Biochemistry Exam 3 Squarecap

At which position on the tRNA is the wobble base of the triplet anticodon? A. 1st position (5' letter) B. 2nd position C. 3rd position (3' letter) D. Wobble can occur at any position.

A. 1st position (5' letter) Explanation: Recall that the wobble base of the codon is in the 3rd position. Because of antiparallel base pairing, it pairs with the first base in the anticodon.

What is the oxidizing agent in the irreversible reaction below? CH3CH2CHO + NADH --> CH3CH2CH2OH + NAD+ A. CH3CH2CHO B. NADH C. CH3CH2CH2OH D. NAD+

A. CH3CH2CHO Explanation: Since the reaction is irreversible we must consider the forward reaction only. The oxidizing agent is the agent that oxidizes another molecule and is therefore reduced. Propanal (aldehyde) is reduced to propanol and NADH is oxidized to NAD+. Therefore propanal is the oxidizing agent.

How do enhancers differ from promoters? A. Enhancers do not bind RNA polymerase. B. Enhancers include the UP element. C. Enhancers bind the sigma factor. D. There is no difference; these terms are synonymous.

A. Enhancers do not bind RNA polymerase. Explanation: Enhancers bind transcription factors that interact with RNA polymerase to elevate transcription. Remember, RNA polymerase binds to the core promoter.

Which of the following is not true about RISC? A. It is only involved in cleaving dsRNA from viruses. B. It is involved with processing of miRNA and siRNA. C. It uses a protein from the argonaut family. D. It unwinds dsRNA and eventually discards the passenger strand.

A. It is only involved in cleaving dsRNA from viruses. Explanation: The RISC or RNA-induced silencing complex operates to control gene expression as a normal function in eukaryotes.

Which of the following is not a function of elongation factors in E. coli? A. Keep the ribosomal subunits bound together. B. Lead the incoming tRNA to its correct position on the ribosome. C. Carry GTP to provide energy for certain steps in the elongation process. D. Regenerate the elongation factors by replacing GTP for GDP. E. The elongation factors do all of these.

A. Keep the ribosomal subunits bound together. Explanation: The large and small ribosomal subunits have an affinity for each other and spontaneously associate. Recall that there is an initiation factor (IF3) that must prevent their premature association.

Which of the following is true concerning the first nucleotide incorporated into an RNA chain? A. The first nucleotide incorporated into the RNA chain retains its 5'-triphosphate. B. The first nucleotide is always GMP. C. The first nucleotide is always cleaved off post-transcriptionally. D. The first nucleotide is always modified after transcription.

A. The first nucleotide incorporated into the RNA chain retains its 5'-triphosphate. Explanation: This is the consequence of building the chain without attaching to a primer. The first nucleotide is not attached to another on its 5' end - so it carried all 3 phosphate groups.

How do the core enzyme and the holoenzyme of RNA polymerase differ in E. coli? A. The holoenzyme includes the sigma subunit, the core enzyme does not. B. The core enzyme includes the sigma subunit, the holoenzyme does not. C. The holoenzyme transcribes from an RNA template, the core enzyme from a DNA template. D. The core enzyme transcribes from an RNA template, the holoenzyme from a DNA template

A. The holoenzyme includes the sigma subunit, the core enzyme does not. Explanation: The core enzyme is responsible for RNA synthesis in terms of elongation and termination - but the sigma factor is required for promoter recognition and initiation.

Hybridomas, which produce monoclonal antibodies, are made by fusing cells of the immune system with A. cancerous cells. B. viruses. C. bacterial cells. D. ribosomes.

A. cancerous cells. Explanation: You'll want to keep in mind the goal in producing the hybridomas - to establishment a reproducible cell line that produces one type of antibody. We need, then, a B cell that will product the antibody but we need a way of perpetuating indefinitely in culture. This is what the cancer or myeloma cell provides.

A virus life cycle that involves the incorporation of the viral DNA into the host chromosome is A. lysogeny. B. lytic. C. oncogenic. D. non-existent.

A. lysogeny. Explanation: You can think of lysogeny as a cycle in which there is the potential for ("ogeny") cell lysis ("lys"). In contrast, the lytic cycle results in direct cell lysis. It's also important to recognize that some viruses use only one or the other of these cycles - others have options depending on the host cell status.

The eukaryotic TATA-binding protein (TBP) functions in a manner similar to A. sigma factor in E. coli. B. rho factor in E. coli. C. lac I in E. coli. D. CAP in E. coli.

A. sigma factor in E. coli. Explanation: Recall that sigma factor is responsible for directing RNA polymerase to a promoter. In eukaryotic systems, this role is performed by TBP - which is a subunit of TFIID. TBP is also required for Pol I and Pol II enzymes, and is therefore considered a universal transcription factor.

Consider this rxn which has a ΔG° = +0.4 kJ/mol. A + B --> C + D 1 M A, 1 M B, 0.1 M C and 0.1 M D are added to a container at room temperature. Which of the following statements is true? A. ΔG < 0 B. ΔG = 0 C. ΔG > 0 D. Cannot be determined from the information provided.

A. ΔG < 0 Explanation: Since the substrates are at so much higher concentrations than products - we know that the forward reaction is favored and therefore represents a negative change in free energy. In other words, it is favored by the law of mass action.

Formation of an RNA hairpin cannot be the sole factor in the termination of transcription in prokaryotes. Explain why.

As transcription proceeds, the nascent RNA forms a variety of secondary structures as portions of the transcript form complementary base pairs. The formation of these secondary structures may cause transcription to pause but not necessarily terminate.

The first tRNA (bearing fmet) binds to the A. "A" site of the ribosome B. "P" site on the ribosome C. "E" site on the ribosome

B. "P" site on the ribosome Explanation: It's important to recognize that the first tRNA MUST bind to the P site, whereas all subsequent tRNAs will bind to the A site. If the initiator tRNA was bound at the A site - there would be nowhere for the next tRNA to bind in order to form the first peptide bond.

Which of following best describes the two-step process for adding amino acids to tRNAs? AA = amino acid, Pi = phosphate, PPi = pyrophosphate A. AA + ATP → AA−P + ADP; 2) AA−P + tRNA → AA−tRNA + Pi B. AA + ATP → AA−AMP + PPi; 2) AA−AMP + tRNA → AA−tRNA + AMP C. tRNA + ATP → tRNA−P + ADP; 2) tRNA−P + AA → AA−tRNA + Pi D. tRNA + ATP → tRNA−AMP + PPi; 2) tRNA−AMP + AA → AA−tRNA + AMP

B. AA + ATP → AA−AMP + PPi; 2) AA−AMP + tRNA → AA−tRNA + AMP Explanation: The important thing to recognize here is that the amino acid is charged before being attached to the tRNA. Also, it's important to recognize that it is AMP that is transferred to the carboxyl group with the release of PPi. Hydrolysis of PPi by pyrophosphatase is what makes this process irreversible.

Positive selection results from A. DP cells that react to self-antigens with high affinity. B. DP cells that react to self-antigens with low affinity. C. DP cells that don't react in any degree to self-antigens.

B. DP cells that react to self-antigens with low affinity. Explanation: DP cells that don't recognize anything die by neglect. Those that bind to self-antigens with high affinity are eliminated by negative selection. Those that bind with low affinity are allowed to persist - hence the positive selection. These require a second signal for activation and hence are prevented from acting prematurely or inappropriately.

More than one type of antigen can bind to an antibody, but there is only one type of antibody for each antigen. A. True B. False

B. False Explanation: Each antibody is specific for one epitope. The same epitope may be present on different antigens and the same antibody could bind to either. Each antigen carries multiple epitopes and therefore binding sites for antibodies with multiple and variable specificities.

The capsid of a virus is composed of A. DNA or RNA B. Protein C. Lipid D. Carbohydrate

B. Protein Explanation: The capsid is the protein coat that protects the genome. All viruses have a capsid but not all also have a membrane envelope.

All of the following apply to attenuation mechanisms, except: A. They are used most often for operons for amino acid synthesis. B. The rate of RNA synthesis is regulated by the conformation of the protein being synthesized. C. They involve the synthesis of proteins in the regulation of RNA synthesis. D. They require the presence of partially completed mRNA molecules. E. All of these apply to attenuation mechanisms.

B. The rate of RNA synthesis is regulated by the conformation of the protein being synthesized. Explanation: Regulations depends on coupled transcription and translation - but only a short leader peptide is produced. It lacks any real conformation that could affect transcription.

Dendritic cells A. secrete cytokines, recruiting other cells to destroy microbes. B. display antigens associated with MHC proteins. C. produce antibodies. D. bind to antigens, which leads to their proliferation.

B. display antigens associated with MHC proteins. Explanation: The main function of dendritic cells is to recognize foreign invaders and alert those cells that can destroy them. It engulfs the antigen (or cell) and digests it. It then displays a portion of the antigen on its surface in order to alert the immune system.

Genetic engineers must modify eukaryotic genes so that they can be expressed in bacterial host cells. The DNA from a eukaryotic gene cannot be placed directly into the bacteria but is first transcribed to mRNA and then reverse-transcribed back to cDNA. Explain why this is necessary.

Bacterial genes contain no introns, so the cells, which lack splicing machinery, cannot express eukaryotic genes containing introns and exons. Mature eukaryotic mRNA, which has already been spliced, contains only exons, which, after they have been converted back to DNA, can be transcribed and translated by the bacteria.

What happens when a stop codon is reached by a ribosome (in the A site)? A. A termination tRNAter binds to the codon and is used to release the growing peptide from the P site tRNA. The ribosome then dissociates. B. A termination tRNAter binds to the codon and the growing peptide is transferred to it. When the peptidyl-tRNAter reaches the P site, the ribosome is signaled to release the protein. The ribosome then dissociates. C. A termination protein binds to the codon and is used to release the growing peptide from the P site tRNA. The ribosome then is likely to dissociate. D. A termination protein binds to the codon and the ribosome dissociates. A separate peptidyl transferase then releases the protein from the last tRNA to which was attached.

C. A termination protein binds to the codon and is used to release the growing peptide from the P site tRNA. The ribosome then is likely to dissociate. Explanation: There is no tRNA that binds to the stop codons - so both A) and B) are false. The release factor binds to the stop codon and stimulates the ribosome to hydrolyze the ester bond connecting the polypeptide chain to the tRNA in the P site. We need a separate factor - the RRF or ribosome recycling factor - to break the interactions between the large and small subunit and release the message.

How does an expression vector differ from a regular cloning vector? A. An expression vector does not have an origin of replication. B. An expression vector is always a linear molecule, while a cloning vector may be circular. C. An expression vector has the ability to have the inserted DNA be transcribed. D. An expression vector must be of viral origin, so that it can infect a cell naturally.

C. An expression vector has the ability to have the inserted DNA be transcribed. Explanation: A cloning vector is used to insert and amplify foreign DNA - not necessarily its expressed product. In the case of an expression vector, we want to clone the gene so that it can be expressed in the host.

What is the main function of TH cells? A. They function to stimulate B cells to become antibody-producing cells. B. They function to stimulate TC cells to destroy infected cells. C. Both A and B are true. D. Neither A nor B are true.

C. Both A and B are true. Explanation: Helper T cells function in two capacities - though this requires differentiation separately and in response to different cytokines. They help to stimulate B cells as well as TC cells to perform their separate roles.

Which of the following is the most direct reason that AIDS is deadly? A. HIV is a retrovirus. B. HIV enters helper T-cells. C. Budding HIV particles damage helper T-cell membranes lowering the helper T-cell count. D. HIV has an RNA genome. E. None of these explain why HIV is deadly.

C. Budding HIV particles damage helper T-cell membranes lowering the helper T-cell count. Explanation: It is not the nature of the genome that makes HIV life threatening. It's true that it targets immune cells that could help destroy it - but just entering the cell is not the problem. It's the fact that it replicates and continues to lower helper T cell count that makes it deadly.

Which of the following cell types of innate immunity kills cells that have been infected by viruses or that are cancerous? A. dendritic cells B. T cells C. NK cells D. B cells

C. NK cells Explanation: Remember, dendritic cells target foreign cells. Natural killer cells target normal cells that have been hijacked by some invader or that have become a threat (cancerous).

What provides the energy for the conversion from the open complex to chain elongation? A. ATP hydrolysis distinct from any incorporation into the chain B. Nucleotide hydrolysis associated with incorporation into the chain C. Torsional stress built into the separating DNA strands D. Binding of rho factor to the holoenzyme E. None of these

C. Torsional stress built into the separating DNA strands Explanation: RNA polymerase essentially launches itself from the promoter by scrunching the DNA and thereby creating torsional stress.

A viral genome is A. always double-stranded B. always single-stranded C. may be double or single stranded D. always made of DNA E. always made of RNA

C. may be double or single stranded Explanation: The basic lesson here is that there is no "always" when it comes to viral genomes. They may be either DNA or RNA and single-stranded, double-stranded, or partially double-stranded.

A mutation in the lacA gene would result in A. continuous production of the proteins encoded by the three structural genes. B. continuous production of the lac repressor. C. normal operation of the lac operon, but with an alteration in the proteins encoded by the lacA gene. D. no transcription from the lac operon.

C. normal operation of the lac operon, but with an alteration in the proteins encoded by the lacA gene. Explanation: The lacA gene encodes the transacetylase and has no effect on regulation of the operon.

In DNA sequencing, fragments to be analyzed are produced by A. acid hydrolysis B. base hydrolysis C. selective interruption of DNA synthesis D. exposure to 32P

C. selective interruption of DNA synthesis Explanation: The Sanger-Coulson method of DNA sequencing is referred to as the chain termination method. Incorporation of a dideoxynucleotide halts synthesis - since there is no O atom on the 3' carbon to act as the nucleophile.

Using the genetic code provided, identify the amino acid sequence (using 1- or 3-letter codes) of the peptide encoded by the template DNA sequence shown below. 5'-T G C T A C G A G A T C G C T T G G T C G G A C A T T A T C G-3'

Coding strand: 5'-C G A U A A U G U C C G A C C A A G C G A U C U C G U A G C A-3' After getting the sequence of the coding strand in the 5' to 3' direction, I then looked for the first AUG as being the start codon for translation. This sets the reading frame for the 3-base codons. I then noticed that the UAG stop codon marks the end of the reading frame. The sequence of 3-letter codes specifies the following amino acid sequence: M S D Q A I S (Met-Ser-Asp-Gln-Ala-Ile-Ser).

I am performing a reaction, A --> B, with ΔG°' = −0.3 kJ/mol. I start the reaction with 10 mM A and no B. After allowing the reaction to proceed for 24 hrs at room temperature and atmospheric pressure, I analyze a sample of the reaction mix to find I now have 1 mM A and 9 mM B. Which of the following conclusions should I make? ∆Gº = -RTlnKeq R = 8.3145 J/mol•K T = ºC +273 A. The reaction has reached equilibrium. B. I should come back again later; equilibrium has not yet been reached. C. The formation of B from A is thermodynamically unfavorable, so I should find another starting material to make B. D. I must've screwed up; there's no way I could get that result with that ΔG°'

D. I must've screwed up; there's no way I could get that result with that ΔG°' Explanation: ∆Gº = -RTlnKeq -300 J/mol ÷ (-8.3145 J/mol•K) ÷ (298 K) = 0.121079 e^-0.121079 = 1.1287 = 1.1 = Keq Your results suggest this value is 9 (9/1) - that can't be right!

Considering the conditions below, indicate the amount of transcription of the lac operon from least to greatest. I. High glucose ; high lactose II. Low glucose ; low lactose III. High glucose ; low lactose IV. Low glucose ; high lactose A. I < II < IV < III B. III < I < IV < II C. IV < I < II < III D. III < II < I < IV

D. III < II < I < IV Explanation: It may help to know that induction of the operon (high lactose) can increase expression up to 1000-fold. Activation of the operon (low glucose) can increase expression by 20-50 fold. If glucose is high, catabolite repression is working and if lactose is also low there is no induction - this is the lowest amount of expression of the operon (no induction; no activation). When glucose and lactose concentrations drop - catabolite repression no longer applies (activation increases expression ~50-fold) but there is no induction. If both glucose and lactose concentrations are high - there is no activation of the operon but it is induced by the presence of allolactose (1000-fold increase). The highest expression of the operon occurs when lactose is high (induction) and glucose is low (activation).

Which of the following is not an advantage of degeneracy in the genetic code? A. Each tRNA can bind to more than one codon. B. Cells need to invest less energy in the synthesis of tRNAs. C. Fewer tRNA molecules are needed. D. More than one amino acid can bind to a tRNA.

D. More than one amino acid can bind to a tRNA. Explanation: The fact that more than one codon can specify the same amino acid, means that the same tRNA may bind more than one codon. This also means that fewer tRNA molecules are needed - making it more efficient. However, the linking of the amino acid to its cognate tRNA MUST be specific.

The protein called p53 is especially important in cancer because: A. Protein p53 has the ability to prevent cells from dividing. B. Protein p53 has the ability to repair mutations. C. Protein p53 can cause damaged cells to enter programmed cell death. D. Protein p53 can prevent cells from dividing and cause them to enter programmed cell death. E. All of these.

D. Protein p53 can prevent cells from dividing and cause them to enter programmed cell death. Explanation: Recall that p53 is a tumor suppressor - in that it halts cell division if DNA damage is detected and promotes apoptosis if repair mechanisms fail.

Which of the following best describes the assembly process to initiate translation in prokaryotes? A. Small ribosome subunit associates with large subunit; mRNA associates next; IFs bring in first f-Met-tRNA; bound GTP hydrolyzed. B. IF binds f-Met-tRNA; small ribosome subunit associates next; mRNA associates next; then large subunit; bound GTP hydrolyzed. C. IF w/ bound GTP binds f-Met-tRNA; mRNA associates next; small subunit associates next; bound GTP hydrolyzed; large subunit associates. D. Small subunit associates with mRNA; IF w/ bound GTP brings in first f-Met-tRNA; large subunit associates; IF then hydrolyses bound GTP.

D. Small subunit associates with mRNA; IF w/ bound GTP brings in first f-Met-tRNA; large subunit associates; IF then hydrolyses bound GTP. Explanation: The association of the small ribosomal subunit with the mRNA positions it at the correct location in order to initiate translation. IF-2 then escorts the initiator tRNA to the ribosome.

How do TC cells bind and recognize antigen on APCs? A. The TCR binds to MHC I and CD8 binds antigen. B. The TCR binds to MHC II and CD8 binds antigen. C. The TCR binds to antigen and CD8 binds B7. D. The TCR binds to antigen and CD8 binds to MHC I. E. The TCR binds to antigen and CD8 binds to MHC II.

D. The TCR binds to antigen and CD8 binds to MHC I. Explanation: This is a good example of the various mechanisms that ensure proper recognition and response and prevent premature or inappropriate responses. The T cell receptor is constructed in such a way as to be specific for a particular antigen and this will vary with each T cell. However ALL TCcells carry the CD8 protein - so that all can recognize the same MHC I molecule on the APC.

What is the need for a primer in transcription? A. It ensures the fidelity of the newly synthesized RNA strand. B. RNA polymerase requires a preexisting strand with a nucleotide having a 3'-OH. C. RNA polymerase requires a preexisting strand with a nucleotide having a 5'-OH. D. There is none.

D. There is none. Explanation: This is one of the major differences between DNA and RNA polymerases. RNA polymerase can carry out de novo synthesis - can start synthesis "from scratch."

What is the function of protein spikes on a virus? A. They protect the genome. B. They are involved with incorporation of the genome into that of host (lysogeny). C. They transcribe the RNA genome into a DNA copy. D. They help the virus attach to the host.

D. They help the virus attach to the host. Explanation: This is an important consideration - as it will determine which cells it can infect. Only those cells that carry cell-surface proteins that the viral protein can bind to, can be infected.

Long terminal repeats (LTRs) are involved in which part of the retroviral lifecycle? A. replicating the viral DNA B. creating the nucleocapsid C. production of spike proteins D. incorporating the viral genome into the host genome E. none of these

D. incorporating the viral genome into the host genome Explanation: LTRs are part of the viral DNA and enables it to be recognized by the enzyme involved in integrating it into the host DNA.

In general, catabolism A. is a reductive process that releases energy B. is an oxidative process that requires energy C. is a reductive process that requires energy D. is an oxidative process that releases energy E. none of these

D. is an oxidative process that releases energy Explanation: Oxidative pathways break down larger molecules into smaller ones, thereby releasing energy.

The usefulness of blotting techniques in molecular biology is that A. spills of hazardous chemicals are minimized B. only the substance of interest is transferred to a nitrocellulose disk C. it directly gives rise to a genetic map D. transferred material is in the same relative position on the disk as on the original sample

D. transferred material is in the same relative position on the disk as on the original sample Explanation: The benefit of direct transfer from gel to blot, is that DNA bands can be easily identified in terms of restriction fragment identity and size.

During transpeptidation, nucleophilic attack occurs between the α-amine group in the ____ site and the esterified carbon in the ____ site. A. P;E B. E;A C. P;A D. A;E E. A;P

E. A;P Explanation: The significance here is that the direction of attack determines the order of the amino acid sequence - from the N to the C terminus.

Which of the following are methods used to determine where the DNA bands are located on an electrophoresis gel? A. Radioactivity B. Fluorescence C. Dyes which bind to DNA D. Luminescence E. All of these can visualize the DNA

E. All of these can visualize the DNA Explanation: All of these methods allow us to distinguish the bands of DNA and to measure their relative migration distances.

Which of the following parameters affects the distance DNA molecules migrate during electrophoresis, at pH = 8? A. The mass of the DNA B. The total ionic charge on the DNA molecule C. The fact that each nucleotide contributes one negative charge at this pH D. The concentration of agarose or polyacrylamide in the gel E. All of these features control the distance the DNA migrates

E. All of these features control the distance the DNA migrates Explanation: DNA molecules are separated based on their charge-to-mass ratio - which means both size and charge matter - and therefore the fact that each nucleotide adds one negative charge owing to the presence of the phosphate groups. The separation also depends on the concentration of the matrix component - as this determines the pore size and the degree of diffusion. Generally speaking, a larger pore size is used to separate larger fragments and smaller pore sizes for smaller fragments.

Restriction enzymes are especially useful for genetic recombination work for all of the following reasons, except: A. They cut DNA in the middle of specific sequences. B. They cut DNA independent of the source of the DNA. C. They often generate single stranded tails or "sticky ends." D. There are a large variety of them commercially available. E. All of these traits make restriction enzymes useful.

E. All of these traits make restriction enzymes useful. Explanation: The specificity of restriction endonucleases means that we can predict where they will cut - it will not be random. So long as the DNA is not methylated as the host DNA, the enzyme will cut any DNA molecule that contains an unmethylated sequence - regardless of source. There is, in fact, a large variety of enzymes - in terms of recognition sites and how they cut.

Which of the following is not a reason that AIDS has been difficult to eradicate? A. It is slow acting. B. The HIV reverse transcriptase is inaccurate. C. The gp120 protein makes a conformational change when it binds to the CD4 receptor causing antibodies to be ineffective. D. HIV evades the innate immune system. E. It gives the infected individual multiple symptoms within weeks of infection.

E. It gives the infected individual multiple symptoms within weeks of infection. Explanation: All of these factors contribute to the difficulty in treating and eradicating AIDS - except the last choice. In fact, it is the absence of symptoms in the early stages that make it so difficult to treat. Early symptom presentation would mean treatment might begin earlier and therefore be more effective.

Some viruses can undergo lysis or lysogeny within the same host. Under what conditions might the virus favor lysis instead of lysogeny?

If the host cell is healthy and the virus is well supplied with nutrients then the virus has sufficient material to enable it to move completely through the lytic cycle. In contrast, if the cell is starved or unhealthy - the virus will likely utilize its lysogenic cycle until conditions improve.

Describe the positive and negative effects of the innate immune system on cancer cells.

The innate immune system is instrumental in fighting cancer cells. Cells that turn cancerous display antigens on their surfaces that act as a help signal. Cells of the innate immune system such as macrophages and natural killer cells attack cells that display these cancer-linked antigens on their surfaces. Often they destroy the cancerous cell, ending the threat. However, if they do not, the presence of the innate immune cell can lead to inflammation. More and more research is showing that inflammation is the switch that takes a precancerous cell and turns it into a full-fledged cancer cell. Thus, innate immune cells that attack a cancer cell but fail to kill it may just make it stronger.

In bacteria, the core RNA polymerase binds to DNA with a dissociation constant of 5 x 10^-12 M. In complex with its sigma factor, the polymerase has a dissociation constant of 10^-7. Explain.

The presence of the sigma factor decreases the affinity of RNA polymerase for DNA. This allows the polymerase-sigma factor complex to quickly scan long segments of DNA for promoter sequences. Once transcription has begun, the sigma factor is no longer needed and dissociates from the enzymes. Now the RNA polymerase has a high affinity for DNA, which helps keep it associated with the template during transcription.

The following pair of PCR primers has a problem with it. Why will the primers not work well? Forward primer: 5'-TCGAATTGCCAATGAAGGTCCG-3' Reverse primer: 5'-CGGACCTTCATTGGCAATTCGA-3'

The primers are complementary and will anneal to each other faster than they anneal to the DNA templates.

Given below is a table of the free energies of hydrolysis of several compounds. Using this data, calculate the value of ∆Gº' for the following reaction in kJ/mol: Creatine phosphate + Glycerol --> Creatine + Glycerol-3-phosphate Express answer to one decimal in units of kJ/mol.

The two steps are hydrolysis of creatine phosphate (-43.1 kJ/mol) and then phosphoryl transfer to glycerol - which is the reverse of hydrolysis of Glycerol-3-phosphate (-9.7 kJ/mol). The combined reactions would therefore be: -43.1 kJ/mol + 9.7 kJ/mol = -33.4 kJ/mol). This is a good example of how coupling the hydrolysis of a high-energy bond with an unfavorable step results in a reaction that is favorable overall.