Biochemistry : Anabolic Pathways and Synthesis

Study concepts, example questions & explanations for Biochemistry

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Example Questions

Example Question #51 : Anabolic Pathways And Synthesis

Which of the following is true regarding both prokaryotic and eukaryotic transcription?

Possible Answers:

Both involve polyadenylation and addition of methyl cap at the 5’ end

Both involve polyadenylation and addition of methyl cap at the 3’ end

None of the these are true

Polyadenylation and addition of methyl cap occurs only in prokaryotic transcription

Correct answer:

None of the these are true

Explanation:

Prokaryotic and eukaryotic transcription are similar in many ways; however, they are also different from one another. One main difference is that in eukaryotic transcription there are several events that occur after the completion of transcription. These events are called post-transcriptional modifications.

There are three main post-transcriptional modifications: polyadenylation, capping, and splicing. Polyadenylation involves the addition of multiple adenine molecules at the 3’ end of the newly synthesized RNA molecule. This segment of RNA with multiple adenine molecules is called the poly A tail. Capping involves the addition of a methyl cap to the 5’ end of the RNA molecule. Splicing involves the excision of segments of the mRNA molecule that aren’t used in translation. These segments are called introns and the usable segments are called exons. The exons are ligated back together and the end product is released into the cytoplasm where it can undergo translation.

Products of prokaryotic transcription don’t undergo any of these post-transcriptional modifications and are immediately translated into proteins.

Example Question #1186 : Biochemistry

Consider the following events in transcription:

1. RNA polymerase adds new nucleotides

2. Sigma factor binds to inactive RNA polymerase

3. Transcription bubble forms

Which of the following is the correct order of these events?

Possible Answers:

2, 3, 1

2, 1, 3

3, 2, 1

3, 1, 2

Correct answer:

2, 3, 1

Explanation:

Transcription is the process of producing RNA molecules from a parent strand of DNA. The first step in transcription involves the binding of a sigma factor to inactive RNA polymerase, which in turn binds to the promoter region on the DNA molecule. After the binding of RNA polymerase to the DNA promoter, RNA polymerase gets activated and removes the hydrogen bonds between nucleotides of the double stranded DNA. Finally, RNA polymerase adds complementary nucleotides to the growing RNA molecule.

After completion of transcription, RNA molecules undergo post-transcriptional modifications and exit the nucleus and enter cytoplasm, where it can undergo translation.

Example Question #51 : Anabolic Pathways And Synthesis

Which RNA Polymerase is associated with the transcription of DNA to synthesize mRNA?

Possible Answers:

RNA polymerase III

All of these answers

RNA polymerase II

None of these answers

RNA polymerase I

Correct answer:

RNA polymerase II

Explanation:

RNA Polymerase II is used to catalyze the polymerization of mRNA during transcription. RNA polymerase I catalyzes the polymerization of rRNA, and RNA polymerase III catalyzes the polymerization of tRNA.

Example Question #52 : Anabolic Pathways And Synthesis

If a given piece of coding DNA is 5’ AATGACGTC 3’, which of the following is true?

Possible Answers:

tRNA corresponding to this DNA would have the anticodons: 3’ AAU 5’ 3’ GAC 5’ 3’ GUC 5’

tRNA corresponding to this DNA would have the anticodons: 5’ UUA 3’ 5’ CUG 3’ 3’ CAG 5’

The amino acid chain matching this DNA sequence is Leu-Leu-Gln

tRNA corresponding to this DNA would have the anticodons: 3’ UUA 5’ 3’ CUG 5’ 3’ CAG 5’

Correct answer:

tRNA corresponding to this DNA would have the anticodons: 3’ UUA 5’ 3’ CUG 5’ 3’ CAG 5’

Explanation:

If the coding DNA reads 5’ AATGACGTC 3’, then the template strand would read 3’ UUACTGCAG 5’. The mRNA transcription would read 5’ AAUGACGUC 3’. The corresponding tRNA anti-codons would be 3’ UUA 5’ 3’ CUG 5’; 3’ CAG 5’, which produce the amino acids Asn-Asp-Val. To determine the amino acid sequence, you find the portion of the genetic code table corresponding to the DNA or mRNA, not the tRNA, nucleotides. (That's why ”Leu-Leu-Gln" is incorrect.)

Example Question #1 : Transcription

Which enzyme could be considered to violate the "central dogma" of biology?

Possible Answers:

Carboxylase

Reverse transcriptase

Catalase

Ligase

Helicase

Correct answer:

Reverse transcriptase

Explanation:

The "central dogma" of biology says that information goes from DNA via transcription to RNA via translation to proteins. Reverse transcriptases, however, employed by retroviruses, synthesize DNA from RNA. As for the other enzymes: one function of helicases (among others) is to pull apart double helix strands. Catalase breaks down hydrogen peroxide. Carboxylase adds a carboxyl group to a substrate, and a ligase creates a bond between two molecules, for example, via a phosphodiester bond.

Example Question #52 : Anabolic Pathways And Synthesis

The post-transcriptional spliceosome reaction, which removes the intron in a lariat structure, takes place via __________.

Possible Answers:

two transacetylations

one transacetylation

two transesterifications

two isomerizations

one transesterification and one transacetylation

Correct answer:

two transesterifications

Explanation:

The spliceosome first releases the 5' exon by forming a lariat structure (2'-5' phosphodiester) bond between two introns in a transesterification reaction. Then, the exons are spliced together with another transesterifaction reaction, and the intron lariat is released.

Example Question #9 : Transcription

Which of the following is a ribozyme?

Possible Answers:

RNase H

RNA polymerase II

All of these are ribozymes

Guanyltransferase

RNase P

Correct answer:

RNase P

Explanation:

RNase H is involved in reverse transcriptase, but it is not a ribozyme. RNase P is a ribonuclease that cleaves/processes rRNA and generates 5' ends. The RNA component of RNase P is its catalytic subunit.

Example Question #6 : Transcription

Within a eukaryotic chromosome, which of the following pairs of processes can occur at the same time?

Possible Answers:

Transcription and translation

Replication and splicing

Translation and splicing

Transcription and splicing

Replication and transcription

Correct answer:

Transcription and splicing

Explanation:

In eukaryotes, transcription and splicing could occur simultaneously. Both of these processes take place in the nucleus of the cell, while translation takes place in the cytoplasm. Therefore, translation could not happen at the same time as either transcription or splicing. Replication occurs totally independently from all of the other processes listed.

Example Question #11 : Transcription

Oftentimes, biochemists say that the genetic code is degenerate. What are they referring to when they say this?

Possible Answers:

A change in the reading frame alters expression of all subsequent codons

Some amino acids can be coded for by more than just one codon

Some codons are able to code for more than just one amino acid

The sequence of one type of molecule is able to code for the sequence of another type of molecule

The same genetic code is not shared by all species

Correct answer:

Some amino acids can be coded for by more than just one codon

Explanation:

While some of these answer choices are true, others are false. But even for the choices that are true, only one of them directly answers the question. Let's go ahead and look at each choice.

  • Some amino acids can be coded for by more than just one codon

This is a true statement, and is also the correct answer. The degeneracy of the code is due to the fact that, for some amino acids, a number of different codons can result in the same amino acid. For example, the amino acid tyrosine can be coded for by either UAU or by UAC.

  • A change in the reading frame alters expression of all subsequent codons

While this is a true statement, it does not answer the question. Nonetheless, it's important to know that mutations which either insert or delete a nucleotide will change the entire rest of the reading frame. Consequently, there are likely to be many errors and the resultant polypeptide will likely not be functional.

  • Some codons are able to code for more than just one amino acid

This is a false statement. But be careful. This answer choice looks a lot like the correct one. The difference is that, for this choice, we're talking about a single codon being able to code for more than one amino acid. This is not the case.

  • The sequence of one type of molecule is able to code for the sequence of another type of molecule

This is a true statement with regards to gene expression. For transcription, the DNA (sequence of deoxyribonucleotides) serves as a template for the formation of mRNA (sequence of ribonucleotides). And for translation, that same mRNA serves as a precursor for the formation of a polypeptide (sequence of amino acids).

  • The same genetic code is not shared by all species

From what we know, this is a false statement. The genetic code is universal, meaning that all living things have been found to have the same codon-amino acid relationship.

Example Question #1191 : Biochemistry

When biochemists says that genes can be in either direction of the DNA, what do they mean?

Possible Answers:

The different alleles for a given gene within chromosomes can be partitioned to either of two cells during mitosis

The template for transcription can be found on either strand of DNA

Genes can be transcribed in both directions, from  and 

Genes can be found both in the middle and at ends of a chromosome

Correct answer:

The template for transcription can be found on either strand of DNA

Explanation:

The saying that genes can run in either direction is referring to how they can be expressed. Genes can be found on either strand of the DNA molecule. Since these two strands run antiparallel to one another, the genes being read on one strand would be read in the opposite direction on the other strand.

Genes can also be located on various loci within a given chromosome. Also, it is true that genes assort independently during cell division, but this doesn't answer the question. Also, genes can only be read in the  direction. They can never be read in the  direction. This is because DNA polymerase, the enzyme responsible for elongating the chain, can only add nucleotides to a chain that has a hydroxyl group at the  position.

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