

D. insulin (Humulin (recombinant human insulin) was first produced by cloning the human gene for insulin and expressing it in bacterial cells. Humulin was licensed for human use by the FDA in 1982.)

Which component of the Cas9 expression plasmid ensures that Cas9 can access the DNA of a mammalian cell? A. 20-nucleotide targeting sequence B. promoter C. Cas9 coding sequence D. nuclear localization signals (NLSs)

D. nuclear localization signals (NLSs) (Cas9 is a protein that is guided to cut DNA at specific sequences. Since Cas9 is derived from bacteria, which do not have a nucleus, nuclear localization signals (NLSs) have to be added when using Cas9 in mammalian cells. These signals ensure that the Cas9 expression plasmid reaches the nucleus, where mammalian genomic DNA is located.)

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with the following circumstance: A. sex-linked inheritance with 30% crossing over. B. independent assortment. C. linkage with 50% crossing over. D. 100% recombination. E. linkage with approximately 33 map units between the two gene loci.

E. linkage with approximately 33 map units between the two gene loci.

DNA ligase ________. A. cuts the DNA to produce sticky or blunt ends B. reconnects the bases together between the DNA strands C. removes bases from a DNA molecule D. adds bases into a growing DNA molecule E. reconnects the phosphodiester linkage between bases on the same strand of DNA

E. reconnects the phosphodiester linkage between bases on the same strand of DNA

The cross GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data one can conclude that ________. A. the recombinant progeny are GE/GE and ge/ge B. the recombinant progeny are GE/ge and GE/ge C. the recombinant progeny are Ge/Ge and gE/gE D. the recombinant progeny are gE/GE and GE/ge E. the recombinant progeny are gE/ge and Ge/ge

E. the recombinant progeny are gE/ge and Ge/ge

T/F: Two genes that are separated by 10 map units show a recombination percentage of 10%.

True

Genes X, Y, and Z are linked. Crossover gametes between genes X and Y are observed with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a frequency of 5%. What is the expected frequency of double crossover gametes among these genes? A. 1.25% B. 50% C. 5% D. 30%

A. 1.25% (The probability of a double crossover is the product of the probabilities of the single crossovers: 0.25 x 0.05 = 0.0125, or 1.25%)

Phenotypically wild-type F1 female Drosophila, whose mothers had light eyes (lt) and fathers had straw (stw) bristles, produced the following offspring when crossed to homozygous light-straw Compute the map distance between the light and straw loci with the image below. A. 2 map units B. 8 map units C. 98 map units D. 4 map units E. 12 map units

A. 2 map units

What is the difference between a knockout animal and a transgenic animal? A. A knockout animal has a piece of DNA missing, whereas a transgenic animal usually has a piece of DNA added. B. A knockout animal can't pass the investigated feature by inheritance, whereas a transgenic animal usually can. C. A knockout animal has definite inactivated piece of DNA, whereas a transgenic animal usually has an unknown activated piece of DNA. D. A knockout animal has a piece of foreign DNA missing, whereas a transgenic animal usually has a piece of such DNA.

A. A knockout animal has a piece of DNA missing, whereas a transgenic animal usually has a piece of DNA added.

What is the normal function of a tumor-suppressor gene? A. A tumor-suppressor gene is a gene whose normal function is to suppress cell division. B. A tumor-suppressor gene is a gene whose normal function is to suppress transcription factors C. .A tumor-suppressor gene is a gene whose normal function is to suppress cell growth. D. A tumor-suppressor gene is a gene whose normal function is to suppress the process of apoptosis.

A. A tumor-suppressor gene is a gene whose normal function is to suppress cell division.

What is the difference between saying that cancer is inherited and saying that the predisposition to cancer is inherited? A. Inheritance conveys the assumption that when a particular genetic circumstance is present, a particular trait will be revealed in the phenotype. An inherited predisposition usually refers to situations in which a particular phenotype is expressed in families in some consistent pattern and may manifest itself in different ways. B. Inheritance conveys the assumption that a particular trait will always be revealed in the phenotype. An inherited predisposition is conveys the assumption that a particular trait has a 50% chance of being revealed in the phenotype. C. Inheritance is all of the genetic characters or qualities transmitted from parents to offspring. An inherited predisposition is all of the genetic characters or qualities that might be transmitted from parent to offspring.

A. Inheritance conveys the assumption that when a particular genetic circumstance is present, a particular trait will be revealed in the phenotype. An inherited predisposition usually refers to situations in which a particular phenotype is expressed in families in some consistent pattern and may manifest itself in different ways.

What happens in cases where the ras gene is mutated? A. It continually signals cell division. B. It signals cell death. C. It signals suppression of cell division D. .It has no effect on the cell.

A. It continually signals cell division.

What is the relationship between the degree of crossing over and the distance between two genes? A. It is direct: as the distance increases, the frequency of recombination increases. B. There is no correlation. C. It is indirect: as the distance increases, the frequency of recombination doubles. D. It is indirect: as the distance decreases, the frequency of recombination doubles. E. It is direct: as the distance decreases, the frequency of recombination increases.

A. It is direct: as the distance increases, the frequency of recombination increases.

Why do mapping experiments become less accurate when the distances between genes become large? A. Multiple crossovers are more common. B. Interference is greater when the distance between genes is large. C. Crossover gametes become less common. D. Recombination occurs less frequently in long chromosomes.

A. Multiple crossovers are more common.

Genetic tests that detect mutations in the BRCA1 and BRCA2 oncogenes are widely available. These tests reveal a number of mutations in these genes - mutations that have been linked to familial breast cancer. Assume that a young woman in a suspected breast cancer family takes the BRCA1 and BRCA2 genetic tests and receives negative results. That is, she does not test positive for the mutant alleles of BRCA1 or BRCA2. Can she consider herself free of risk for breast cancer? A. No, she will still have the general population risk of about 10 percent. B. No, she will still have the general population risk of about 30 percent. C. Yes, she will not have the general population risk.

A. No, she will still have the general population risk of about 10 percent.

While CRISPR originated as a mechanism in prokaryotes to defend against viral attack, moving the system to eukaryotes poses some challenges given the differences between prokaryotic and (larger) eukaryotic genomes. One challenge is that while a prokaryotic chromosome is easily accessible in the cytoplasm, eukaryotic chromosomes are contained within a nucleus. Scientists have circumvented this by adding nuclear localization signals to Cas9 to ensure it can access the nucleus in eukaryotic cells. Another challenge is that CRISPR-Cas9 can sometimes cause unintended consequences elsewhere in the genome. A particular 20-bp target sequence chosen by researchers—and encoded in the sgRNA--may appear more than once in the genome. This is especially likely to happen in eukaryotes, which have much larger genomes than prokaryotes. If Cas9 cuts those other sequences, known as off-target sequences, there may be unintended edits that affect the experiment. Cas9 itself also has modest infidelity, which means that it does not always cut at sequences that exactly match the 20-bp target sequence encoded in the sgRNA. Other hurdles can arise due to how long the genome-editing machinery is active in individual cells. Depending on the method of introducing Cas9 and sgRNA into cells, they can either be heritable or transiently expressed. Supplying the components as transgenes can make them present in later cell generations in a stable manner if they are integrated into the genome. However, injecting the components into the cell as RNA or protein means their presence and active cutting will be transient, with later daughter cells eventually no longer containing the components. This is important because gene edits are not made with 100% efficiency, meaning that not all cells that are targeted will actually be edited. If Cas9's presence is transient, some cells may never be cut and edited. However, if Cas9 is stable, there is a higher chance of off-target effects. It is important to note that once a eukaryotic cell's genome has been edited, those changes will be inherited by its daughter cells, even if CRISPR-Cas9 is no longer active. Now, consider a scenario in which researchers are trying to make a mouse model of a disease that is suspected to be caused by simultaneous mutations in two genes affecting kidney development. They develop two sgRNAs, each specific for one of the target genes, and introduce the sgRNAs and Cas9 components into early mouse embryos. When analyzing the resulting mice, they found that some mice have new mutations in more than those two genes, an indication that CRISPR-Cas9 performed off-target cuts. What steps could help refine the researchers' experiment to modify only the two target genes? Select all that apply. A. Screen the rest of the mouse genome to be sure the 20-bp targets do not appear elsewhere in the genome. B. Make stable transgenic lines so the cutting by Cas9 is not transient. C. Insert more Cas9 and sgRNA during the initial introduction of the components. D. Engineer a version of Cas9 that demonstrates higher fidelity to the target sequence. E. Inject the plasmids into younger mouse embryos so more cells are modified.

A. Screen the rest of the mouse genome to be sure the 20-bp targets do not appear elsewhere in the genome. D. Engineer a version of Cas9 that demonstrates higher fidelity to the target sequence. (In addition to off-target effects and less-than-perfect editing efficiency, additional efficacy, specificity, and safety concerns face researches who want to apply CRISPR-Cas9 gene editing to human therapeutics. Successful editing does not usually occur in 100% of cells targeted by CRISR-Cas9. Now consider that many human diseases are only diagnosed in a patient after symptoms develop in childhood or adulthood. In those cases, doctors may be faced with trying to introduce CRISPR-Cas9 into many cells. The percentage of cells that need to be edited to improve symptoms will vary. Editing becomes more complex when one considers the challenges of introducing the Cas9 and sgRNA components into larger numbers of cells and different types of cells, potentially low editing efficiency in cells that do contain the components, and the potential for dangerous off-target effects. In addition to technical hurdles, many ethical questions face researchers who want to apply CRISPR-Cas9 genome editing to human cells.)

What is the overall goal of PCR? A. To make many copies of a short nucleotide sequence B. To identify mutations in a sequence C. To learn the sequence of a stretch of nucleotides D. To make many copies of a genome

A. To make many copies of a short nucleotide sequence

What is the main purpose of a DNA probe? A. hybridizes to a target sequence B. cuts DNA targets C. binds to proteins D. extend the growing polynucleotide

A. hybridizes to a target sequence

As more is learned about cancer, it has become clear that cancer, with few exceptions, ________. A. is a result of genetics and environmental factors B. has no genetic basis C. is only dependent on the inherited genetics D. is 100% dependent on inherited genetics

A. is a result of genetics and environmental factors

Which of the following proteins function as a cell-cycle regulator and transcription factor that can result in cell death (apoptosis) to a damaged cell? A. p53 B. p34 C. p102 D. cyclin E. phosphokinase

A. p53

The cross GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data, one can conclude that ________. A. the G and E loci are linked B. the G and E loci reside on the same chromosome over 50 map units apart C. the G and E loci show complete linkage D. the G and E loci assort independently E. the G and E loci are segregating independently

A. the G and E loci are linked

Amniocentesis and chorionic villus sampling can be used for all of the following EXCEPT __________. A. to detect adult carriers B. to karyotype a fetus C. to look for chromosomal aberrations D. to determine the sex of a fetus

A. to detect adult carriers

Which of the following is not a common system for generating knock-out organisms? A. transformation B. cre-lox C. homologous recombination D. CRSPR-CAS

A. transformation

One complication of making a transgenic animal is that the transgene may integrate at random into the coding region, or the regulatory region, of an endogenous gene. What might be the consequences of such random integrations? How might this complicate genetic analysis of the transgene? A. The random integration will have no effect on the function of the endogenous DNA or the transgene. B. The random integration may alter the function of the endogenous DNA as well as the transgene. C. The random integration will alter the function of the endogenous DNA, but not the transgene. D. The random integration will alter the function of the transgene, but not the endogenous DNA.

B. The random integration may alter the function of the endogenous DNA as well as the transgene.

The cross GE/ge × ge/ge produces the following progeny: GE/ge 404; ge/ge 396; gE/ge 97; Ge/ge 103. From these data, one can conclude that there are ________ map units between the G and E loci. A. 10 B. 20 C. 5 D. 25 E. 15

B. 20

A researcher is studying a gene related to vesicle formation. She identifies several exons within the gene that she thinks are responsible for a phenotype. She wants to predict how many copies of her target sequence she will have after each cycle of PCR. If she begins with a single double-stranded target sequence, after 3 cycles she will have 8 product molecules. After 5 cycles of PCR, how many product molecules will she have? A. 25 B. 32 C. 40 D. 64 E. 256

B. 32 (The formula for calculating the number of product molecules from PCR is 2n, where n equals the number of cycles. If starting with a single double-stranded target sequence, 5 rounds of PCR amplification would yield 32 product molecules (25). After only 20 cycles, there would be over 1 million product molecules (220 = 1.05 million). After 30 cycles, there are over a billion product molecules. PCR lets researchers quickly make many copies of genetic regions of interest.)

Assume that there are 12 map units between two loci in the mouse and that you are able to microscopically observe meiotic chromosomes in this organism. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between the two loci mentioned above? A. 96 B. 48 C. 12 D. 24 E. 6

B. 48

The second part of a PCR cycle involves annealing the primers. Which conditions will encourage annealing of the primer to the genetic region of interest? Select all that apply. A. A long primer - about half of the target sequence length up to 250 nucleotides B. A primer that is complementary to the target sequence C. A short primer - about 20 nucleotides D. A temperature of 92°C to 95°C E. A temperature of 45°C to 65°C F. A primer that is the same sequence as the target sequence

B. A primer that is complementary to the target sequence C. A short primer - about 20 nucleotides E. A temperature of 45°C to 65°C

Which of the following describes a "knock-out" organism? A. An organism that is capable of out-competing another organism. B. An organism that is negative for alleles coding a gene of interest. C. An organism that is positive for alleles coding a gene of interest. D. An organism that loses consciousness randomly.

B. An organism that is negative for alleles coding a gene of interest.

How can information from GWAS be used to inform scientists and physicians about genetic diseases? A. GWAS attempt to identify genes that influence mutation risk. B. GWAS attempt to identify genes that influence disease risk.

B. GWAS attempt to identify genes that influence disease risk.

The sequencing of the human genome lead to the realization that chromosome 19 contains many genes while chromosomes 13 and Y contain relatively few. This finding implies what about the density of genes in a genome? A. Genes are uniformly distributed throughout the genome. B. Genes are not uniformly distributed and appear in clusters separated by large noncoding regions. C. Genes have no organization and randomly appear throughout the genome and chromosomes. D. Genes are uniformly distributed on chromosomes but not through the entire genome.

B. Genes are not uniformly distributed and appear in clusters separated by large noncoding regions.

What is the role of the p53 protein in the cell cycle in normal cells? A. It prevents cells from exiting the cell cycle and undergoing apoptosis. B. It temporarily arrests the cell cycle in G1 before entering S. C. It causes cells to enter G0 and stop dividing. D. It binds to cyclin-dependent kinases to selectively phosphorylate certain proteins.

B. It temporarily arrests the cell cycle in G1 before entering S. C. It causes cells to enter G0 and stop dividing.

Which of the following best describes a goal of using the CRISPR-Cas9 genome editing system? A. Making edits at random locations in mammalian DNA B. Making edits at specific locations in mammalian DNA C. Editing bacterial DNA with a mammalian protein D. Editing mammalian DNA with bacteria

B. Making edits at specific locations in mammalian DNA (Cas9 is a protein that is guided to cut DNA at specific sequences. Sequences including the Cas9 coding sequence and the tracrRNA, which are originally from bacteria, can be combined in two small expression plasmids that are introduced into mammalian cells. Several modifications are needed to ensure the expression plasmids reach the nucleus and are expressed properly in mammalian cells.)

How might hypermethylation of the TP53 gene promoter influence tumorigenesis? A. The concentration of p53 will be increased, the process of tumorigenesis will be stimulated. B. The concentration of p53 will be decreased, the process of tumorigenesis will be stimulated. C. The concentration of p53 will be increased, the process of tumorigenesis will be suppressed. D. The concentration of p53 will be decreased, the process of tumorigenesis will be suppressed.

B. The concentration of p53 will be decreased, the process of tumorigenesis will be stimulated.

Evaluate the difference between whole-genome sequencing and whole-exome sequencing. Select all that apply. A. Whole-exome sequencing sequences the genome of an organism without exons. B. Whole-exome sequencing generates deeper reads of only the exons of an organism's genome. C. Whole-genome sequencing sequences the genome of an organism without exons. D. Whole-genome sequencing generates deeper reads of only the transcribed regions of an organism's genome. E. Whole-genome sequencing sequences the entire genome of an organism.

B. Whole-exome sequencing generates deeper reads of only the exons of an organism's genome. E. Whole-genome sequencing sequences the entire genome of an organism.

A linkage group _______. A. is a chromosomal region in which crossing over cannot occur B. represents all of the genes located on the same chromosome C. is a collection of genes that never produce recombinant gametes D. can occur only on an autosome

B. represents all of the genes located on the same chromosome

During the hybridization/annealing phase of a PCR reaction, the primers bind to the target DNA sequence at a specific temperature. What effect would an increase in the GC content of the primer have on the optimal temperature for annealing the primer to a target DNA sequence? A. the annealing temperature would remain the same B. the annealing temperature would increase C. the annealing temperature would decrease

B. the annealing temperature would increase

Many of the known cancers are a result of ________. A. exercising regularly B. the genetic instability of the human genome C. the non-carcinogenic nature of our environment D. the genetic stability of the human genome

B. the genetic instability of the human genome

Transcriptomics is ________. A. determining epigenetic markers B. the quantification of gene expression C. determining phenotype D. measuring how much DNA is in the genome

B. the quantification of gene expression

Which term describes plants or animals that carry a foreign gene? A. transmuted B. transgenic C. transduced D. transformed

B. transgenic

There are multiple cloning vector types in modern recombinant DNA technology ranging from plasmids to viral vectors. Which vector type is most useful when cloning an insert of approximately 500kb? A. bacterial artificial chromosome B. yeast artificial chromosome C. bacterial plasmid D. human artificial chromosome E. viral vector

B. yeast artificial chromosome

When a 5-kb circular plasmid is digested with a restriction enzyme that has three recognition sites on the plasmid, how many bands can be visualized on an agarose gel? A. 1 B. 2 C. 3 D. 4

C. 3

Applying Hardy-Weinberg equilibrium, how many genotypes are predicted for a gene that has three alleles? A. 3 B. 4 C. 6 D. 9

C. 6 (If a gene has three alleles, then six genotypes are predicted by applying the Hardy-Weinberg equilibrium equation. If the alleles are represented by the variables p, q, and r, then the sum of the genotype frequencies is represented by p 2 + 2pq + q 2 + 2pr + r 2 + 2qr = 1. The distribution of alleles in the aforementioned genotypes is determined by the trinomial expansion (p + q + r)^2.)

Assume that a cross is made between AaBb and aabb plants and that the offspring occur in the following numbers: 106 AaBb, 48 Aabb, 52 aaBb, 94 aabb. These results are consistent with which scenario described? A. The genes are unlinked. B. In the AaBb parent, each homolog has one dominant allele and one recessive allele. C. In the AaBb parent, the dominant alleles are on one homolog and the recessive alleles are on the other. D. There is a very high coefficient of coincidence. E. It is impossible to tell.

C. In the AaBb parent, the dominant alleles are on one homolog and the recessive alleles are on the other.

How can mutations in noncoding segments of DNADNA contribute to the development of cancers? A. Mutations in noncoding regulatory sequences (such as exons and introns) of genes such as proto-oncogenes, tumor-suppressor genes, or cell cycle genes could alter conformation of the correspondent proteins. B. Mutations in noncoding regulatory sequences (such as spacer DNADNA and telomeres) of genes such as homeotic genes or receptor-coding genes could alter conformation of the correspondent proteins. C. Mutations in noncoding regulatory sequences (such as promoters, enhancers, and transcription binding sites) of genes such as proto-oncogenes, tumor-suppressor genes, or cell cycle genes could alter the timing or level of their expression leading to cancer. D. Mutations in noncoding regulatory sequences (such as enhancers, silencer, and insulators) of genes such as homeotic genes or receptor-coding genes could alter the timing or level of their expression leading to cancer.

C. Mutations in noncoding regulatory sequences (such as promoters, enhancers, and transcription binding sites) of genes such as proto-oncogenes, tumor-suppressor genes, or cell cycle genes could alter the timing or level of their expression leading to cancer.

The advent of genomic sequencing and annotations using bioinformatics has led to the realization that some of the long-held beliefs about prokaryotic genome structure were incorrect. Which of the following has not been refuted by modern bioinformatic discoveries? A. Bacterial genomes are always smaller than eukaryotic genomes. B. Prokaryotes have a single chromosome. C. Prokaryote genomes do not contain introns. D. All bacterial genomes are circular.

C. Prokaryote genomes do not contain introns.

How are transgenic animals different from knockout animals? A. In transgenic animals, a single gene is disrupted or eliminated, while knockout animals have several of their genes disrupted. B. Both transgenic and knockout animals have disruptions in one or more of their genes. C. Transgenic animals express or overexpress a particular gene of interest, while knockout animals have one of their genes eliminated or disrupted. D. Transgenic animals express or overexpress a particular gene of interest, while knockout animals have several of their genes disrupted.

C. Transgenic animals express or overexpress a particular gene of interest, while knockout animals have one of their genes eliminated or disrupted.

Choose a simple definition of a carcinogen. A. a mutant form of the cancer gene's product B. a cell division inactivating agent C. a cancer-causing agent D. a cell division promoting agent

C. a cancer-causing agent

Using RNA-seq generates data in situ. What does RNA-seq not provide data for? A. gene expression levels B. sequence of the RNA C. amount of protein produced by RNA D. location of transcript in the genome

C. amount of protein produced by RNA

A DNA microarray (also called a DNA chip) can be used to ________. A. determine protein translation levels B. generate mutations in the gene of interest C. detect mutations that may indicate a risk of disease D. detect RFLPs

C. detect mutations that may indicate a risk of disease

Which of the following general mechanisms appear to be involved in the formation of cancer cells? A. suppression, tabulation, projection B. inversions, operon formation, methylation C. genomic instability, DNA repair failure, chromatin modifications D. RNA failure, DNA phosphorylation, phosphorylation of adenyl cyclase E. transdetermination, mutation, allosteric interactions

C. genomic instability, DNA repair failure, chromatin modifications

The generation of a knock-out organism generally requires all the following EXCEPT ________. A. a targeting vector B. knowledge of the target sequence C. knowledge of the sequence product D. a selectable marker E. embryonic stem cells

C. knowledge of the sequence product

What term is applied when two genes fail to assort independently, that is, they tend to segregate together during gamete formation? A. tetrad analysis B. Mendelian inheritance C. linkage D. discontinuous inheritance E. dominance and/or recessiveness

C. linkage

Once it is expressed inside the mammalian nucleus, how does the Cas9 protein know where to cut? A. promoter B. tracrRNA sequence of the sgRNA C. targeting sequence of the sgRNA D. Cas9 coding sequence

C. targeting sequence of the sgRNA (The targeting sequence of the sgRNA is designed to recognize a 20-nucleotide length of DNA in the genome. This sequence can be chosen by the scientist depending on the gene they are studying. The sgRNA helps guide the Cas9 protein to that specific sequence, where Cas9 will make a double-stranded cut in the DNA. After cutting, repair mechanisms in the cell will then try to fix the break, which allows the possibility of edits being introduced.)

Which of the following scenarios would erroneously lead you to believe two genes reside on different chromosomes when they in fact are on the same chromosome? A. the genes are very close together on the chromosome, such that there is never a crossover between them B. too large a sample size C. the genes are very far apart on the same chromosome, such that there is always a crossover between them D. the genes are very far apart on the same chromosome, such that there is never a crossover between them E. a mating between consanguineous individuals

C. the genes are very far apart on the same chromosome, such that there is always a crossover between them

Assume that the genes from the previous example are located along the chromosome in the order X, Y, and Z. What is the probability of recombination between genes X and Z? A. 50% B. 20% C. 5% D. 30%

D. 30% (Recombination frequencies between linked genes along a chromosome are additive, so the recombination frequency between genes X and Z is 25 + 5 = 30.)

In a population of 100 individuals, 49% are of the NN blood type. What percentage is expected to be MN assuming Hardy-Weinberg equilibrium conditions? A. 9% B. 51% C. 21% D. 42% E. There is insufficient information to answer this question.

D. 42%

How does the Ras protein transmit a signal from outside the cell into the cytoplasm? A. Inactivated Ras proteins transduce a signal, which activates the transcription of genes that start cell division. B. Inactivated Ras proteins transduce a signal, which inactivates the transcription of genes that suppress cell division. C. Activated Ras proteins transduce a signal, which inactivates the transcription of genes that suppress cell division. D. Activated Ras proteins transduce a signal, which activates the transcription of genes that start cell division.

D. Activated Ras proteins transduce a signal, which activates the transcription of genes that start cell division.

The first step of a PCR cycle is denaturing the double-stranded DNA into single-stranded DNA. How is DNA denaturation accomplished during a PCR reaction? A. By adding a primer that is complementary B. By removing all but a 20-nucleotide sequence C. By heating to 45°C - 65°C D. By breaking hydrogen bonds

D. By breaking hydrogen bonds

Which of the following best describes why mapping is most accurate when genes are close together on a chromosome? A. This is not true; mapping genes is most accurate when genes are far away from each other. B. Double crossover events yield a result that looks the same as one crossover in a two gene mapping experiment and this throws off the calculations. C. The centromere gets in the way. D. Double crossover events yield a result that looks the same as no crossover in a two gene mapping experiment and this throws off the calculations. E. This is not true; relative distance of two loci on a chromosome has no effect on accuracy of mapping.

D. Double crossover events yield a result that looks the same as no crossover in a two gene mapping experiment and this throws off the calculations.

The human insulin gene contains a number of sequences that are removed in the processing of the mRNA transcript. In spite of the fact that bacterial cells cannot excise these sequences from mRNA transcripts, explain how a gene like this can be cloned into a bacterial cell and produce insulin. A. Plasmids containing insulin genes derived from genomic DNA would be void of introns. Therefore, intron processing would not be necessary. B. Plasmids containing insulin genes derived from genomic DNA would contain introns, but because the intron sequences are relatively short, the newly synthesized insulin would still likely be functional. C. Bacterial cells can be easily modified so that they are able to excise introns. D. Plasmids containing insulin genes derived from cDNA, rather than genomic DNA, would be void of introns. Therefore, intron processing would not be necessary.

D. Plasmids containing insulin genes derived from cDNA, rather than genomic DNA, would be void of introns. Therefore, intron processing would not be necessary.

In which ways could radiotherapy control or cure cancer? A. Radiotherapy is a cancer treatment only for external treatment. It works by damaging the cancer cells' plasma membrane and killing the cell. B. Radiotherapy is a cancer treatment only for internal applications. It starts the process of apoptosis in cancer cells by damaging the cells' proteins. C. Radiotherapy is a universal treatment for tumors caused by viruses. The radiation destroys the virus and products the cell from further damage. D. Radiotherapy is often administered externally or internally to kill cancer cells by damaging their DNA.

D. Radiotherapy is often administered externally or internally to kill cancer cells by damaging their DNA.

Choose the cellular and molecular function of the ras gene family and the consequences of mutations in ras. A. The ras gene family encodes a protein that represses or stimulates transcription of more than 50 different genes. Point mutations may cause changes in the function that promotes abnormal transcription, thus stimulating uncontrolled expression of genes B. The ras gene family encodes a protein that is involved in repression of apoptosis. Point mutations may cause changes in the function that promotes the process of programmed cell death. C. The ras gene family encodes a protein that is involved in regulation of the cell cycle at checkpoints. Point mutations may cause changes that promote cell cycle arrest, thus stimulating apoptosis. D. The ras gene family encodes a protein that is involved with signal transduction in the cell membrane. Point mutations may cause changes in the function that promotes abnormal signaling, thus stimulating uncontrolled cell growth.

D. The ras gene family encodes a protein that is involved with signal transduction in the cell membrane. Point mutations may cause changes in the function that promotes abnormal signaling, thus stimulating uncontrolled cell growth.

Conditional knock-out organisms provide which advantage over full knock-out organisms? A. They allow scientists to turn on the gene at any time during the organism's development. B. They allow scientists to render the organism unconscious at any time. C. They allow scientists to study genome size. D. They allow scientists to turn the gene off at any time during the organism's development.

D. They allow scientists to turn the gene off at any time during the organism's development.

Choose the true statement below. A. When two genes are over 50 map units apart, a crossover would theoretically occur 0% of the time. B. The percentage of tetrads involved in exchange between two genes is equal to the number of recombinant progeny seen. C. Linkage (viewed from results of typical crosses) always occurs when two loci are on the same chromosome. D. When two genes are over 50 map units apart, a crossover can theoretically occur 100% of the time. E. A single crossover on a chromosome will always yield the recombinant phenotype of the genes you are interested in.

D. When two genes are over 50 map units apart, a crossover can theoretically occur 100% of the time.

A knock-out mouse is made with respect to the PFKL gene, and this mouse is now described as which of the following? A. a gain-of-function PFKL mouse B. an enhancement-of-function PFKL mouse C. a retention-of-function PFKL mouse D. a loss-of-function PFKL mouse

D. a loss-of-function PFKL mouse

Cancer is best described as a __________. A. genetic disease at the gametic cell level B. bacterial disease at the somatic cell level C. viral disease D. genetic disorder at the cellular level

D. genetic disorder at the cellular level

What is the name of the protein that appears to regulate the entry of cells into an S phase? This protein is also known as the "guardian of the genome." A. phosphokinase B. p34 C. p102 D. p53 E. cyclin

D. p53

Evolving genome sequencing methods continue to be used in ever more time-sensitive situations. Which of the following is an emerging situation in which the CDC is using WGS, NGS, and TGS? A. tracking impact of pathogens on host genomes B. isolating pathogens after they have spread C. vaccine generation D. pathogen identification

D. pathogen identification

When considered as a root cause for cancer, which of the following is linked to at least 17 types of human cancer? A. eating a low-fat diet B. not exercising regularly C. eating a high-fat diet D. smoking

D. smoking

Next, you would like study the gene that you ligated into the plasmid in Part B. From that experiment, which colonies should you select to culture for further experimentation? A. white colonies that grew on complete media + ampicillin B. white colonies that grew on complete media C. blue colonies that grew on complete media + ampicillin + X-gal D. white colonies that grew on complete media + ampicillin + X-gal

D. white colonies that grew on complete media + ampicillin + X-gal (Colonies that are white when grown on complete media + ampicillin + X-gal contain the plasmid with the gene ligated into the MCS and are the colonies that should be selected to culture for further experiments.The ampicillin in the media is used to select for cells that contain the plasmid which has an AmpR gene. The X-gal in the media is used to select for cells that contain the plasmid and the gene of interest.If the gene of interest is present in the MCS, the LacZ gene will be disrupted and will not be able to produce a blue color in the presence of X-gal, and the colonies will be white.)

The genes for purple eyes and curved wings are approximately 21 map units apart on chromosome II in Drosophila. Assume that a purple-eyed female was mated to a curved wing male and that the resulting F1 phenotypically wild-type females were mated to purple, curved males. Of 1000 offspring, what would be the expected number of flies with purple eyes and curved wings? A. 790 B. 210 C. 395 D. 540 E. 105

E. 105

Assume that two genes are 80 map units apart on chromosome II of Drosophila and that a cross is made between a doubly heterozygous female and a homozygous recessive male. What percent recombination would be expected in the offspring of this type of cross? A. 5 B. 100 C. 80 D. 30 E. 50

E. 50

What is one cause for the 21,000 protein encoding sequences in the human genome producing between 200,000 and 1 million different proteins? A. CRISPR-Cas systems B. SNPs C. restriction enzymes D. CNVs E. alternative splicing

E. alternative splicing

Assume that a cross is made between AaBb and aabb plants and that the offspring fall into approximately equal numbers of the following groups: AaBb, Aabb, aaBb, aabb. These results are consistent with the following circumstance: A. alternation of generations. B. hemizygosity. C. complete linkage. D. incomplete dominance. E. independent assortment.

E. independent assortment.

A sequence of DNA was analyzed with a bioinformatics program that translated the sequence in all 6 possible reading frames. Below is the outcome of this analysis. If this sequence contains one gene, which frame is most likely the correct reading frame for this gene? (Remember that Met represents a potential start codon and Stop represents a stop codon.)

Frame with the largest Met - Stop region

Now that you have identified the number of genotypes predicted for a gene that has three alleles, can you calculate phenotype frequencies when just allele frequencies are known? For the ABO blood group in humans, there are three alleles (IA , IB , and i), six possible genotypes (IAIA , IBIB , IAIB , IAi, IBi, and ii), and four possible phenotypes (A, B, AB, and O). Recall that IA is dominant to i, IB is dominant to i, and IA and IB are codominant. In a given population, the allele frequencies are as follows: IA = 0.15 IB = 0.25 i = 0.60 Assuming this population is in Hardy-Weinberg equilibrium, determine the expected phenotype frequencies. Enter your answers to four decimal places.

Frequency of A phenotype: 0.2025 Frequency of B Phenotype: 0.3625 Frequency of AB Phenotype: 0.075 Frequency of O Phenotype: 0.36 (If a gene has three alleles, then six genotypes are predicted by applying the Hardy-Weinberg equilibrium equation. In the case of ABO blood groups, four phenotypes are possible: A, B, AB, and O. To determine the frequency of each phenotype, let p designate the frequency of allele IA , let q designate the frequency of allele IB , and let r designate the frequency of allele i. The A phenotype is composed of genotypes IAIA and IAi. Thus, the frequency of individuals with phenotype A will be p 2 + 2pr. The B phenotype is composed of genotypes IBIB and IBi. Thus, the frequency of individuals with phenotype B will be q 2 + 2qr. The AB phenotype is determined by one genotype: IAIB . Thus, the frequency of individuals with phenotype AB will be 2pq. The O phenotype is determined by one genotype: ii. Thus, the frequency of individuals with phenotype O will be r^2.)

What is the difference between nonhomologous end-joining (NHEJ) and homology-directed repair (HDR) in the context of genome editing? Fill in the Blanks Repair of CRISPR-Cas9 editing by ______, when a suitable donor template is provided, allows precise substitutions as well as additions or deletions. This type of repair would be appropriate if the goal were to ________ Repair of CRISPR-Cas9 editing by _______, an error prone pathway, would likely disrupt the function of the target gene. This would be appropriate if the goal were to _______

Repair of CRISPR-Cas9 editing by HDR, when a suitable donor template is provided, allows precise substitutions as well as additions or deletions. This type of repair would be appropriate if the goal were to correct a homozygous recessive mutation. Repair of CRISPR-Cas9 editing by NHEJ, an error prone pathway, would likely disrupt the function of the target gene. This would be appropriate if the goal were to inactivate a harmful dominant allele in a heterozygous cell.