Genetics

In humans, an average-sized chromosome contains about 100 million bp of DNA. If the DNA in such a chromosome were stretched out in a linear manner, how long would it be?

1 bp = 0.34 nm (100,000,000)(0.34 nm) = 34,000,000 nm = 3.4 x 107nm or 0.034 m

What are the expected results for the Meselson and Stahl experiment after 4 generations (i.e, 4 rounds of DNA replication in the presence of light nitrogen)? Note: during generation zero, the DNA is all heavy, and subsequent generations only make light DNA. 1/4 half heavy, 3/4 light 1/8 half heavy, 7/8 light 1/16 half-heavy, 15/16 light 1/32 half-heavy, 31/32 light

1/8 half heavy, 7/8 light

The following list describes events that are required for transcription of a eukaryotic gene. Put them in the correct order. Chromatin remodeling and histone modification Binding of activators Elongation Formation of the pre-initiation complex 2, 4, 1, 3 1, 4, 2, 3 4, 1, 2, 3 2, 1, 4, 3

2, 1, 4, 3

Let's suppose an organism has a G + C content of 64% in its DNA. What are the percentages of A, T, G, and C?

A - 18% T - 18% G - 32% C- 32% This is because A pairs to T and G pairs to C which means A and T will have the same percentage and C and G will also have the same percentage.

Heterochromatin formation can be influenced by various types of drugs. Which of the following would you expect to inhibit heterochromatin formation? A. A drug that inhibits DNA methyltransferase B. A drug that inhibits histone methyltransferase C. A drug that inhibits histone acetyltransferase D. A drug that inhibits histone deacetylase

A B D DNA and Histone methylation are two ways in which heterochromatin can persist in newly divided cells. If these two processes are inhibited, then heterochromatin will no longer be condensed and will revert to euchromatin. Histone deacetylases are required for the nucleation phase of heterochromatin formation.

Below is a DNA molecule, with the ends labeled A, B, C and D. Which ends are not extended by telomerase? A 5'-----------------------------------------------3' B C 3'-----------------------------------------------5' D A and C A and D B and D B and C

A and D

After three and after four rounds of replication, what would be the expected results for the Meselson and Stahl experiment if the correct mechanism were conservative, semiconservative, or dispersive? In your answer, describe the relative proportions of DNA molecules and their densities (e.g., 25% half-heavy, etc.). Note: generation 0 is all heavy, and then only light DNA strands are made in subsequent generations.

After Round 3 After Round 4 Conservative 12.5% fully heavy and 87.5% light 6.25% fully heavy and 93.75% light Semiconservative 25% half-heavy and 75% light 12.5% half-heavy and 87.5% light Dispersive 100% ⅛-heavy 100% 1/16-heavy

Which of the following possibilities could explain how PcG (polycomb group) complexes are able to silence genes? The compaction of nucleosomes The attachment of ubiquitin to histone proteins The direct inhibition of transcription factors, such as TFIID All of the above

All of the above

In a certain bacterial species, the amino acid arginine is synthesized by a particular enzyme so the bacterium does not require arginine in its growth medium. A key enzyme, which we will call arginine synthase, is necessary for arginine biosynthesis. When these bacteria are given arginine in their growth media, they stop synthesizing arginine intracellularly. Based on this observation alone, propose three different regulatory mechanisms to explain why arginine biosynthesis ceases when arginine is added to the growth medium. To better understand the mechanism of regulation, you measure the amount of intracellular arginine synthase protein when cells are grown in the presence and absence of arginine. Under both growth conditions, the amount of this protein is identical. Which mechanism of regulation would be consistent with this experimental observation?

Based on the observation that bacteria stop synthesizing arginine intracellularly when given arginine in the growth media, it is possible that expression of arginine synthase is regulated at the transcriptional, translational, or post-translational levels. If it is regulated at the transcription level, it could be possible that a repressor binds to a silencer for the gene that codes for this enzyme. If it is regulated at the translational level, a translational repressor could be binding next to the Shine-Dalgarno sequence, hindering the ribosome from initiating translation. If it is regulated at the post-translational level, a post-translational modification of the protein, such as phosphorylation, could make the arginine synthase inactive. The mechanism of regulation that would be consistent with the experimental observation is regulation at the post-translational level because the amount of protein is identical under both growth conditions. This suggests that the protein is being created under either condition.

Which of the following are found in the major and minor grooves of double-stranded DNA: sugars, phosphates, and/or bases? Which of the following are found in the DNA backbone of double-stranded DNA: sugars, phosphates, and/or bases? If you knew that a DNA-binding protein does not recognize a specific base sequence, would you expect that it recognizes the major groove, the minor groove, or the DNA backbone? Explain your reasoning.

Bases are found in the major and minor grooves of DNA helices. Sugars and Phosphates are found in the backbone of DNA helices. If a DNA-binding protein was known not to recognize a specific base sequence we would expect it to be found on the backbone because base sequences are recognized within the major and minor grooves, as is the case with histones which aid in chromosomal compaction and regulation.

Which of the following is not one of the steps in RNA interference (RNAi) that occurs with miRNA (microRNA)? Transcription to produce Pri-mRNA Cleavage of pre-miRNA by dicer Binding of miRNA to form a P-body Binding of RISC to a complementary mRNA

Binding of miRNA to form a P-body

When comparing the genome of a simple eukaryote, such as a yeast, to a complex eukaryote, such as a plant, why is the plant genome so much larger? The plant genome has more genes. The plant genome has more repetitive DNA. The plant genome has more G's and C's, compared to A's and T's. Both a and b are correct.

Both a and b are correct.

Which of the following is/are ways that a retroelement (retrotransposon) can proliferate? Using reverse transcriptase and integrase Using transposase Target-site primed reverse transcription Both a and c

Both a and c

The following statements below concern DNA replication. One of them is not true. Identify the false statement and explain why it is false. A. A DNA double helix obeys the AT/GC rule. B A DNA double helix could contain one strand that is 4 generations older than its complementary strand. C.A DNA double helix may contain two strands of DNA that were made at the same time. D. A DNA strand can serve as a template strand on many occasions. E. Following semiconservative DNA replication, one strand is a newly made daughter strand and the other strand is a parental strand.

C, False, DNA is semiconservative so one strand will be the parental strand and the other will be the newly synthesized complementary strand.

DNA replication of the entire genome in Drosophila embryos is known to happen about every 6 minutes. The Drosophila genome is composed of approximately 1.8 x 108 base pairs. If Drosophila DNA polymerases synthesize DNA strands at a rate of approximately 40 nucleotides per second, how many origins of replication would be required to replicate the entire genome?

Case 1: Assume that DNA polymerase synthesis rate is for one replication fork: 6 min = 360 sec; 40 nucleotides/sec*360 sec = 14400 nucleotides; 14400 nucleotides*2 replication forks per origin of replication = 28800 nucleotides after 6 minutes have passed; 1.8 x 10^8 bp/28800 bp = 6,250 origins of replication. Case 2: Assume that DNA polymerase synthesis rate is aggregated for one origin of replication: 6 min = 360 sec; 40 nucleotides/sec*360 sec = 14400 nucleotides synthesized in 6 minutes per origin of replication; 1.8 x 10^8 bp/14400 bp = 12,500 origins of replication.

ATP-dependent chromatin remodeling enzymes are a diverse set of multi-protein machines that harness the energy of ATP hydrolysis to alter nucleosome organization in local regions of chromatin. List three distinct types of changes in chromatin structure or composition that ATP-dependent chromatin-remodeling enzymes are known to implement. Pick one of your listed remodeling activities and offer an explanation for how the resulting change in nucleosome arrangements could contribute to gene activation.

Change in positions of nucleosomes Eviction of histone octamers from DNA - This would create gaps where nucleosomes are not found, and this could create space for RNA polymerase II to move along the DNA, and for activators to bind to the DNA. Replacing standard histones with histone variants

Define chromosome territory? Take a look at Figure 10.23 (7th edition). In a nondividing cell, what is the predominant level of structure within a chromosome territory: a, b, c, or d?

Chromosome territory is a region in the cell's nucleus during interphase occupied by a chromosome. The predominant level of structure within a chromosome territory would be C, the loop domain. The reason for this being the predominant level is due to the loop domains and adjacent nucleosomes being closer together and compacted, allowing chromatin to be in its predominant or strongest form.

List the structural differences between DNA and RNA.

DNA is double stranded while RNA can be single stranded. DNA uses the base thymine, but RNA uses the base uracil instead. DNA has sugar deoxyribose and RNA has sugar ribose.

In a short, well-written (and inspiring) paragraph, describe the underlying features of DNA replication that are the most important.

DNA is opened by DnaA proteins that bind to DnaA boxes and cause hydrogen bonds to break apart in A-T rich regions nearby, this allows DNA helicase to bind to the two newly formed replication forks. DNA helicase causes supercoiling as it works its way through the DNA separating the strands. This supercoiling is relieved by topoisomerase, which cuts one DNA segment ahead of the replication fork allowing it to spin every so often. DNA polymerase III binds to the strands after the replication forks and synthesizes complementary DNA (according to AT/GC Rule) in the 5'-3' direction, one strand in a constant leading manner and the other in a lagging manner with Okazaki fragments, which are adjoined by DNA ligase.

Due to imprinting, the Igf2 gene is silenced because ______________ of the ICR and DMR _________ occur during _____________. covalent histone modification, does, oogenesis DNA methylation, does not, spermatogenesis DNA methylation, does, spermatogenesis DNA methylation, does not, oogenesis

DNA methylation, does not, oogenesis

Describe the different ways that primers are removed during DNA replication in bacteria versus DNA replication in eukaryotes.

DNA replication in bacteria has primers removed by DNA polymerase I. The DNA polymerase I is able to digest away the primers in the 5' to 3' direction. With this vacant area created from the removal of primers, the DNA polymerase will then synthesize DNA to fill the vacant region. DNA polymerase I has 5' to 3' exonuclease activity which gives it the ability to digest the primers. DNA replication in eukaryotes has primers removed by an enzyme, flap endonuclease which is responsible for RNA primer removal. Flap endonuclease remove RNA flaps that are produced when DNA polymerase elongating the left Okazaki fragment, causing the right Okazaki fragment to have a flap. However, if the flap is too long, it is then cleaved by Dna2 nuclease/helicase, which generates a short flap that can now be removed by the flap endonuclease.

Two circular DNA molecules are topoisomers of each other; they contain the same amount of DNA. Let's call them DNA-1 and DNA-2. Under the electron microscope, DNA-1 appears more compact compared to DNA-2. The level of gene transcription is much lower for DNA-2. Which of the following three possibilities would best explain these observations? A. DNA-1 has 4 positive supercoils and DNA-2 has 2 negative supercoils. B. DNA-1 has 2 positive supercoils and DNA-2 has 4 negative supercoils. C. DNA-1 has 4 negative supercoils and DNA-2 has 2 positive supercoils.

DNA-1 and DNA-2 have the same amount of DNA, but DNA-1 is more compact. This means that supercoiling in DNA-1 is tighter than supercoiling in DNA-2. Also, despite the compact structure, DNA-1 has higher levels of gene transcription as compared to DNA-2. Option C would best explain this phenomenon because negative supercoiling promotes strand separation, which can allow higher levels of transcription.

Let's call two different DNA molecules DNA-A and DNA-B. DNA-B is more compact, and its level of gene transcription is higher compared to DNA-A. Based on these observations, which of the following statements is likely to be correct? DNA-A has 2 positive supercoils and DNA-B has 2 negative supercoils. DNA-A has 2 positive supercoils and DNA-B has 4 negative supercoils. DNA-A has 4 positive supercoils and DNA-B has 2 negative supercoils. DNA-A has 4 negative supercoils and DNA-B has 2 positive supercoils.

DNA-A has 2 positive supercoils and DNA-B has 4 negative supercoils.

Which of the following sites in an E. coli origin of replication would be used first to begin replication? DnaA boxes AT-rich region GATC methylation sites None of the above

DnaA boxes

Which of the following is not found in a replisome? two DNA polymerases primase DnaA protein helicase

DnaA protein

Explain how transposition via cut-and-paste transposons and retrotransposition always produces direct repeats in the chromosomal DNA.

During insertion of transposons and retrotransposons, the direct repeats are created due to the insertion method. The insertion element of the TE creates staggered cuts in the DNA which then come apart to insert the insertion. The missing sections of DNA are then filled in using the complementary bases thus causing the same sequence to be repeated in the same direction on each side of the insertion.

In the CRISPR-Cas9 system, which RNA(s) function as a guide? Explain how?

During the expression phase, an RNA that functions as a guide is the tracrRNA. In this phase a cRNA is attached to a tracrRNA. A portion of the tracrRNA is recognized by the Cas9 protein, by using the tracrRNA as a guide it allows the cRNA-tracrRNA to bind to Cas9. In the interference phase, an RNA that functions as a guide is cRNA. When the cRNA-tracrRNA-Cas9 complex is formed a spacer within the cRNA is complementary to one of the strands in the bacteriophage DNA. This allows the cRNA to guide the complex to the bacteriophage DNA strand, allowing it to bind.

A liver-specific gene was cloned and then subjected to promoter bashing as described in More Genetic TIPS (page 383, 6th edition, or page 391-392, 7th edition). Five regions adjacent regions (let's call them regions A, B, C, D, and E) were deleted using this strategy. (Note: region A is farthest away from the start codon of this gene). The DNA was transformed into liver cells. The data shown here are from this experiment. Percentage of Region Deleted β-Galactosidase Activity None 100 A 17 B 100 C 745 D 5 E <1 Discuss where enhancers, silencers, and/or the core promoter may be located for this gene.

E must be where the core promoter is located. If the core promoter is deleted, very little expression occurs, because the core promoter is needed for transcription. Enhancers must be in regions D and A, because if enhancers are deleted, then the activity will be lessened as binding of activators to enhancers increases transcription, and this would not occur. Region C must be where silencers are located, because the activity of a gene should be above 100% if these are deleted, since these elements allow the binding of repressors that prevent transcription, and this would not occur. Region B must not contain regulatory elements, because the deletion has no effect on the percentage of the activity.

What is meant by the term epigenetics? Are all types of epigenetic changes passed from parent to offspring? Explain.

Epigenetics is the study of changes in gene expression that can be passed from cell to cell, that are reversible, and do not involve a change in the DNA sequence. Not all types of epigenetic changes are passed from parent to offspring. For example, an outside environmental exposure could cause a change in a somatic cell that would then spread to other cells and possibly form cancer; because it occurred in somatic cells, the cancer would not be passed from parent to offspring. However, some epigenetic changes can be passed from parent to offspring through genomic imprinting. An example of this is the Igf2 gene.

Let's suppose a bacterium is part of a lineage that has been previously exposed to bacteriophage lambda. The ancestors to this bacterium were able to mount an attack against bacteriophage lambda via the CRISPR-Cas9 system. If this bacterium is now exposed to bacteriophage lambda, which phase(s) of the CRISPR-Cas9 system would be necessary for it to destroy bacteriophage lambda? Adaptation only Adaptation and Interference Expression and Interference Adaptation, Expression, and Interference

Expression and Interference

Which of the following is not one of the steps of X inactivation that occurs during embryonic development in female mammals? Binding and spreading of Xist RNA along the inactivated chromosome Expression of the Tsix gene on the inactivated X chromosome X chromosome pairing Shifting of the pluripotency factors and CTCFs to the active X chromosome

Expression of the Tsix gene on the inactivated X chromosome

In 100 words or less, discuss the advantages of gene regulation in bacteria and eukaryotes.

Gene regulation prevents the under or overproduction of energy costly materials at any given time with respect to cell conditions. This allows the cell to function optimally and efficiently across a wide array of conditions in each stage of its life. Gene regulation is important for metabolism, cell division and responding to environmental stress.

3. Figure 11.5 shows the sequences of the DnaA boxes that are found in the origin of replication in E. coli. Actually, these five sequences are similar to each other if you flip some of them in the opposite orientation. If you don't flip the first one, which one(s) would you have to flip so that all five of them have a similar sequence running in the same direction? What is the consensus sequence for the DnaA box? TGTGGATAA TGTGAATGA Normal TTATACACA - Flipped TGTGTATAA TTTGGATAA Normal TTATCCACA - Flipped TGTGGATAA

If you flip the third and the fifth sequence, then all five sequences will be similar. The consensus sequence is shown below. These nucleotides are most consistent at each spot of the sequence across all five sequences. Consensus Sequence 5'-TGTGGATAA-3' 3'-ACACCTATT-5'

Let's suppose you isolated a mutant strain of E. coli in which the lac operon is not expressed even in the presence of lactose. To understand the effect of the mutation, you make a merozygote in which the mutant strain contains an F' factor with a normal lac operon and a normal lacI gene. When you compare the mutant strain and the merozygote with regard to their β-galactosidase activities in the presence and absence of lactose, you obtain the following results: Amount β-Galactosidase Strain Addition of lactose (percentage of a normal strain in the presence of lactose) Normal Yes 100 Mutant No 0 Mutant Yes 0 Merozygote No 0 Merozygote Yes 0 Explain the nature of the mutation in the mutant strain.

It appears that the nature of the mutation in the mutant strain is at the site at which allolactose binds to the repressor protein. The operon will have the repressor protein binding to the operator. When there is lactose present, allolactose is supposed to bind to the repressor protein to inhibit its activity. However, if there is a mutation in the binding site of the repressor protein, the allolactose is not able to exert its effect because it cannot bind to the site. As a result, the repressor will be active and inhibit transcription of β-Galactosidase in the presence or absence of lactose. Even though the merozygote contains an F' factor with a normal lac operon and a normal lacI gene, it is important to note that the mutated repressor can diffuse from the bacterial chromosome region to the F' factor region and bind to the F' factor DNA. Therefore, transcription of the β-Galactosidase in the presence or absence of lactose is inhibited.

In a particular eukaryotic species, the linker DNA averages 54 bp in length. How many molecules of H2A would you expect to find in a DNA sample that is 46,000 bp in length?

Let the length of DNA wrapped around the histones (H) be equal to 146 bp and the average length of linker DNA (L) be equal to 54 bp. Observe that H+L=200bp. Further observe that if the total length of the DNA sample (T) is 46,000 bp, then T/(H+L)=230molecules. There are two H2A histones in an octamer. Therefore, we expect to find 460 H2A histones.

In order to replicate an entire eukaryotic chromosome, such as a human chromosome, how many replication forks are used? One Two Four Many

Many

Which of the following is not an advantage of gene regulation? Mutability: increase the rate of mutation of genes associated with metabolism. Efficiency: only make proteins when they are needed. Specialization: only make proteins in particular cell types. Development: only make proteins at particular stages of development.

Mutability: increase the rate of mutation of genes associated with metabolism.

Which of the following is not found in a bacterial chromosome? Many genes One centromere Some repetitive sequences One origin of replication

One centromere

Which of the following are components that HOTAIR ncRNA can bind to? PRC2 complex, LSD1 complex, and GA-rich regions PRC2 complex, LSD1 complex, and miRNA PRC2 complex, miRNA, and GA-rich regions PRC2 complex only

PRC2 complex, LSD1 complex, and GA-rich regions

The glucocorticoid receptor and the CREB protein form homodimers and activate transcription. Other transcription factors form heterodimers. For example, myogenic bHLH is a transcription factor that forms a heterodimer with a protein called the E protein and thereby activates the transcription of genes required for muscle cell differentiation. In contrast, when myogenic bHLH forms a heterodimer with a protein called the Id protein, activation does not occur. Which of the following scenarios would best explain this observation? Only one possibility is correct. Myogenic bHLH E Protein Id Protein Possibility 1 DNA-binding domain Yes No No Leucine zipper: Yes No Yes Possibility 2 DNA-binding domain: Yes Yes No Leucine zipper: Yes Yes Yes Possibility 3 DNA-binding domain: Yes No Yes Leucine zipper: Yes No No

Possibility 2 would best explain the observation seen. This is because in possibility 2 there is no DNA-binding domain in the Id protein suggesting that if a heterodimer was formed with the Id protein and myogenic bHLH the heterodimer wouldn't bind as well, therefore inhibiting activation of transcription. Whereas if leucine zippers are present, the E protein and myogenic bHLH forms heterodimers, and it promotes dimerization, which is important for transcription.

When an activator interacts with mediator, what is the result? TFIID can bind to the core promoter. RNA polymerase can bind to the core promoter. RNA polymerase can switch to the elongation phase of transcription. Histones are displaced from the core promoter.

RNA polymerase can switch to the elongation phase of transcription.

What does the term RNA interference (RNAi) mean? Describe how RNAi occurs and how it leads to a silencing of gene expression.

RNAi is when double stranded RNA molecules interfere with mRNA to inhibit gene expression. It's mediated by two methods: small interfering RNAs and microRNAs. pre-miRNA is transcribed or pre-siRNA enters the cell from exogenous sources and is cleaved by dicer to achieve si/miRNAs. The si/miRNA associates with proteins to form a RISC complex, and one of the RNA strands is degraded. When RISC binds to the mRNA using the remaining si/miRNA strand, it may inhibit translation, which is common for miRNAs, remain in a P-body to be stored, or degrade the mRNA, which is common for siRNAs

Which of the following is not a function of heterochromatin formation? Gene silencing Prevention of transposable element movement Removal of histones Prevention of viral proliferation

Removal of histones

Which of the following statements regarding retrotransposons (also called retroelements) is false? Retrotransposons are commonly found in prokaryotic genomes. Some retrotransposons are similar in structure to known viruses. Retrotransposons are most abundant in the genomes of complex, multicellular eukaryotes. Retrotransposons require reverse transcriptase to be inserted into a new site.

Retrotransposons are commonly found in prokaryotic genomes.

Three common ways to modulate the activities of transcription factors are (1) the binding of an effector molecule, (2) protein-protein interactions, and (3) covalent modifications. Which of these mechanisms is/are used by steroid receptors and the CREB protein?

Steroid receptors use the binding of an effector molecule and protein-protein interactions. A glucocorticoid molecule acts as an effector molecule for steroid receptors, and the steroid receptors complex with HSP90, which is an example of a protein-protein interaction. CREB protein uses protein-protein interactions and covalent modifications. The CREB protein interacts with CBP, which is an example of a protein-protein interaction. The phosphorylation of the CREB protein by protein kinase A is an example of a covalent modification because the CREB protein is phosphorylated.

A diagram of a linear chromosome is shown here. The end of each strand is labeled with an A, B, C, or D. Which ends could not be replicated by DNA polymerase? Why not? Note: The question is asking you which ends cannot be copied. Look at the two GC base pairs on the left and the two AT base pairs on the right. Would it not be possible to synthesize the GG (at the A end), the AA (at the B end), the CC (at the C end) and/or the TT (at the D end)? 5′-A—GG—————————————————————————————AA—B-3′ 3′-C—CC————————————————————————————— TT—D-5′

The B and C ends could not be replicated by DNA polymerase. It would not be possible to replicate the AA at the B end and the CC at the C end. This is because DNA polymerase creates strands from 5' to 3' using each template, sliding from 3' to 5'. At the 3' ends of the template DNA, there is no more area upstream for a primer to be made. DNA polymerase requires a primer - it cannot add nucleotides without one, so if no primer can be placed before these ends, the 3' ends will not be copied.

Let's suppose a mutation deletes the ICR that is located between the H19 and Igf2 genes. If this mutation was inherited from the mother, would the Igf2 gene (from the mother) be silenced or expressed? Explain.

The Igf2 gene is only expressed when inherited from the father. This is due to methylation of the ICR and DMR segments which allows enhancers to express the Igf2 gene. In females, there is a CTCF dimer that binds to the unmethylated ICR and DMR regions to form a loop. In this case, the mother does not have the ICR segment, which means that the CTCF dimer will not bind to it in order to form a loop. Therefore, since the loop is not present, the Igf2 gene will be stimulated by the enhancer and the gene (from the mother) will be expressed.

Which of the following does not happen for CREB protein to activate a gene? Protein-protein interaction (e.g., dimer formation) Covalent modification of the CREB protein The binding of a small effector molecule to the CREB protein All of the above happen.

The binding of a small effector molecule to the CREB protein

Take a look at the experimental results shown in Figure 16.13. What is the relationship between coat color and DNA methylation? How is coat color related to the diet of the mother?

The coat colors of the offspring were correlated with the degree of methylation that occurred in CpG islands in the transposable element. Offspring with greater DNA methylation had darker coats. Offspring of mothers that were fed a supplemental diet with folic acid, vitamin B12, betaine, and choline chloride (nutrients that increase the synthesis of S-adenosyl methionine in cells), had darker coats than those fed a normal diet. This diet resulted in an increase in methylation resulting in darker coats of the mice.

Let's suppose you have isolated a mutant strain of E. coli in which the lac operon is constitutively expressed. You create a merozygote in which the mutant strain also contains an F' factor with a normal lac operon and a normal lacI gene. You then compare the mutant strain and the merozygote with regard to their β-galactosidase activities in the presence and absence of lactose. You obtain the following results: Strain Addition of lactose (no glucose) Amount β-Galactosidase Normal Yes 100 Mutant No 100 Mutant Yes 100 Merozygote No 1 Merozygote Yes 200 Which of the following is consistent with these data? The mutation is in the lacI gene and results in a lac repressor that cannot bind to the operator. The mutation is in the lac operator and prevents the lac repressor from binding to the operator. The mutation is in the lacI gene and results in a lac repressor that cannot bind allolactose. All of the above.

The mutation is in the lacI gene and results in a lac repressor that cannot bind to the operator.

Let's suppose that a species of fruit flies has two different types of DNA transposons, which we will call S elements and M elements. The S elements are so named because they are quite stable. When analyzing a population of 47 flies, every fly has seven S elements, and they are always located in the same chromosomal locations among different individuals. In contrast, the M elements seem to be mobile. Within the same 47 flies, the number of M elements ranges from 2 to 14, and the locations of the M elements tend to vary considerably among different individuals. Explain how one simple transposon can be stable and another simple transposon can be mobile, within the same group of flies.

The recognition of inverted repeats by transposase allows for the removal and reintegration of transposable elements. It is possible that the fruit flies do not have a transposase for the S elements, which means they would not be removed and transposed. This would make it a nonautonomous element. On the other hand, M elements must have a transposase that allows them to be mobile. They must be autonomous elements.

The ferritin mRNA in mammals is controlled by the iron regulatory protein (IRP) that binds to an iron response element (IRE) near the 5' end of the mRNA. What do you expect to happen when iron levels are low? The translation of the mRNA would be high because the IRP would not bind to the IRE. The translation of the mRNA would be low because the IRP would not bind to the IRE. The translation of the mRNA would be high because the IRP would bind to the IRE. The translation of the mRNA would be low because the IRP would bind to the IRE.

The translation of the mRNA would be low because the IRP would bind to the IRE.

The Avy allele of the Agouti gene involves the insertion of a transposable element upstream from the normal Agouti promoter. The transposable element carries a promoter that causes the over expression of the Agouti gene. Mice carrying this allele tend to have coat colors that are more yellow than mice that don't have this transposable element. If pregnant female mice are fed a diet that contains chemicals that increase DNA methylation, how would you expect that this diet would affect the coat color of offspring carrying the Avy allele? Their fur would be more yellow because the Agouti gene would tend to be over expressed. Their fur would be more yellow because the Agouti gene would tend to be under expressed. Their fur would be less yellow (more dark brown) because the Agouti gene would tend to be over expressed. Their fur would be less yellow (more dark brown) because the Agouti gene would tend to be under expressed.

Their fur would be less yellow (more dark brown) because the Agouti gene would tend to be under expressed.

In the lagging strand, DNA is made in the direction _________ the replication fork and is made as __________. toward, one continuous strand away from, one continuous strand toward, Okazaki fragments away from, Okazaki fragments

away from, Okazaki fragments

The proofreading function of DNA polymerase involves the recognition of a ________ and the removal of a short segment of DNA in the __________ direction. missing base, 5' to 3' base mismatch, 5' to 3' missing base, 3' to 5' base mismatch, 3' to 5'

base mismatch, 3' to 5'

The ___________ is the feature that allows DNA to store information. spiral structure relative locations of the major and minor groove base sequence 5' to 3' direction of the strands

base sequence

During transcription, the removal of histone proteins from the DNA is important for the phosphorylation of CTD. the formation of the preinitiation complex. transcriptional elongation. both B and C.

both B and C.

Which component(s) of a nucleotide is/are found along the DNA backbone? bases deoxyribose phosphate both b and c

both b and c

The enzyme called phosphodiesterase hydrolyzes cAMP into AMP. Researchers have identified an E. coli strain in which this phosphodiesterase works 10-times faster than a normal E. coli strain. How would this overactive phosphodiesterase affect diauxic growth?

cAMP is a small effector molecule that is necessary to induce binding of the activator protein CAP. The cAMP-CAP complex binds to the CAP site near the lac promoter, which enhances the transcription of the lac operon. A higher rate of cAMP hydrolyzation would prevent CAP from binding, so the genes in the lac operon would be transcribed at a lower rate than if CAP was bound. This overactive phosphodiesterase would not enhance the transcription of genes that help with lactose metabolism in diauxic growth.

Which of the following could be the components of a single nucleotide found within a DNA strand? deoxyribose, thymine, and phosphate deoxyribose, guanine, and three phosphates ribose, thymine, and phosphate deoxyribose, uracil, and phosphate

deoxyribose, thymine, and phosphate

In the image below, the ncRNA is acting as a decoy guide scaffold blocker

guide

For a riboswitch that controls transcription, the binding of a small molecule such as TPP controls whether the RNA has an antiterminator or terminator stem-loop. has a Shine-Dalgarno antisequestor or the Shine-Dalgarno sequence in a stem-loop. is degraded from its 5' end. has both a and b.

has an antiterminator or terminator stem-loop.

A chromosome territory is a region of a chromosome that carries a cluster of genes. of a chromosome that has particular sets of DNA-binding proteins. in a cell nucleus that is occupied by a single chromosome. in a dividing cell where a chromosome will travel via the spindle apparatus.

in a cell nucleus that is occupied by a single chromosome.

A small effector molecule that enhances transcription binds to a regulatory protein and causes it to not bind to the DNA. The regulatory protein is a repressor. is an activator. could be a repressor or an activator. is neither a repressor or an activator.

is a repressor.

The two things that DNA polymerase cannot do is to synthesize DNA in the 5' to 3' direction and begin synthesis on a bare template strand. synthesize DNA in the 3' to 5' direction and begin synthesis on a bare template strand. use an RNA primer and proofread the DNA. proofread the DNA and synthesize the DNA in the 5' to 3' direction.

synthesize DNA in the 3' to 5' direction and begin synthesis on a bare template strand.

What is the underlying structural reason why DNA can be replicated to produce two double helices with the identical base sequences? base stacking because the strands are antiparallel the AT/GC rule because DNA polymerase can make DNA only in the 5' to 3' direction

the AT/GC rule

Chargaff's results suggest that (in DNA), the amount of G = C and A = T. the amount of G = A = C = T. the amount of G = T and A = C. all of the above.

the amount of G = C and A = T.

During diauxic growth involving glucose and lactose, bacterial cells use up lactose first, then they metabolize glucose. use up glucose first, then they metabolize lactose. metabolize both sugars at maximal rates at the same time. None of the above

use up glucose first, then they metabolize lactose.