MCAT Biology : Structure of DNA and RNA

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

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Example Question #1 :Nucleic Acids

Which of the following statements regarding the double-helix of DNA is true?

Possible Answers:

Both strands are identical to one another

Each strand is parallel with 5' to 3' direction

All hydroxyl groups of pentoses are involved in linkages

The nitrogenous bases are perpendicular to the axis

Correct answer:

The nitrogenous bases are perpendicular to the axis

Explanation:

The DNA double helix is a structure of double stranded nucleic acids, and is held together by base pairing of these nucleic acids perpendicular to the helical axis. The 5’ carbons and 3’ carbons between the pentoses are involved in phosphodiester bonds, while the 1’ carbons are involved in N-glycosidic bonds with the bases. Not all hydroxyl groups are involved in bonding, however, as there is a free 3’ carbon hydroxyl group on the extreme 3’ end of linked nucleoside monophosphate monomers. Both strands are sequentially unique as base pairing occurs between a purine and pyrimidine only. Finally, the strands are antiparallel, meaning they run in opposition to one another. Thus, the correct answer is that the nitrogenous bases are perpendicular to the axis.

Example Question #2 :Nucleic Acids

Human chromosomes are divided into two arms, a long q arm and a short p arm. Akaryotypeis the organization of a human cell’s total genetic complement. A typical karyotype is generated by ordering chromosome 1 to chromosome 23 in order of decreasing size.

When viewing a karyotype, it can often become apparent that changes in chromosome number, arrangement, or structure are present. Among the most common genetic changes are Robertsonian translocations, involving transposition of chromosomal material between long arms of certain chromosomes to form onederivative chromosome.Chromosomes 14 and 21, for example, often undergo a Robertsonian translocation, as below.

1

A karyotype of this individual for chromosomes 14 and 21 would thus appear as follows:

Pic2

Though an individual with aberrations such as a Robertsonian translocation may be phenotypically normal, they can generate gametes through meiosis that have atypical organizations of chromosomes, resulting in recurrent fetal abnormalities or miscarriages.

Which of the following is true of the DNA component of a standard chromosome 14?

I. The intra-strand bonds are covalent

II. The inter-strand bonds are covalent

III. During replication, Okazaki fragments are found on the leading strand

Possible Answers:

II and III

I and III

I only

I, II, and III

I and II

Correct answer:

I only

Explanation:

Intra-strand bonds are the covalent bonds between sugars and phosphates in the sugar-phosphate backbone. The other choices are incorrect, as inter-strand bonds in nucleic acids are hydrogen bonds between nitrogenous bases. Additionally, Okazaki fragments are found on the lagging strand of DNA during replication, because DNA on both the leading and lagging strands must be synthesized by DNA polymerase in the 5' to 3' direction.

Example Question #1 :Structure Of Dna And Rna

Human chromosomes are divided into two arms, a long q arm and a short p arm. Akaryotypeis the organization of a human cell’s total genetic complement. A typical karyotype is generated by ordering chromosome 1 to chromosome 23 in order of decreasing size.

When viewing a karyotype, it can often become apparent that changes in chromosome number, arrangement, or structure are present. Among the most common genetic changes are Robertsonian translocations, involving transposition of chromosomal material between long arms of certain chromosomes to form onederivative chromosome.Chromosomes 14 and 21, for example, often undergo a Robertsonian translocation, as below.

1

A karyotype of this individual for chromosomes 14 and 21 would thus appear as follows:

Pic2

Though an individual with aberrations such as a Robertsonian translocation may be phenotypically normal, they can generate gametes through meiosis that have atypical organizations of chromosomes, resulting in recurrent fetal abnormalities or miscarriages.

Which of the following are important differences between the DNA and histone components of chromosome 21?

I. DNA contains carbohydrates; histones do not

II. Histones contain amino acids; DNA does not

III. DNA contains nitrogenous bases, while histones are nitrogen-free

Possible Answers:

I and III

I only

II only

I and II

I, II, and III

Correct answer:

I and II

Explanation:

The first two choices are correct. DNA contains carbohydrate in its sugar-phosphate backbone, and histones are made entirely of amino acids. The third choice is deliberatly worded to lead you astray. It is true that DNA contains nitrogen in its bases, and that histones do not have these bases. Histones are not nitrogen-free, however, because amino acids contain nitrogen.

Example Question #1 :Nucleic Acids

Of the following groups of nitrogenous bases, which does not contain a purine?

Possible Answers:

Cytosine, thymine, uracil

Cytosine, thymine, guanine

Guanine, uracil, adenine

Adenine, guanine, thymine

Thymine, adenine, cytosine

Correct answer:

Cytosine, thymine, uracil

Explanation:

We can use the mnemonic "PurAG" to remember that the purines are adenine and guanine. The only choice that does not contain a purine, therefore, is "cytosine, thymine, and uracil." Remember, pyrimidines contain a single ring, while purines have two.

Example Question #1 :Structure Of Dna And Rna

Scientists use a process calledFlourescentIn-SituHybridization, orFISH, to study genetic disorders in humans. FISH is a technique that uses spectrographic analysis to determine the presence or absence, as well as the relative abundance, of genetic material in human cells.

To use FISH, scientists apply fluorescently-labeled bits of DNA of a known color, calledprobes, to samples of test DNA. These probes anneal to the sample DNA, and scientists can read the colors that result using laboratory equipment. One common use of FISH is to determine the presence of extra DNA in conditions ofaneuploidy, a state in which a human cell has an abnormal number ofchromosomes.Chromosomes are collections of DNA, the totality of which makes up a cell’sgenome.另一个典型的使用是癌症研究的cells, where scientists use FISH labels to ascertain if genes have moved inappropriately in a cell’s genome.

Using red fluorescent tags, scientists label probe DNA for a gene known to be expressed more heavily in cancer cells than normal cells. They then label a probe for an immediately adjacent DNA sequence with a green fluorescent tag. Both probes are then added to three dishes, shown below. In dish 1 human bladder cells are incubated with the probes, in dish 2 human epithelial cells are incubated, and in dish 3 known non-cancerous cells are used. The relative luminescence observed in regions of interest in all dishes is shown below.

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When probe DNA binds with target DNA, what is the main bonding mechanism likely at play?

Possible Answers:

Hydrogen bonding

Ionic bonding

Van der Waal forces

Coordinate covalent bonding

Covalent bonding

Correct answer:

Hydrogen bonding

Explanation:

When two complimentary strands of nucleic acid bind together, hydrogen bonding is the bonding at play. Guanine-cytosine pairs have three hydrogen bonds between them, and adenine-thymine pairs have two.

Example Question #2 :Nucleic Acids

In which of the following ways does RNA differ from DNA?

Possible Answers:

The sugar components are different

Proteins do not form from RNA

RNA has the base uracil instead of adenine

The phosphate groups are different

Correct answer:

The sugar components are different

Explanation:

RNA contains a ribose sugar, while DNA contains a deoxyribose sugar (missing a hydroxyl group at the second carbon). The phosphate groups are the same in both molecules, and RNA replaces the base thymine, not adenine, with uracil.

Example Question #7 :Nucleic Acids

Which of the following is NOT found in both RNA and DNA?

Possible Answers:

Phosphodiester bonds

Thymine

Nitrogenous bases

A pentose sugar

Correct answer:

Thymine

Explanation:

RNA and DNA share many common attributes. The nucleotides are attached to one another via phosphodiester bonds. They both have a pentose sugar, as well as nitrogenous bases. Only DNA, however, has thymine bases. RNA uses uracil in place of thymine.

Example Question #2 :Nucleic Acids

Which of the following statements is not true about DNA and RNA?

Possible Answers:

Both DNA and RNA have adenine bases

Both DNA and RNA have thymine bases

Both DNA and RNA are polymerized with phosphodiester bonds

Both DNA and RNA have a pentose sugar

Correct answer:

Both DNA and RNA have thymine bases

Explanation:

DNA and RNA differ in the nitrogenous bases used in their nucleotides; DNA uses thymine, while RNA uses uracil. Both nucleic acids are joined into polymers by phosphodiester bonds and use adenine as a nitrogenous base. Both contain a pentose sugar: DNA uses deoxyribose, while RNA uses ribose.

Example Question #1 :Nucleic Acids

DNA is comprised of a double-stranded helix in which purine bases are paired with pyrimidine bases. Which base pairing requires more energy to separate?

Possible Answers:

腺嘌呤和胸腺嘧啶,因为他们被浸配对alent bonds

Guanine and cytosine because they are paired by covalent bonds

Adenine and thymine because they are paired by three hydrogen bonds

Guanine and cytosine because they are paired by three hydrogen bonds

Correct answer:

Guanine and cytosine because they are paired by three hydrogen bonds

Explanation:

所有purine-pyrimindine对受氢bonds; covalent bonds are only found in the DNA backbone. Guanine-cytosine pairs are bound by thee hydrogen bonds, while adenine-thymine pairs are bound by two hydrogen bonds. The additional bond adds additional stability and energy to the guanine-cytosine linkage, making it harder to separate this base pair.

Example Question #10 :Nucleic Acids

过敏反应发生在当身体组织are affected by an abnormal immune reaction. The result is damage to normal tissues and clinical illness. A peanut allergy is an example of a hypersensitivity reaction, but there are three additional broad classes.

一个类涉及到生产或depo异常sition of antibodies. Antibodies are B-cell derived molecules that normally adhere to pathogens, rendering them unable to continue an infection. When antibodies are produced against normal tissues, however, disease can result. Figure 1 depicts a schematic structure of an antibody.

Antibodies can be divided into two peptide chains: heavy and light. Heavy chains form the backbone of the antibody, and are attached to light chains via covalent bonding. Each heavy and light chain is then further divided into constant and variable regions. Variable regions exhibit molecular variety, generating a unique chemical identity for each antibody. These unique patterns help guarantee that the body can produce antibodies to recognize many possible molecular patterns on invading pathogens.


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Humans must generate an enormous array of antibodies to account for all the possible patterns that they must recognize on pathogens. In fact, the diversity needed in antibody chains cannot be explained by DNA sequence variation. As a result, segments of DNA are physically rearranged in a process called VDJ rearrangement in order to create unique antibody chains. In order to carry out this process, DNA bonds must be broken. What kinds of intra-strand DNA bonds are most likely broken in this process?

Possible Answers:

van der Waals interactions

Phosphodiester covalent bonds

Peptide bonds

Hydrogen bonds

Base stacking interactions

Correct answer:

Phosphodiester covalent bonds

Explanation:

Intra-strand DNA存在于债券是债券each of the two complementary strands that make up a DNA molecule. The two strands are held together by hydrogen bonds between nitrogenous bases and base stacking interactions between the aromatic rings of the bases. Phosphodiester bonds tie together the adjacent ribose sugars in the DNA backbone, and thus give structure to each DNA strand.

Essentially, phosphodiester bonds are the primary intra-strand linkage, while hydrogen bonds are the primary inter-strand linkage.

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