Operons and DNA Technology and other study quizlet

3. In the absence of glucose, when lactose is present it combines with the repressor, allowing RNA polymerase to carry on transcription A. True B. False

A

9. Muscle cells differ from nerve cells mainly because they A. express different genes. B. contain different genes. C. use different genetic codes. D. have unique ribosomes. E. have different chromosomes.

A

5. Both liver cells and lens cells have the genes for making the proteins albumin and crystalline. However, only liver cells express the blood protein albumin and only lens cells express crystalline, the main protein in the lens of the eye. Both of these genes have enhancer sequences associated with them. The claim that gene regulation results in differential gene expression and influences cellular products (albumin or crystalline) is best supported by evidence in which of the following statements? A. Liver cells possess transcriptional activators that are different from those of lens cells. B. Liver cells and lens cells use different RNA polymerase enzymes to transcribe DNA. C. Liver cells and lens cells possess the same transcriptional activators. D. Liver cells and lens cells possess different general transcription factors.

A. Liver cells possess transcriptional activators that are different from those of lens cells.

Eukaryotes transcribe RNA from DNA that contains introns and exons. Alternative splicing is one posttranscriptional modification that can create distinct mature mRNA molecules that lead to the production of different proteins from the same gene. Figure 1 shows a gene and the RNA produced after transcription and after alternative splicing. A cell needs to metabolize the substrate illustrated in Figure 1 for a vital cellular function. Which of the following best explains the long-term effect on the cell of splicing that yields only enzyme CC mRNAmRNA? A. The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown. B. The cell will remain healthy because all three of the above enzymes can metabolize the substrate, as they are from the same gene. C. The cell will remain healthy because the enzyme C mRNA will undergo alternative splicing again until it transformed into enzyme A mRNA. D. The cell will remain healthy because enzyme-substrate interactions are nonspecific and enzyme C will eventually metabolize the substrate. *Alternative splicing allows all three enzymes to be produced. If only enzyme C mRNA is available for translation, then only enzyme C (which is not specific for the substrate) will be produced. Thus, the cell will not be able to perform a vital cellular function and it will die.*

A. The cell will die because it is unable to metabolize the substrate without enzyme A, which is structurally specific for the substrate shown.

24. Antigens are foreign proteins that invade the systems of organisms. Vaccines function by stimulating an organism's immune system to develop antibodies against a particular antigen. Developing a vaccine involves producing an antigen that can be introduced into the organism being vaccinated and which will trigger an immune response without causing the disease associated with the antigen. Certain strains of bacteria can be used to produce antigens used in vaccines. Which of the following best explains how bacteria can be genetically engineered to produce a desired antigen? A. The gene coding for the antigen can be inserted into plasmids that can be used to transform the bacteria. B. The bacteria need to be exposed to the antigen so they can produce the antibodies. C. The DNA of the antigen has to be transcribed in order for the mRNA produced to be inserted into the bacteria D. The mRNA of the antigen has to be translated in order for the protein to be inserted into the bacteria.

A. The gene coding for the antigen can be inserted into plasmids that can be used to transform the bacteria.

6. Arsenic is a toxic element found in both aquatic and terrestrial environments. Scientists have found genes that allow bacteria to remove arsenic from their cytoplasm. Arsenic enters cells as arsenate that must be converted to arsenite to leave cells. Figure 1 provides a summary of the arsenic resistance genes found in the operons of three different bacteria. *E. coli R773* is found in environments with low arsenic levels. *Herminiimonas arsenicoxydans* and *Ochrobactrum tritici* are both found in arsenic‑rich environments. Pretend Figure 1 since image unfound= -*E. coli R773*= only one line of operon devoted to arsenic removal -*Herminiimonas arsenicoxydans*= four lines of operons devoted to arsenic removal -*Ochrobactrum tritici*= two lines of operons devoted to arsenic removal. Researchers claim that bacteria that live in environments heavily contaminated with arsenic are more efficient at processing arsenic into arsenite and removing this toxin from their cells. Justify this claim based on the evidence shown in Figure 1. A. There are multiple operons controlling the production of proteins that process and remove arsenite from cells in both *H. arsenicoxydans* and *O. tritici*. In contrast, *E. coli* has only one operon devoted to arsenic removal. B. Both *H. arsenicoxydans* and *O. tritici* contain the arsR gene that codes for a repressor that turns on the operon to eliminate arsenite from the cell. C. Both *O. tritici* and *E. coli* contain the arsD gene, which codes for a protein that helps remove arsenite from the cell. D. Both *H. arsenicoxydans* and *O. tritici*. have more arsenic resistance genes than has *E. coli*.

A. There are multiple operons controlling the production of proteins that process and remove arsenite from cells in both *H. arsenicoxydans* and *O. tritici*. In contrast, *E. coli* has only one operon devoted to arsenic removal.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. ***********TYR**************TYR Tyrosine -------> L-DOPA ------- **********************TRP2******** ----> DOPAquinone ----> DHICA **TRP1***************************** ---------> Melanin The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Mice have melanocytes in the skin on their ears and show a tanning response to UV radiation. Researchers were studying a mutant population of mice that do not show a tanning response. Genetic testing of these mutant mice showed that the pathway causing the production of α-MSH by keratinocytes in response to UV radiation was fully functional. Thus, the researchers claimed that the lack of tanning response was due a nonfunctional MC1R. Which of the following pieces of evidence would best support the researchers' claim above? Which of the following pieces of evidence would best support the researchers' claim above? A. When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased. B. When researchers viewed sections of mutant mouse ears under the microscope, they found melanocyte numbers comparable to nonmutant mice. C. When researchers exposed the mutant mice to UV radiation, the amount of POMC mRNA in keratinocytes did not change. D. When researchers exposed the mutant mice to UV radiation, the level of melanin production did not change. *The fact that mutant mice did not produce melanin in response to UV radiation demonstrates a lack of a tanning response. However, this evidence does not narrow down the molecular mechanism for this lack of a tanning.*

A. When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased.

2. The lac repressor... A. binds to the operator and prevents transcription B. binds to the CAP site and prevents transcription C. binds to the CAP site and facilitates transcription D. binds to the operator and facilitates transcription

A. binds to the operator and prevents transcription

23. How does a bacterial cell protect its own DNA from restriction enzymes? A. by adding methyl groups to adenines and cytosines. B. by using DNA ligase to seal the bacterial DNA into a closed circle. C. by adding histones to protect the double-stranded DNA D. by forming "sticky ends" of bacterial DNA to prevent the enzyme from attaching E. by reinforcing the bacterial DNA structure with covalent phosphodiester bonds

A. by adding methyl groups to adenines and cytosines.

1. The lac operon in E. coli is involved in... A. regulating the expression of a gene B. regulating the translation of mRNA C. controlling the formation of ribosomes D. controlling DNA replication E. preventing the transfer of the F plasmid.

A. regulating the expression of a gene

17. Restriction enzymes A. work at recognition sites. B. function only at "sticky ends." C. produce uniform lengths of DNA. D. function only in genetic laboratories. E. none of these

A. work at recognition sites.

4. In response to chemical signals, prokaryotes can do which of the following? A. turn off translation of their mRNA B. alter the level of production of various enzymes C. increase the number and responsiveness of their ribosomes D. inactivate their mRNA molecules E. alter the sequence of amino acids in certain proteins.

B

Which of the following best explains how this model illustrates DNADNA replication of both strands as a replication fork moves? A. I and IV are synthesized continuously in the 5′ to 3′ direction. B. II and III are synthesized in segments in the 3' to 5′ direction. C. I is synthesized continuously in the 5′ to 3′ direction, and III is synthesized in segments in the 5′ to 3′ direction. D. II is synthesized in segments after DNA polymerase is released from synthesizing strand IV. Although II and III are lagging strands and are synthesized in segments, they are not replicated in the 3′ to 5′ direction. DNA is always synthesized in the 5′ to 3′ direction.

B. II and III are synthesized in segments in the 3' to 5′ direction.

Huntington's disease, an autosomal dominant disorder, is caused by a mutation in the HTT gene. The HTT gene contains multiple repeats of the nucleotide sequence CAG. A person with fewer than 35 CAG repeats in the HTT gene is unlikely to show the neurological symptoms of Huntington's disease. A person with 40 or more CAG repeats almost always becomes symptomatic. Due to errors in meiosis, an individual without symptoms of Huntington's disease can produce gametes with a larger number of CAG repeats than there are in their somatic cells. A woman develops Huntington's disease. Her father had the disorder. Her mother did not, and there is no history of the disorder in the mother's family. Which of the following best explains how the woman inherited Huntington's disease? A. She inherited an allele with fewer than 40 CAG repeats in the HTT gene because her mother did not have Huntington's disease. B. She inherited an allele with more than 40 CAG repeats in the HTT gene from her father. C. Her mother produced eggs that all have more than 40 repeats in the HTT gene. D. Her mother produced eggs that all have fewer than 40 CAG repeats in the HTT gene. *If her father was heterozygous for Huntington's disease, half of his sperm would carry an HTTHTT allele with more than 40 CAGCAG repeats. His children would have a fifty-fifty chance of inheriting the allele responsible for Huntington's disease.*

B. She inherited an allele with more than 40 CAG repeats in the HTT gene from her father.

Which claim is most consistent with the information provided by the diagram and current scientific understanding of gene regulation and expression? A. Reversible changes in the DNA sequence may influence how a gene is expressed in a cell. B. Some sequences of DNA can interact with regulatory proteins that control transcription. C. This is an inducible operon controlled by several regulatory factors. D. The transcription factor may produce mutations in the binding site at the promoter sequence inhibiting the synthesis of the protein. *The regulatory base sequence (called a silencer) interacts with regulatory proteins such as this transcriptional repressor, which inhibits transcription.*

B. Some sequences of DNA can interact with regulatory proteins that control transcription.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. ***********TYR**************TYR Tyrosine -------> L-DOPA ------- **********************TRP2******** ----> DOPAquinone ----> DHICA **TRP1***************************** ---------> Melanin The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Which of the following claims about the TYR, TRP2, and TRP1 mammalian genes is most likely to be accurate? A. The TYR, TRP2, and TRP1 genes are located next to each other on a single chromosome and are organized into an operon. B. The TYR, TRP2, and TRP1 genes may be located on different chromosomes but are activated by the same transcription factor. C. The TYR, TRP2, and TRP1 genes are identical genes since they are activated by the same transcription factor. D. The TYR, TRP2, and TRP1 genes may be located on different chromosomes but with identical operator sequences. *Operons are groups of genes in prokaryotes that share an operator and promoter and are turned on or off together depending on if a repressor is bound to the operator. Mammals are eukaryotes.*

B. The TYR, TRP2, and TRP1 genes may be located on different chromosomes but are activated by the same transcription factor.

Lynch syndrome is an inherited condition associated with an increased risk for colon cancer, as well as certain other cancers. Mutations in one of several genes involved in DNA repair during DNA replication have been associated with Lynch syndrome. DNA sequencing was performed for an individual. The results indicated that the individual carries one of the dominant alleles that has been associated with Lynch syndrome. Which of the following best explains how the results should be interpreted? A. The individual does not have an increased risk of developing cancer because one dominant allele is insufficient to cause the disease. B. The individual has an increased risk of developing colon cancer. C. Because the person's DNA has the mutation, other family members must have cancer. D. Results cannot be interpreted until testing determines if additional mutated alleles are present. *The individual may have an increased risk of developing colon cancer. Not all people with alleles associated with Lynch syndrome develop cancer.*

B. The individual has an increased risk of developing colon cancer.

16. The "natural" use of restrictions enzymes by bacteria is to A. integrate viral DNA. B. destroy viral DNA. C. repair "sticky ends." D. copy the bacterial genes. E. clone DNA.

B. destroy viral DNA.

20. The method used to produce single strands of DNA A. is known as amplification. B. is heating a solution of DNA. C. uses restriction enzymes. D. isolates DNA molecules from the nucleus.

B. is heating a solution of DNA.

11. The obvious advantage of the lactose operon is that A. lactose is not needed as energy for bacteria. B. lactose-metabolizing enzymes need not be made when lactose is not present. C. the bacteria will make lactose only in the presence of the proper enzymes. D. milk is not needed for adult humans' diet. E. glucose can substitute for lactose in the diet of intolerant persons.

B. lactose-metabolizing enzymes need not be made when lactose is not present.

12. The model of the prokaryotic operon explains the regulation of which of the following? A. replication B. transcription C. induction D. Lyonization E. none of these

B. transcription

18. Which of the following enzymes joins the paired sticky ends of DNA fragments? A. reverse transcriptase B. restriction enzymes C. DNA ligase D. DNA polymerase E. transferase

C. DNA ligase

Cystic fibrosis (CF) is a progressive genetic disease that causes persistent lung infections and affects the ability to breathe. CF is inherited in an autosomal recessive manner, caused by the presence of mutations in both copies of the gene for the cystic fibrosis transmembrane conductance regulator (CFTR) protein. Partial nucleotide sequences and the corresponding amino acid sequences for an unaffected individual and an affected individual are modeled in Figure 1. Based on the information in Figure 1, which type of mutation explains the nature of the change in DNA that resulted in cystic fibrosis in the affected individual? A. Substitution, because the amino acid tryptophan is replaced with glycine. B. Insertion, because an extra guanine is present, which changes the reading frame. C. Deletion, because a thymine is missing, which changes the reading frame. D. Duplication, because the amino acid leucine occurs twice, which changes the reading frame. *This is not an example of a substitution mutation. A substitution of one nucleotide for another may result in a change in the amino acid coded for but may also be a silent mutation that results in no change.*

C. Deletion, because a thymine is missing, which changes the reading frame.

Scientists conducted a transformation experiment using E. coli bacteria and the pTru plasmid. Samples of the pTru plasmid (lane A) and the chromosomal DNA from two different E. coli strains that the scientists attempted to transform (lane B and lane C) were compared using gel electrophoresis. The results are shown in Figure 1. Which of the following statements best explains the experimental results observed in Figure 1 ? A. E. coli in both lanes B and C have been successfully transformed and contain additional genetic information. B. E. coli in lane B have been successfully transformed and contain additional genetic information. C. E. coli in lane C have been successfully transformed and contain additional genetic information. D. Which E. coli have been transformed cannot be determined from this gel. *A successful transformation would contain a band of DNA from the pTru plasmid and a band of DNA from the E. coli chromosome. Lane C has both bands correctly identified and is experimental evidence of a successful bacteria transformation of E. coli with the pTru plasmid.*

C. E. coli in lane C have been successfully transformed and contain additional genetic information.

Nucleotide base pairing in DNA is universal across organisms. Each pair (T−A; C−G) consists of a purine and a pyrimidine. Which of the following best explains how the base pairs form? A. Ionic bonds join two double-ringed structures in each pair. B. Hydrogen bonds join two single-ringed structures in each pair. C. Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair. D. Covalent bonds join a double-ringed structure to a single-ringed structure in each pair. Each pair consists of one single-ringed structure (pyrimidine) joined to a double-ringed structure (purine). The hydrogen bonds allow the strands to separate for DNA replication and for transcription from DNA to RNA.

C. Hydrogen bonds join a double-ringed structure to a single-ringed structure in each pair.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. ***********TYR**************TYR Tyrosine -------> L-DOPA ------- **********************TRP2******** ----> DOPAquinone ----> DHICA **TRP1***************************** ---------> Melanin The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Which of the following claims best explains why keratinocytes do not produce melanin? A. Keratinocytes do not contain the TYR, TRP2, and TRP1 genes. B. Keratinocytes do not contain the MC1R gene. C. Keratinocytes do not express the MITF gene. D. Keratinocytes do not express the POMC gene. *Most non‑sex cells in mammals contain the same chromosomes with the same genes (genomic equivalence). Cell differentiation occurs when different cells express genes for proteins that are specific to that cell type.*

C. Keratinocytes do not express the MITF gene.

Students subjected three samples of five different molecules to gel electrophoresis as shown in Figure 1. + 1 2 3 _______________________________ A.| -- | B.| -- | | [ ] [ ] [ ] | *Wells* C.| *--* | D.| *--* | E.| *--* | |_______________________________| - Which of the following statements best explains the pattern seen on the gel with regard to the size and charge of molecules A and B? A. Molecules A and B are positively charged, and molecule A is smaller than molecule B. B. Molecules A and B are positively charged, and molecule A is larger than molecule B. C. Molecules A and B are negatively charged, and molecule A is smaller than molecule B. D. Molecules A and B are negatively charged, and molecule A is larger than molecule B.

C. Molecules A and B are negatively charged, and molecule A is smaller than molecule B.

All cells must transcribe rRNA in order to construct a functioning ribosome. Scientists have isolated and identified rRNA genes that contribute to ribosomal structure for both prokaryotes and eukaryotes. Figure 1 compares the transcription and processing of prokaryotic and eukaryotic rRNA. Which of the following statements provides the best explanation of the processes illustrated in Figure 1 ? A. Introns are removed from the pre-rRNA, and the mature rRNA molecules are joined and then translated to produce the protein portion of the ribosome. B. Introns are removed from the pre-rRNA, and each mature rRNA molecule is translated to produce the proteins that make up the ribosomal subunits. C. Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits. D. Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to bring different amino acids to the ribosome. *Although sections of the pre‑rRNA are removed, these sections are not introns and the mature rRNA molecules are not rejoined. These molecules are not messenger RNA molecules and are not translated to form proteins.*

C. Sections of the pre-rRNA are removed, and the mature rRNA molecules are available to combine with proteins to form the ribosomal subunits.

An evolutionary biologist hypothesizes that two morphologically similar plant species are not closely related. To test the hypothesis, the biologist collects DNA samples from each of the two plant species and then uses restriction enzymes to cut the DNA samples into fragments, which are then subjected to gel electrophoresis. The results are shown in Figure 1. Species********Species******** ****A***************B********* ----------------------------- [ ] [ ] | Dir ---- | ect ---- ---- | tion ---- ---- | | ---- | | of ---- ---- | | ---- | | mov ---- | | eme ---- ---- | V nt ---- | ---- | ---- | ---- | ---- | ---- | ---- ---- | ---- | | ----------------------------- Given the results shown in Figure 1, which of the following correctly describes a relationship between the two species? A. Species B is the ancestor of species A because it has fewer bands. B. Species A is more complex than species B because it has more bands. C. Species B has more short fragments of DNA than species A does. D. Species A has more short fragments of DNA than species B does. *Smaller DNA fragments move further through the gel than larger ones do and are furthest from the wells. Species B has more short fragments than species A does.*

C. Species B has more short fragments of DNA than species A does.

Retroviruses such as HIV and hepatitis B virus use RNA as their genetic material rather than DNA. In addition, they contain molecules of reverse transcriptase, an enzyme that uses an RNA template to synthesize complementary DNA. Which of the following best predicts what will happen when a normal cell is exposed to a retrovirus? A. The reverse transcriptase will cut the host DNA into fragments, destroying the host cell. B. The reverse transcriptase will insert the viral RNA into the host's genome so it can be transcribed and translated. C. The reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated. D. The reverse transcriptase will force the host ribosomes to translate the viral RNA prior to polypeptide assembly. *Reverse transcriptase will transcribe DNA from RNA. The DNA integrates into the host genome and is transcribed and translated for the assembly of new viral progeny.*

C. The reverse transcriptase will produce DNA from the viral RNA, which can be incorporated into the host's genome and then transcribed and translated.

The trp operon in E. coli is an example of a repressible operon that consists of genes coding for enzymes used to synthesize tryptophan. When tryptophan levels are high, the operon is turned off and these genes are not transcribed. However, it is also known that tryptophan does not bind directly to the operator DNA sequence. A regulatory gene called trpR has also been discovered although it is not part of the trp operon. The proposed model of how tryptophan acts as a corepressor is shown in Figure 1. Which of the following evidence best supports a claim that tryptophan functions as a corepressor? A. Normal expression of trpR causes the trp operon to be transcribed regardless of tryptophan levels. B. When the operator sequence is mutated, the trp operon is not transcribed. C. The trpR gene codes for a repressor protein that has a DNA binding domain. D. When trpR is mutated, the trptrp operon is transcribed regardless of tryptophan levels. *This evidence suggests that the trp repressor protein binds directly to DNA but does not specifically relate to the role of tryptophan as a corepressor.*

C. The trpR gene codes for a repressor protein that has a DNA binding domain.

22. The use of RFLPs for "genetic fingerprinting" is based upon A. the type of gel used in electrophoresis. B. identical alleles at loci. C. differences of locations where enzymes make their cuts. D. differences between blood and semen DNA. E. bonding of DNA to RNA.

C. differences of locations where enzymes make their cuts.

15. In order for DNA molecules to undergo recombination they A. they must be from the same species. B. their strands must separate as in replication. C. must be cut and spliced at specific nucleotide sequences. D. undergo lysis. E. must be identical.

C. must be cut and spliced at specific nucleotide sequences.

8. If a particular operon encodes enzymes for making an essential amino acid and is regulated like the trp operon, then the A. amino acid inactivates the repressor. B. enzymes produced are called inducible enzymes. C. repressor is active in the absence of the amino acid. D. amino acid acts as a corepressor. E. amino acid turns on transcription of the operon.

D

Which of the following best explains how the pattern of DNA arrangement in chromosomes could be used, in most cases, to determine if an organism was a prokaryote or a eukaryote? A. *Prokaryotic* DNA= Single ciruclar chromosome *Eukaryotic* DNA= Multiple circular chromosomes B. *Prokaryotic* DNA= Multiple chromosomes *Eukaryotic* DNA= Single chromosome C. *Prokaryotic* DNA= Single linear chromosome *Eukaryotic* DNA= Multiple linear chromosomes D. *Prokaryotic* DNA= Single ciruclar chromosome *Eukaryotic* DNA= Multiple linear chromosomes *This describes the typical arrangement of chromosomes: prokaryotes have DNA arranged in a single circular chromosome and eukaryotes have DNA arranged in multiple linear chromosomes.*

D. *Prokaryotic* DNA= Single ciruclar chromosome *Eukaryotic* DNA= Multiple linear chromosomes

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. ***********TYR**************TYR Tyrosine -------> L-DOPA ------- **********************TRP2******** ----> DOPAquinone ----> DHICA **TRP1***************************** ---------> Melanin The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Which of the following best explains a process occurring between point 1 and point 2 in Figure 3 ? A. α-MSH is produced. B. The TYR gene is transcribed. C. Polypeptides are removed from a protein. D. A poly‑A tail is added to RNA. *Eukaryotic mRNA is modified after transcription and a poly‑A tail is added before it is translated at the ribosome.*

D. A poly‑A tail is added to RNA.

Labeled nucleotides were supplied to a cell culture before the cells began DNA replication. A simplified representation of the process for a short segment of DNA is shown in Figure 1. Labeled DNA bases are indicated with an asterisk (*). Which of the following best helps explain how the process represented in Figure 1 produces DNA molecules that are hybrids of the original and the newly synthesized strands? A. Each template strand is broken down into nucleotides, which are then used to synthesize both strands of a new DNA molecule. B. Each template strand is broken into multiple fragments, which are randomly assembled into two different DNA molecules. C. Each newly synthesized strand is associated with another newly synthesized strand to form a new DNA molecule. D. Each newly synthesized strand remains associated with its template strand to form two copies of the original DNA molecule. *In the process of semiconservative DNA replication, the template strand is not broken into multiple fragments that are reassembled to form unique DNA molecules. Instead, the template strand remains intact to guide the synthesis of a new strand by complementary base pairing.*

D. Each newly synthesized strand remains associated with its template strand to form two copies of the original DNADNA molecule.

A simplified model of a DNA replication fork is represented in Figure 1. The protein labeled Enzyme 1 carries out a specific role in the DNA replication process. Which of the following statements best explains the role of Enzyme 1 in the DNADNA replication process? A. Enzyme 1 is a DNA ligase that joins together the DNA fragments at a replication fork to form continuous strands. B. Enzyme 1 is a DNA primase that catalyzes the synthesis of RNA primers on the lagging strand of a replication fork. C. Enzyme 1 is a DNA polymerase that synthesizes new DNA by using the leading and lagging strands of a replication fork as templates. D. Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork. *Because Enzyme 1 carries out its role in the DNA replication process ahead of the DNA replication fork, it is most likely a topoisomerase. Topoisomerases relieve tension in the overwound DNA in front of replication forks.*

D. Enzyme 1 is a topoisomerase that relieves tension in the overwound DNA in front of a replication fork.

Genetic engineering techniques can be used when analyzing and manipulating DNA and RNA. Scientists used gel electrophoresis to study transcription of gene L and discovered that mRNA strands of three different lengths are constistently produced. Which of the following explanations bests accounts for this experimental result? A. Gel electrophoresis can only be used with DNA (not mRNA), so experimental results are not interpretable. B. RNA polymerase consistently makes the same errors during transcription of gene L. C. Gene L is mutated, so RNA polymerase does not always transcribe the correct sequence. D. Pre-mRNA of gene L is subject to alternative splicing, so three mRNA sequences are possible.

D. Pre-mRNA of gene L is subject to alternative splicing, so three mRNA sequences are possible.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. ***********TYR**************TYR Tyrosine -------> L-DOPA ------- **********************TRP2******** ----> DOPAquinone ----> DHICA **TRP1***************************** ---------> Melanin The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Based on the information provided in Figure 1 and Figure 2, which of the following best predicts the effects of a mutation in the promoter of the TYR gene that prevents it from being transcribed? A. DNA damage due to UV radiation will be strongly inhibited, resulting in a positive selection pressure. B. DNA damage due to UV radiation will be strongly inhibited, resulting in a negative selection pressure. C. Skin pigmentation will not be able to change, resulting in a positive selection pressure. D. Skin pigmentation will not be able to change, resulting in a negative selection pressure. *As shown in Figure 1, TYR is required for two steps in the melanin synthesis pathway. If it is not expressed properly, then this will prevent melanocytes from synthesizing melanin regardless of exposure to UV radiation. This would result in a negative selection pressure.*

D. Skin pigmentation will not be able to change, resulting in a negative selection pressure.

In mammals, the dark color of skin, hair, and eyes is due to a pigment called melanin. Melanin is produced by specialized skin cells called melanocytes. The melanin is then transferred to other skin cells called keratinocytes. Melanocytes synthesize melanin in a multistep metabolic pathway (Figure 1). The amount of melanin produced is dependent on the amount of the enzymes TYR, TRP2, and TRP1 present inside melanocytes. ***********TYR**************TYR Tyrosine -------> L-DOPA ------- **********************TRP2******** ----> DOPAquinone ----> DHICA **TRP1***************************** ---------> Melanin The peptide hormone α-melanocyte stimulating hormone (α-MSH) activates a signal transduction pathway leading to the activation of MITF. MITF is a transcription factor that increases the expression of the TYR, TRP2, and TRP1 genes (Figure 2). Some mammals increase melanin production in response to ultraviolet (UV) radiation. The UV radiation causes damage to DNA in keratinocytes, which activates the p53 protein. p53 increases the expression of the POMC gene. The POMC protein is then cleaved to produce α-MSH. The keratinocytes secrete α-MSH, which signals nearby melanocytes. The increased melanin absorbs UV radiation, reducing further DNA damage. Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Researchers discovered a mutant form of the TYR gene with a deletion of a single guanine nucleotide in the beginning of the coding sequence. Which of the following best predicts the phenotype of an individual who is homozygous for this TYR mutation? A. The mutation will cause a single amino acid change in the TYR protein, which will not be enough to disrupt its function. Therefore, those with this mutation will produce melanin in the hair, skin, and eyes and tan in response to UV radiation. B. The mutation will cause a single amino acid change in the TYR protein, leading to a nonfunctional TYR protein. Therefore, those with this mutation will lack melanin in the hair, skin, and eyes and will not tan in response to UV radiation. C. The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Since the TRP1 and TRP2 genes were not affected, the TRP1 and TRP2 proteins will fill the role of the TYR protein. Therefore, those with this mutation will produce melanin in the hair, skin, and eyes in response to UV radiation. D. The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Individuals with this mutation will lack melanin in their hair, skin, and eyes and will not tan in response to UV radiation. *Without a functional TYR enzyme to catalyze the first two steps of the melanin synthesis pathway, melanin will not be produced. Without melanin, individuals will not have a tanning response to UV radiation. Note that UV radiation exposure will still lead to the production of α-MSH in keratinocytes. The α-MSH will then activate a signal transduction pathway in melanocytes, leading to the production of MITF, which will activate the TYR, TYR2, and TYR1 genes. However, the TYR enzyme will still be nonfunctional because of the major frameshift mutation.*

D. The mutation will change all subsequent amino acids in the TYR protein, leading to nonfunctional TYR protein. Individuals with this mutation will lack melanin in their hair, skin, and eyes and will not tan in response to UV radiation.

Exposure to ultraviolet (UV) radiation is the leading cause of skin cancer in humans. Figure 1 shows a model of how UV exposure damages DNA. Which of the following statements best explains what is shown in Figure 1 ? A. UV exposure triggers DNA replication, which results in rapid cell proliferation. B. Naturally occurring dimers are removed by the UV photons, causing misshapen DNA, which results in replication errors. C. The hydrogen bonds between base pairs absorb the UV photons, separating the two DNA strands, which results in rapid DNA replication. D. UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors. *Although hydrogen bonding is an important aspect of both replication and transcription, DNA replication requires a series of coordinated enzymatic reactions. In the absence of these enzymes, strands separated by UV-induced dimer formation would re-form into a misaligned helix, as shown in Figure 1.*

D. UV photons cause dimers to form, leading to misshapen DNA, which results in replication and transcription errors.

Small single-stranded RNA molecules called microRNAs (miRNAs) are capable of base pairing with specific binding sites in the 3′ untranslated region of many mRNA transcripts. Transcription of gene Q yields an mRNA transcript that contains such an miRNA binding site, which can associate with miRNA‑delta, a specific miRNA molecule. Which of the following best supports the claim that binding of miRNA‑delta to the miRNA binding site inhibits translation of gene Q mRNA? A. When the promoter for gene Q is altered, transcription is inhibited. B. Translation of Q mRNA is inhibited regardless of whether the miRNA binding site sequence is altered. C. Translation of Q mRNA is inhibited in the absence of miRNA‑delta. D. When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta. *If this were the case, it would suggest that miRNA‑delta does not need to interact directly with Q mRNA in order to inhibit translation. So, this evidence refutes the claim.*

D. When the miRNA binding site sequence is altered, translation of Q mRNA occurs in the presence of miRNA-delta.

10. What would occur if the repressor of an inducible operon were mutated so it could not bind the operator? A. irreversible binding of the repressor to the promoter B. reduce transcription of the operon's genes C. buildup of a substrate for the pathway controlled by the operon D. continuous transcription of the operon's genes E. overproduction of catabolite activator protein (CAP)

D. continuous transcription of the operon's genes

13. During the early part of a young mammal's life, the E. coli in the young offspring's intestinal tract are exposed to high levels of which of the following that later generations of E. coli will never be exposed to? A. glucose B. ribose C. cellulose D. lactose E. fructose

D. lactose

7. This can inhibit transcription by blocking the binding of positively acting transcription factors to the DNA: A. enhancer B. promoter C. activator D. repressor E. terminator

D. repressor

21. Separation of DNA fragments by gel electrophoresis A. requires priming. B. is controlled by the size of the fragment. C. is based upon the negative charges of phosphate groups. D. is easily accomplished. E. all are correct except "requires priming."

E. all are correct except "requires priming."

14. Plasmids A. are self-reproducing circular molecules of DNA. B. are sites for inserting genes for amplification. C. may be transferred between different species of bacteria. D. may confer the ability to donate genetic material when bacterial conjugate. E. all of these

E. all of these

19. For the polymerase chain reaction to occur, A. insolated DNA molecules must be primed. B. all DNA fragments must be identical. C. the DNA must be separated into single strands. D. a sticky end must be available for the ligaase enzyme to function. E. isolated DNA molecules must be primed and the DNA must be separated into single strands.

E. isolated DNA molecules must be primed and the DNA must be separated into single strands.

Most non‑sex cells in mammals contain the same chromosomes with the same genes (genomic equivalence). Cell differentiation occurs when different cells express genes for proteins that are specific to that cell type.

specific