Biochemistry : Macromolecule Structures and Functions

Study concepts, example questions & explanations for Biochemistry

varsity tutors app store varsity tutors android store

Example Questions

Example Question #1 : Protein Degradation

If a protein is bonded to ubiquitin, this tells the cell that the protein should be __________.

Possible Answers:

inactivated

activated

shortened

elongated

degraded

Correct answer:

degraded

Explanation:

When a protein is damaged, it can be tagged with the molecule, ubiquitin. This signals to the cell that the protein is no longer functioning properly and needs to be degraded.

Example Question #91 : Macromolecule Structures And Functions

HMGCoA reductase (3-hydroxy-3-methyl-glutaryl-CoA reductase) is the rate-limiting enzyme in cholesterol synthesis. Which of the following are true about the ubiquitination of this enzyme?

I. When cholesterol levels in the cell are high, the reductase binds to insulin-induced gene 1 proteins.

II. Binding to insulin induced gene 1 proteins leads to ubiquitination and proteasomal degradation of reductase.

III. Ubiquitination occurs through the binding of the C-terminal glycine of ubiquitin to the amino group of a lysine on the reductase.

IV. The enzyme tagged with ubiquitin is recognized by the proteasome where proteolysis occurs.

Possible Answers:

I and II

II, III, and IV

I, II, III, and IV

II and IV

II and III

Correct answer:

I, II, III, and IV

Explanation:

HMGCoA reductase is the rate-limiting enzyme in cholesterol synthesis. The reductase is present on the endoplasmic reticulum membrane. When levels of its product, cholesterol, are high, the enzyme gets ubiquitinated and degraded in smaller peptides and amino acids. It first binds to insulin-induced gene 1 protein before ubiquitination.

Example Question #102 : Biochemistry

Amino terminal - Ala - Lys - Glu - Phe - Phe - Ala - Leu - carboxyl terminal.

If the above primary sequence is cleaved by trypsin, on which amino acid will the new amino terminal be?

Possible Answers:

Leu

Lys

Glu

Ala

Phe

Correct answer:

Glu

Explanation:

Trypsin will cleave the primary sequence after the lysine residue (on its carboxyl side). Thus, Lys will be the new carboxyl terminal and Glu will be the new amino terminal. Remember that a protein's primary sequence is written from N to C.

Example Question #103 : Biochemistry

Which of the following proteases would cleave lysine at the carbonyl side?

Possible Answers:

Chymotrypsin

Trypsin

None of these

Pepsinogen

Pepsin

Correct answer:

Trypsin

Explanation:

Trypsin will cleave lysine and arginine at the carbonyl side. Chymotrypsin will cleave phenylalanine, tyrosine, and tryptophan at the carbonyl side. Pepsin will cleave leucine, phenylalanine, tryptophan, and tyrosine at the amino side. Pepsinogen is the inactive form of pepsin, which gets activated via cleavage by hydrochloric acid in the stomach.

Example Question #104 : Biochemistry

What is the result of chymotrypsin being added to the peptide shown?

Gly-Ala-Pro-Tyr-His-Cys-Gly-Phe-Gly-Gly-Asn

Possible Answers:

Gly-Ala-Pro-Tyr, His-Cys-Gly, Phe-Gly-Gly-Asn

Gly-Ala-Pro-Tyr, His-Cys-Gly-Phe, Gly-Gly-Asn

Gly-Ala, Pro-Tyr, His-Cys-Gly, Phe-Gly-Gly-Asn

Gly-Ala-Pro, Tyr-His-Cys-Gly, Phe-Gly-Gly-Asn

Gly-Ala, Pro-Tyr, His-Cys, Gly-Phe, Gly-Gly-Asn

Correct answer:

Gly-Ala-Pro-Tyr, His-Cys-Gly-Phe, Gly-Gly-Asn

Explanation:

Chymotryspin cleaves Phe, Trp, and Tyr at the carbonyl side. This means that there will be a cleavage after any of these three amino acids appears. This results in Gly-Ala-Pro-Tyr, His-Cys-Gly-Phe, Gly-Gly-Asn.

Example Question #1 : Protein Hydrolysis

In biochemistry, turnover is a term that refers to the rate at which a compound is produced and subsequently degraded. Within the cell, many compounds are continuously being synthesized and degraded, although at a range of different rates. In a typical cell, how would the turnover rates for DNA, mRNA, and protein be expected to differ from largest to smallest?

Possible Answers:

Protein > DNA > mRNA

mRNA > Protein > DNA

DNA > Protein > mRNA

mRNA > DNA > Protein

Protein > mRNA > DNA

Correct answer:

mRNA > Protein > DNA

Explanation:

In this question, we're provided with a description of turnover rate. We're then asked to identify the relative turnover rates for protein, mRNA, and DNA.

To answer this question, it's important to have a general understanding of the role each of these molecules has in the cell. DNA resides in the nucleus and functions to provide the blueprint for producing mRNA. This mRNA, in turn, is processed and exported to the cytoplasm, where it interacts with ribosomes to be translated into protein. Finally, these proteins can have a wide variety of functions, including structural proteins, enzymes, antibodies, etc.

Based on the function of each of these molecules, we can reason our way to see what their relative turnover rates are expected to be.

Since DNA essentially provides the blueprint for the production of all proteins from a cell, its role is extremely important. While it's true that genes can be turned on and off, there's always going to be activity going on in the form of transcription within the nucleus; genes are being converted into mRNA all the time. Because it serves such an important role, and is even needed for cells that divide, we would expect DNA to have a practically non-existent turnover rate.

Next, let's take a look at mRNA. Though DNA is the blueprint that directs the production of protein, mRNA acts as the intermediate between the two. The advantage of this is that it allows for more levels of control. In addition to regulating transcription of genes through various means, post-transcriptional control is also possible. This level of control involves modifying the mRNA transcript to make it more resistant to degradation, or sometimes even degrading it to turn off gene expression. Overall, mRNA doesn't have a very long half-life within the cells. Because it serves as the intermediate between protein and DNA, once the protein has been translated, the mRNA is not really needed anymore. Thus, mRNA tends to have a high turnover rate, being produced whenever the cell needs it (gene expression turned on) and degraded whenever it isn't needed (gene expression turned off).

Lastly, let's look at protein. As was said previously, proteins have a vast array of functions. Whereas DNA and mRNA are responsible for producing protein, it is the protein that serves as the actual effectors; they're the ones that take action to get things done, either inside or outside the cell. In addition to being produced from DNA and mRNA, proteins can also be degraded via a process called ubiquitination. But, overall, since proteins are the actual effectors in the whole process, their turnover is lower than mRNA but higher than DNA.

Overall, the turnover rate is greatest for mRNA, followed by protein, with DNA having the lowest.

Example Question #92 : Macromolecule Structures And Functions

A polypeptide when treated with trypsin yielded the following fragments: 

(AM) (SAK) (YMPLWGIR)

The same polypeptide treated with chymotrypsin yielded the following fragments:

(MPLW) (GIRAM) (SAKY)

Which of the following displays the original sequence of the polypeptide?

Possible Answers:

AMSAKYMPLWGIR

MPLWGIRAMSAKY

WGIRAMSAKYMPL

SAKYMPLWGIRAM

GIRAMSAKYMPLW

Correct answer:

SAKYMPLWGIRAM

Explanation:

This problem requires knowledge of the endopeptidase properties. Trypsin cleaves after the amino acids lysine and arginine, and chymotrypsin cleaves after the amino acids phenylalanine, tyrosine, and tryptophan. 

Using this information, the slashes below indicate where each enzyme cleaved the polypeptide.

(AM) (SAK)/ (YMPLWGIR)/ trypsin

(MPLW)/ (GIRAM) (SAKY)/ chymotrypsin

Because the (AM) and (GIRAM) fragments have no slash after M in both the first and second treatment, one can conclude that this piece is at the end of the polypeptide. The other two pieces are sequenced by realizing that S starts the (SAK) and (SAKY) fragments and is never found in the middle of a fragment. 

Example Question #93 : Macromolecule Structures And Functions

Which of the following describes the unfolded protein response?

Possible Answers:

Stopping protein translation

Triggering cell death

All of these answers

Increasing amount of local chaperones

Targeting misfolded proteins for degradation by the 26S proteasome

Correct answer:

All of these answers

Explanation:

There are 4 main steps in the unfolded protein response: (1) Translation of proteins is stopped, (2) Chaperones are recruited to the location of misfolded proteins, (3) Misfolded proteins are "tagged" with ubiquitin chains for degradation by the 26S proteasome, (4) if the above steps fail, the cell undergoes programmed cell death. Thus, all of the answers choices are affiliated with the unfolded protein response.

Example Question #1 : Sugar Phosphate Groups And Phosphodiester Bonds

What type of bonds are found between the DNA sugar hydroxyl groups? What are their corresponding carbon numbers?

Possible Answers:

Phosphoester; 1', 5'

Phosphoester; 3', 5'

Phosphodiester; 2', 4'

Phosphodiester; 3', 5'

Phosphodiester; 1', 3'

Correct answer:

Phosphodiester; 3', 5'

Explanation:

The phosphodiester bonds in DNA occur between the 3' and 5' hydroxyl groups on deoxyribose. (This is related to DNA's 5' to 3' directionality as DNA polymerase can only synthesize DNA by adding nucleotides to the 3' hydroxyl group).

Example Question #1 : Sugar Phosphate Groups And Phosphodiester Bonds

In DNA, the 5-carbon sugar is attached to the nitrogenous base by a __________.

Possible Answers:

ester

deoxyribose

glycosidic bond

phosphodiester bond

phosphoester bond

Correct answer:

glycosidic bond

Explanation:

The beta-N-glycosidic bond attaches the nitrogen on the purine or pyrimidine base to the 1' anomeric carbon on the deoxyribose sugar. Phosphodiester linkages connect the 3' and 5' sugar hydroxyl groups on adjacent nucleotides.

Learning Tools by Varsity Tutors