Molecular Bio Flashcards by Summer Moore

How many ATP equivalents are required to translate a 100-amino acid protein?

400

It takes 200 ATP for every 50 amino acids made

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Where is the START codon bound in a ribosome?

The P site

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How many types of RNA polymerases are used by eukaryotic cells?

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Which of the following is NOT a required component of replication?

A. Helicase

B. 3’ to 5’ synthesis with DNA polymerase

C. Primers made of RNA

D. DNA template

B. 3’ to 5’ synthesis with DNA polymerase

DNA is synthesized from 5’ to 3’

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Which of the following identifies a point mutation that creates a STOP codon?

A. Missense mutation

B. Nonsense mutation

C. Frameshift mutation

D. Silent mutation

B. Nonsense mutation

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What are the necessary components of a nucleotide?

A pentose, an aromatic base, and phosphates

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Which of the following is NOT considered to be a DNA repair mechanism?

A. Nucleotide excision removes defective bases and replaces them

B. Telomerase lengthens the ends of chromosomes where primase is unable to bind

C. Photoreactivation reverses pyrimidine dimers caused by UV radiation

D. Non-homologous end joining connects broken ends of chromosomes

B. Telomerase lengthens the ends of chromosomes where primase is unable to bind

This simply allows for more effective replication of the end of the chromosome. Connecting broken ends of chromosomes, reversing UV damage, and removing defective bases are all forms of DNA repair.

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What is The Shine-Dalgarno sequence?

The Shine-Dalgarno sequence is the ribosome binding site in prokaryotes (similar to the Kozak sequence in eukaryotes) and without this the gene product cannot be translated.

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Which of the following is NOT necessary for prokaryotic translation?

A. fMet

B. GTP

C. Shine-Dalgarno sequence

D. 80S ribosome

Eukaryotes have 80S ribosomes.

In prokaryotes, the 30S and 50S subunits of the ribosome must come together for translation to occur but when they do they make a 70S ribosome.

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What is a fMET?

fMet (formylmethionine) is a modified methionine used as the first amino acid in all prokaryotic proteins.

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What is GTP?

GTP is very similar to ATP but it contains a different purine base (guanine replaces adenine). Prokaryotic transcription uses GTP as its energy source, not ATP.

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Fluoroquinolones are a class of antibiotics that inhibit DNA gyrase. Which of the following would be inhibited?

A. Condensing of DNA

B. Releasing of ribosomal subunits after transcription

C. Connecting the Okazaki fragments during DNA replication

D. unwinding of DNA at the origin of replication

A. Condensing of DNA

Gyrase is an enzyme specific to prokaryotes that maintains bacterial DNA in its supercoiled helical state.

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When does ribosomal seperation occur?

This occurs after translation.

Ribosomal separation results when a release factor enters the A site, and peptidyl transferase hydrolyzes the bond between the last tRNA and the completed peptide chain.

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Which of the following produces a strand of DNA in the 5’ to 3’ direction?

I. Eukaryotic DNA polymerase

II. Prokaryotic DNA polymerase III

III. Reverse transcriptase

A. I and II only

B. III only

C. I, II, and III

D. I only

C. I, II, and III

All three polymerases listed use a template strand to create a new strand of DNA. They do so by adding new nucleotides to the 3’ end of the new strand (hence, the strand grows in the 5’ to 3’ direction).

Reverse transcriptase differs from the DNA polymerases only in that is uses an RNA template instead of a DNA template to synthesis its new strand of DNA. However, it too adds new DNA nucleotides in the 5’ to 3’ direction.

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Which of the following enzymes is NOT required for eukaryotic DNA replication?

A. RNA polymerase

B. Topoisomerase

C. Gyrase

D. Ligase

C. Gyrase

Gyrase is used in prokaryotes

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In terms of ATP, approximately how many glucose molecules would it take to translate a 60 amino acid polypeptide chain in a eukaryote undergoing aerobic respiration?

A. 8

B. 10

C. 6

D. 12

A. 8

Aerobic respiration in a eukaryote results in 30 ATP per glucose. Formula is 4n-1

60 x 2 = 120 tRNA needed

Translation requirements are as follows:

2 ATP per AA-tRNA loading = 240 ATP needed

240/30 = 8

Thus, for a 60 AA peptide it would require 239 ATP and ~ 8 molecules of glucose.

Alternative splicing permits somatic cells to contain the same genome while maintaining the capability to express widely differing proteins, based on the tissue in which the cell is located. Which experimental technique would be LEAST useful in detecting the differing cellular products created by alternative splicing?

A. ELISA (enzyme-linked immunosorbent assay)

B. Western blotting

C. Southern blotting

D. Northern blotting

C. Southern blotting

Southern blotting is used to detect DNA, and a point made by the question is that the genomes are the same (therefore the least useful technique).

Northern blotting is used to detect RNA and both Western blotting and ELISAs can be used to detect proteins.

A codon is a segment of an mRNA molecule that codes for one amino acid in a polypeptide chain formed during protein synthesis. Which of the following correctly describes the chain of events that occurs in the synthesis of a polypeptide?

A. DNA generates mRNA; mRNA moves to the ribosomes, where a tRNA anticodon binds to an mRNA codon, causing amino acids to join together in their appropriate order.

B. DNA generates tRNA; the tRNA anticodon attaches to the mRNA codon in the cytoplasm; tRNA is carried by mRNA to the ribosomes, causing amino acids to join together in a specific order.

C. Specific RNA codons cause amino acids to line up in a specific order; tRNA anticodons attach to mRNA codons; rRNA codons cause protein molecules to cleave into specific amino acids.

D. DNA generates mRNA in the nucleus; mRNA moves to the cytoplasm and attaches to a tRNA anticodon; an operon regulates the sequence of events that causes amino acids to line up in their appropriate order.

A. DNA generates mRNA; mRNA moves to the ribosomes, where a tRNA anticodon binds to an mRNA codon, causing amino acids to join together in their appropriate order.

How do chromosomal translocations end up potentially creating new gene products or enhancing the activity of existing gene products?

A. Recombination occurs only between the arms of the same chromosome.

B. Recombination occurs between homologous chromosomes, but the exchange of gene segments is unequal.

C. Recombination occurs only between somatic and sex chromosomes.

D. Recombination occurs between non-homologous chromosomes placing previously unconnected sequences in proximity.

What are the requirements for a complex transposon?

A. An IS element that triggers a frameshift mutation

B. Two IS elements

C. An IS element and one or more genes

D. Two IS elements with intervening sequence between them

C. An IS element and one or more genes

A complex transposon contains an IS element (the transposase and its accompanying inverted repeat sequences) and one or more genes.

A composite transposon has two IS elements and intervening sequence.

Eukaryotic DNA is initially transcribed as hnRNA, then spliced to form mature mRNA. Which of the following is true of RNA splicing?

A. Introns remain in the mature mRNA.

B. Splicing of hnRNA from a single gene can be variable.

C. Splicing of hnRNA occurs in the cytosol.

D. Exons are excluded from the mature mRNA.

B. Splicing of hnRNA from a single gene can be variable.

During RNA splicing, introns are removed and exons are connected together. This process occurs in the nucleus, and can be variable.

UV light can trigger the formation of pyrimidine dimers, which then cause malformed loops of DNA. Visible light can then trigger repair enzymes via photoreactivation. This describes a DNA repair mechanism known as:

A. excision repair.

B. direct reversal.

C. homologous recombination.

D. nonhomologous endjoining.

B. direct reversal.

The ratio of guanine-cytosine (G—C) pairs to adenine-thymine (A—T) pairs is useful in laboratory manipulation of double-stranded DNA. If a segment of DNA has a low G—C : A—T ratio, it would be reasonable to assume that this segment would:

A. require less energy to separate the two DNA strands than would a comparable segment of DNA having a high G—C : A—T ratio.

B. contain more guanine than cytosine.

C. require more energy to separate the two DNA strands than would a comparable segment of DNA having a high G—C : A—T ratio.

D. contain more adenine than thymine.

A. require less energy to separate the two DNA strands than would a comparable segment of DNA having a high G—C : A—T ratio.

G—C base pairs are linked by three hydrogen bonds in the double helix, while A—T base pairs are joined by only two hydrogen bonds. It takes more energy to separate G—C base pairs, and the less G—C rich a piece of double-stranded DNA is, the less energy that is required to separate the two strands of the double helix.

Note that the statements “contain more guanine than cytosine” and “contain more adenine than thymine” can be eliminated since in double-stranded DNA, there must be equal quantities of all nitrogenous bases.

Question 14

Which of the following is a difference between eukaryotic and prokaryotic translation?

A. The first translated codon

B. The function of the codon UAA

C. The mechanism by which ribosomes recognize the 5’ end of mRNA

D. The number of times a single mRNA transcript can be translated

C. The mechanism by which ribosomes recognize the 5’ end of mRNA

Prokaryotic mRNA is recognized by the ribosome using the Shine-Dalgarno sequence (-10) while eukaryotic mRNA is recognized via the 5’ cap that is added during post-transcriptional modification.

Which of the following would lengthen Okazaki fragments? A. Separating stop transcription sequences to a greater degree during replication B. Increasing the number of origins on the DNA strand C. Increasing the rate of all aspects of replication D. Decreasing the number of primers generated on the lagging strand during replication

D. Decreasing the number of primers generated on the lagging strand during replication This would result in DNA polymerase III traveling uninterrupted for a longer period of time and generating longer Okazaki fragments.

Which of the following is the most likely effect of a eukaryotic transcription factor? A. Provide the energy necessary for nascent RNA polymerization. B. Serve to inhibit translation to allow for completion of transcription. C. Increase the encounter rate of DNA with RNA polymerase. D. Bind at specific AUG sequences to start transcription.

C. Increase the encounter rate of DNA with RNA polymerase. Transcription factors influence many processes in transcription including recruiting RNA polymerase, regulating transcription rate, and many others.

It is known that the developing frog embryo requires greater protein production than the adult organism. **If cells from a developing frog embryo and from a mature frog were examined, would the investigator find the greater rate of translation in cells of the embryo or of the adult?** A. The adult, because a mature organism has more complex metabolic requirements. B. The adult, because ribosomal production is more efficient in a mature organism. C. The embryo, because ribosomal production is not yet under regulatory control by DNA. D. The embryo, because a developing organism requires a higher rate of translation than does an adult.

D. The embryo, because a developing organism requires a higher rate of translation than does an adult.

Which of the following is the most accurate, from least organized to most organized? A. Deoxyribose, nucleoside, nucleotide, DNA helix, chromatin, nucleosome B. Deoxyribose, nucleotide, nucleoside, DNA helix, nucleosome, chromatin C. Deoxyribose, nucleotide, nucleoside, nucleosome, DNA helix, chromatin D. Deoxyribose, nucleoside, nucleotide, DNA helix, nucleosome, chromatin

D. Deoxyribose, nucleoside, nucleotide, DNA helix, nucleosome, chromatin **Deoxyribose** is the sugar component of the DNA backbone. A **nucleoside** is the next level of organization and is a deoxyribose with the nitrogenous base attached A **nucleotide** is more organized than a nucleoside and includes phosphate groups . **Double stranded DNA** winds around histone octamers to form nucleosomes, which are further packaged and condensed to form **chromatin**.

Which of the following can affect a ligand's ability to bind to or intercalate into DNA? I. Structural or conformational changes to DNA II. Protonation state of the ligand III. Increase in bulk salt concentration of solvent around DNA A. I and II only B. III only C. II only D. I, II, and III

D. I, II, and III Item I is true: structural and/or conformational changes in DNA can affect which binding sites are accessible to ligands. For example, a binding site can be inaccessible when DNA is in a coiled state (choices B and C can be eliminated). Since both remaining answer choices include Item II, Item II must be true: the passage directly states that the protonation state can affect the ligand's ability to bind with DNA. Item III is also true: the passage states that increasing the bulk salt concentration can make the intercalation process less energetically favorable

When studying DAPP-binding to DNA, researchers face a potential problem. Although DAPP binds more readily to A–T rich sequences, it can also bind less favorably to other regions of the DNA. **If a biochemist is attempting to create intercalated DAPP from unbound DAPP and calf DNA, the initial solution should have which property to ensure that the DAPP only binds to A–T rich sequences?** A. Large ligand to DNA ratio B. Large DNA to ligand ratio C. Equal ligand to DNA ratio D. Random DNA to ligand ratio

B. Large DNA to ligand ratio To achieve the most specific binding possible (and to avoid undesired binding), the biochemist should aim to have many DNA molecules for each ligand. **If there are many ligand molecules and DNA is the limiting factor, the ligand might bind less specifically**. A large DNA to ligand ratio will ensure that the ligand only binds to A–T rich sequences because there are sufficient A–T rich regions for DAPP to bind.

How would raising the pH of the solvent affect the affinity of positively charged ligand for DNA? A. The affinity for DNA would decrease due to less protonated ligand. B. The affinity for DNA would increase since the phosphate backbone would be protonated. C. The affinity for DNA would stay the same since solvent effects are not relevant. D. The affinity cannot be determined for differing pH levels without specific pKa values.

A. The affinity for DNA would decrease due to less protonated ligand. Raising the pH of the solvent will make the solvent more basic, therefore it is less likely that the ligand will be protonated.

Which of the following would facilitate a DNA backbone parallel to that of a standard double helix? I. Adenine and thymine II. Guanine and adenine III. Cytosine and thymine A. I and III only B. II and III only C. I only D. II only

C. I only This question is asking you (in a confusing sort of way) to find a purine-pyrimidine pair, since this is how nucleotides must pair across the double helix to ensure the backbone is parallel.