Ap unit 6

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. Given the results shown in Figure 1, which of the following correctly describes a relationship between the two species?

Species BB has more short fragments of DNADNA than species AA does. Correct. Electrophoresis separates DNADNA fragments by size and charge. The smaller fragments will move more quickly through the gel. In this example, species BB has more fragments that move further through the gel. Thus, species BB has more short fragments than species AA does.

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?

*Prokaryotic DNADNA ^Eukaryotic DNADNA *Single circular chromosome ^Multiple linear chromosomes This describes the typical arrangement of chromosomes: prokaryotes have DNADNA arranged in a single circular chromosome and eukaryotes have DNADNA arranged in multiple linear chromosomes.

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 poly‑A tail is added to RNARNA. Eukaryotic mRNAmRNA is modified after transcription and a poly‑A tail is added before it is translated at the ribosome.

Figure 1. CFTR protein sequences in unaffected and affected individuals 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?

Deletion, because a thymine is missing, which changes the reading frame. Deletion of the thymine nucleotide results in a frameshift that alters the triplet code sequence and changes the rest of the amino acid sequence.

Which of the following statements best explains the experimental results observed in Figure 1 ?

E. coli in lane CC have been successfully transformed and contain additional genetic information. A successful transformation would contain a band of DNADNA from the pTrupTru plasmid and a band of DNADNA from the E. coli chromosome. Lane CC has both bands correctly identified and is experimental evidence of a successful bacteria transformation of E. coli with the pTrupTru plasmid.

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 (*).Figure 1. A simplified representation of the DNA replication process 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?

Each newly synthesized strand remains associated with its template strand to form two copies of the original DNADNA molecule. Figure 1 shows a simplified representation of semiconservative DNADNA replication. The process of semiconservative replication produces DNADNA molecules that consist of one template strand and one newly synthesized strand.

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. Figure 1. A simplified model of a DNA replication fork Which of the following statements best explains the role of Enzyme 1 in the DNA replication process?

Enzyme 1 is a topoisomerase that relieves tension in the overwound DNADNA in front of a replication fork. Because Enzyme 1 carries out its role in the DNADNA replication process ahead of the DNADNA replication fork, it is most likely a topoisomerase. Topoisomerases relieve tension in the overwound DNADNA in front of replication forks.

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?

Hydrogen 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 DNADNA replication and for transcription from DNADNA to RNARNA.

Figure 1. DNA replication bubble. Which of the following best explains how this model illustrates DNA replication of both strands as a replication fork moves?

II is synthesized continuously in the 5′5′ to 3′3′ direction, and IIIIII is synthesized in segments in the 5′5′ to 3′3′ direction. As a replication fork moves, DNADNA polymerase can continuously replicate one strand in the 5′5′ to 3′3′ direction as the double stranded DNADNA is separated. The other strand (lagging strand) must be replicated in segments as the fork moves, since DNADNA polymerase only replicates DNADNA in the 5'5′ to 3'3′ direction, and DNADNA ligase later joins these fragments together.

Figure 3. Production of α-MSH in keratinocytes in response to UV radiation Which of the following claims best explains why keratinocytes do not produce melanin?

Keratinocytes do not express the MITFMITF gene.MITFMITF stands for melanocyte inducing transcription factor, and its expression during embryonic development is critical to the differentiation of melanocytes. The genes activated by MITFMITF include the TYRTYR, TRP2TRP2, and TRP1TRP1 genes that are involved in producing melanin. In keratinocytes, MITFMITF is not expressed, so it cannot activate transcription of genes involved in melanin production.

Figure 1. Comparison of rRNA processing in prokaryotes and eukaryotes Which of the following statements provides the best explanation of the processes illustrated in Figure 1 ?

Sections of the pre-rRNArRNA are removed, and the mature rRNArRNA molecules are available to combine with proteins to form the ribosomal subunits. In both prokaryotes and eukaryotes, the mature rRNArRNA molecules are removed from a longer transcript. These mature rRNArRNA molecules, together with proteins, form the large and small ribosomal subunits.

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?

She inherited an allele with more than 40 CAGCAG repeats in the HTTHTT gene from her father. 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.

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?

Skin pigmentation will not be able to change, resulting in a negative selection pressure. As shown in Figure 1, TYRTYR 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 UVUV radiation. This would result in a negative selection pressure.

Figure 1. Model of selected features of DNA transcription Which claim is most consistent with the information provided by the diagram and current scientific understanding of gene regulation and expression?

Some sequences of DNADNA can interact with regulatory proteins that control transcription. The regulatory base sequence (called a silencer) interacts with regulatory proteins such as this transcriptional repressor, which inhibits transcription.

Which of the following claims about the TYR, TRP2, and TRP1 mammalian genes is most likely to be accurate?

The TYRTYR, TRP2TRP2, and TRP1TRP1 genes may be located on different chromosomes but are activated by the same transcription factor.Related eukaryotic genes are often located on different chromosomes. In humans, TYRTYR is on chromosome 11, TRP2TRP2 is on chromosome 13, and TRP1TRP1 is on chromosome 9. The melanocytes control these genes together, as the same transcription factor, MITFMITF, activates the transcription of all three genes.

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. Figure 1. Model of posttranscriptional alternate splicing of mRNA 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 C mRNA?

The cell will die because it is unable to metabolize the substrate without enzyme AA, which is structurally specific for the substrate shown. Alternative splicing allows all three enzymes to be produced. If only enzyme CC mRNAmRNA is available for translation, then only enzyme CC (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.

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?

The individual has an increased risk of developing colon cancer. Correct. The individual may have an increased risk of developing colon cancer. Not all people with alleles associated with Lynch syndrome develop cancer.

Which of the following best predicts the phenotype of an individual who is homozygous for this TYR mutation?

The mutation will change all subsequent amino acids in the TYRTYR protein, leading to nonfunctional TYRTYR protein. Individuals with this mutation will lack melanin in their hair, skin, and eyes and will not tan in response to UVUV radiation. Without a functional TYRTYR 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 UVUV radiation. Note that UVUV radiation exposure will still lead to the production of α-MSHα-MSH in keratinocytes. The α-MSHα-MSH will then activate a signal transduction pathway in melanocytes, leading to the production of MITFMITF, which will activate theTYRTYR, TYR2TYR2, and TYR1TYR1 genes. However, the TYRTYR enzyme will still be nonfunctional because of the major frameshift mutation.

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?

The reverse transcriptase will produce DNADNA from the viral RNARNA, which can be incorporated into the host's genome and then transcribed and translated. Reverse transcriptase will transcribe DNADNA from RNARNA. The DNADNA integrates into the host genome and is transcribed and translated for the assembly of new viral progeny.

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 Figure 1. Model of damage to DNA caused by UV exposure Which of the following statements best explains what is shown in Figure 1 ?

UVUV photons cause dimers to form, leading to misshapen DNADNA, which results in replication and transcription errors. The model shows the partial separation of a segment of DNADNA caused by the structural change resulting from dimer formation. The presence of dimers can lead to errors in both transcription and replication

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?

When researchers applied a drug that activates adenylyl cyclase to the mutant mice's ears, the level of melanin increased. Application of a drug that activates adenylyl cyclase would increase the production of cAMPcAMP, which would activate protein kinase AA, which would activate CREBCREB, which would activate MITFMITF, which would activate the TYRTYR, TYR2TYR2, and TRP1TRP1 genes for the enzymes required to synthesize melanin. This evidence demonstrates that the parts of the signal transduction pathway from adenylyl cyclase onward are functional and that the enzymes required for melanin production are functional.

Which of the following best supports the claim that binding of miRNA‑delta to the miRNA binding site inhibits translation of gene Q mRNA?

When the miRNAmiRNA binding site sequence is altered, translation of Q mRNAmRNA occurs in the presence of miRNAmiRNA-delta. If the miRNAmiRNA binding site sequence is altered, it cannot base-pair with miRNAmiRNA‑delta because they will no longer be complementary with one another. If miRNAmiRNA‑delta cannot bind to the miRNAmiRNA binding site, translation of Q mRNAmRNA will proceed.

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.Figure 1. Model of proposed regulation of the trp operon by corepressors trp repressor and tryptophan Which of the following evidence best supports a claim that tryptophan functions as a corepressor?

When trpRtrpR is mutated, the trptrp operon is transcribed regardless of tryptophan levels. As tryptophan accumulates in the cell, it inhibits its own production by activating the trptrp repressor protein (coded by the trpRtrpR gene), which then binds to the operator and blocks transcription of the trptrp operon. If the trpRtrpR gene is mutated, then tryptophan cannot bind to the allosteric binding site to activate trptrp repressor protein. Thus, no matter how high tryptophan levels get, the trptrp operon will not be repressed.

Cycloheximide (CHX) is a eukaryote protein synthesis inhibitor. It is used in biomedical research to inhibit protein synthesis in eukaryotic cells studied in vitro. Its effects are rapidly reversed by simply removing it from the culture medium. In a translation experiment using a fungus culture, radiolabeled amino acids were added to the culture, allowing the researchers to measure the growth of a single polypeptide chain by measuring counts per minute (CPM). As the chain grew, the CPM increased. After a certain amount of time, CHX was added to the mixture, and the experiment continued. After an additional amount of time, the CHX was removed from the culture medium. Which of the following graphs best predicts the data collected during the experiment?

up middle up In the presence of the drug, protein synthesis halted and the polypeptide chain stopped growing. When the drug was removed, however, elongation resumed and the polypeptide chain started growing again.