MCAT Biology : Cell Biology, Molecular Biology, and Genetics

Study concepts, example questions & explanations for MCAT Biology

varsity tutors app store varsity tutors android store

Example Questions

Example Question #71 : Genetics

Pedigree

Consider the pedigree. Is the trait dominant or recessive?

Possible Answers:

Recessive

Dominant

Codominant or incompletely dominant

The mode of inheritance cannot be determined

Correct answer:

Recessive

Explanation:

The trait is recessive because affected individuals do not occur in every generation. Additionally, dominant traits do not result in "carriers". Individuals are either affected or unaffected.

Example Question #1 : Genetic Abnormalities And Mutation

The concept of genomic imprinting is important in human genetics. In genomic imprinting, a certain region of DNA is only expressed by one of the two chromosomes that make up a typical homologous pair. In healthy individuals, genomic imprinting results in the silencing of genes in a certain section of the maternal chromosome 15. The DNA in this part of the chromosome is "turned off" by the addition of methyl groups to the DNA molecule. Healthy people will thus only have expression of this section of chromosome 15 from paternally-derived DNA.

The two classic human diseases that illustrate defects in genomic imprinting are Prader-Willi and Angelman Syndromes. In Prader-Willi Syndrome, the section of paternal chromosome 15 that is usually expressed is disrupted, such as by a chromosomal deletion. In Angelman Syndrome, maternal genes in this section are deleted, while paternal genes are silenced. Prader-Willi Syndrome is thus closely linked to paternal inheritance, while Angelman Syndrome is linked to maternal inheritance.

Figure 1 shows the chromosome 15 homologous pair for a child with Prader-Willi Syndrome. The parental chromosomes are also shown. The genes on the mother’s chromosomes are silenced normally, as represented by the black boxes. At once, there is also a chromosomal deletion on one of the paternal chromosomes. The result is that the child does not have any genes expressed that are normally found on that region of this chromosome.

 

 

Untitled

In addition to the chromosomal deletion on chromosome 15 in the passage, the father is found to have another gene with a mutation, which adds a stop codon prematurely in the base pair sequence. This mutation is best described as a __________.

Possible Answers:

silent mutation

conservative missense mutation

nonsense mutation

non-conservative missense mutation

frameshift mutation

Correct answer:

nonsense mutation

Explanation:

The best answer is a nonsense mutation, which is defined as a point mutation that gives rise to an early stop codon, thus truncating any protein products prematurely. These are typically devastating mutations for protein function.

Example Question #71 : Cell Biology, Molecular Biology, And Genetics

Type II diabetes results from defective pancreatic beta cells and increased insulin resistance, indicating that peripheral tissues (such as skeletal muscle) do not properly respond to insulin.

Mouse models have been developed to model type II diabetes. In addition to global mutations, tissue-specific mutations can be used to delete genes of interest in precise regions of the body. A group of investigators is interested in characterizing the role of the gene Dia in the onset of diabetes.

Four groups of male mice are compared. Group A is a control group, group B has a global deletion of Dia, group C has a beta cell-specific Dia mutation, and group D has a skeletal muscle-specific Dia mutation.

In order to measure the ability of these mice to respond to a glucose challenge, the mice are fasted overnight. Following the fast, their blood glucose levels are measured (in mg/dL). The mice are then injected with two grams of glucose, and blood glucose levels are measured at 30, 60, 90, and 120 minutes post-injection.

 

0 min

30 min

60 min

90 min

120 min

Group A

80

150

120

90

80

Group B

90

220

180

160

140

Group C

100

260

190

150

135

Group D

75

145

110

90

75

Based on the data, what role does Dia play in insulin regulation?

Possible Answers:

Promote insulin production

Initiate beta cell deterioration

Decrease insulin sensitivity

Decrease insulin production

Promote insulin sensitivity

Correct answer:

Promote insulin production

Explanation:

The beta cell-specific Dia mutation in group C causes results similar to the global Dia mutation in group B. From this, we can conclude that Dia is functioning within the beta cells. Essentially, deleting Dia from the beta-cells is equivalent to deleting it from the entire body. Additionally, loss of Dia in skeletal muscle in group D seems to have no phenotypic effect, indicating that Dia is not necessary in skeletal muscle.

Pancreatic beta cells are responsible for insulin production, while skeletal muscle plays a significant role in insulin sensitivity. Since loss of Dia in beta cells leads to high blood glucose, we can conclude that the role of Dia is in the promoting the production of insulin, and that the gene plays no role in insulin sensitivity.

Example Question #71 : Genetics

Some inherited diseases of the liver, including Wilson's Disease, are primarily or entirely genetically determined. Wilson's Disease results when a defect in a copper transporter in the small intestine occurs, leading to copper level disregulation in both the hepatocytes and the systemic circulatory system. Mutations have primarily been found in the copper transporter that helps load copper onto a transport protein, apoceruloplasmin, which normally creates serum-soluble ceruloplasmin with the addition of copper. Given this defect, serum studies of an individual with Wilson's Disease would likely show what kind of change in serum ceruloplasmin compared with a normal individual?

Possible Answers:

Equivalent serum ceruloplasmin

Decreased serum apoceruloplasmin

The comparison cannot be estimated

Decreased serum ceruloplasmin

Increased serum ceruloplasmin

Correct answer:

Decreased serum ceruloplasmin

Explanation:

The question informs us that the mutational defect in the gene involves the enzyme's ability to load copper onto apoceruloplasmin. Healthy individuals are able to load copper to apoceruloplasmin, creating serum-soluble ceruloplasmin. With this process disrupted in an individual with Wilson's Disease, we would expect that less ceruloplasmin would be produced because copper could not be transported. We would expect to see reduced serum levels of the complete protein, and high levels of copper building up in hepatocytes and circulatory serum.

Example Question #72 : Cell Biology, Molecular Biology, And Genetics

Which of the following mutations might lead to the formation of a recessive allele?

I. Silent

II. Frameshift

III. Nonsense

IV. Missense

Possible Answers:

I and IV

I, II, III, and IV

II, III, and IV

II and III

Correct answer:

II, III, and IV

Explanation:

Recessive alleles are often created by mutated versions of functional genes that encode broken/nonfunctional proteins. The question is thus asking us which of the mutations is likely to result in a nonfunctional or abnormal protein.

Silent mutations are types of mutations that result in the insertion of the same amino acid due to the degeneracy of the genetic code and, therefore, will not cause any noticeable change in the protein. Essentially, even organisms with the mutation will still show the dominant allele.

Frameshift mutations change the reading frame used for translation and oftentimes result in premature stop codons. Nonsense mutations are mutations that specifically result in premature stop codons and generally result in nonfunctional proteins. Missense mutations lead to the insertion of a different amino acid at the normal site. This can lead to serious problems for the functionality of the protein, particularly if the mutation is present in the active site or another area that is highly conserved throughout evolution.

Example Question #73 : Cell Biology, Molecular Biology, And Genetics

Which answer choice correctly shows a frameshift mutation?

Possible Answers:

CGGTGAATAGGC  CGGTGAATCAGGC

 

CGGTGAATAGGC  CGGTGAATA

CGGTGAATAGGC  CGGTCAATAGGC

CGGTGAATAGGC  CGGTGAAGACTAGGC

CGGTGAATAGGC  CGAAGTGGC

Correct answer:

CGGTGAATAGGC  CGGTGAATCAGGC

 

Explanation:

A framshift mutation is when an extra base is inserted into (or deleted from) a strand of DNA. When the DNA is transcribed, every codon downstream from the insertion is changed because the reading frame will be different. This will be the case for any mutations that affect a sequence of nucleotides that is not a multiple of three, for example a five-nucleotide deletion. Such mutations affect the three-nucleotide groupings for all codons following the mutation, and often result in a premature stop codon.

The correct answer is CGGTGAATAGGC  CGGTGAATCAGGC. In the original sequence, the codons are CGG-TGA-ATA-GGC. In the second, they become CGG-TGA-ATC-AGG-C. The other answers show either insertions or deletions that do not shift the reading frame, or single nucleotide polymorphism (changing a single nucleotide to another).

Note that the reading frame does not change in the following 3-nucleotide insertion. Though there is an extra codon, the downstream sequence is not affected.

CGG-TGA-ATA-GGC  CGG-TGA-AGA-CTA-GGC

Example Question #74 : Cell Biology, Molecular Biology, And Genetics

Which of the following will least likely affect the length of a protein product?

Possible Answers:

Single-base deletion

Nonsense mutation

Frameshift mutation

Missense mutation

Single-base addition

Correct answer:

Missense mutation

Explanation:

A missense mutation is the substitution of one nucleotide for another in the DNA sequence, resulting in a different resulting amino acid. Essentially, one amino acid is replaced with another. This type of mutation will not alter the overall protein length.

A nonsense mutation results in a pre-mature stop codon, causing early termination of the protein product and a shorter protein. Frameshift mutations commonly cause pre-mature stop codons to arise, and at the very least will result in a highly altered protein product that is likely of a different length from the unaltered protein. Single-base addition and deletions will cause frameshift mutations.

Healthy RNA: 3'-AGG-UCG-UUA-GUC-5'

Missense:     3'-AAG-UCG-UUA-GUC-5'

Nonsense:    3'-AGG-UAG-5' (UAG stop codon)

Frameshift:   3'-AAG-GUC-GUU-AGU-C-5' (single-base addition)

Framshift:     3'-AGU-CGU-UAG-5' (single-base deletion, UAG stop codon)

Example Question #72 : Genetics

Human chromosomes are divided into two arms, a long q arm and a short p arm.  A karyotype is 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 one derivative 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.

During meiosis, gametes result from the isolation of one chromosome from each of a human’s homologous 23 pairs. In Figure 2, der(14,21) is treated as a homologous pair to either 14 or 21. An egg, then, could be formed that had a normal chromosome 21 and the der(14,21) instead of the normal chromosome 14. If a sperm fertilizes this egg, and the sperm has a normal chromosome 14 and 21, what will result?

Possible Answers:

Polyploidy

Unbalanced translocation

Pleiotropy

Triploidy

Trisomy

Correct answer:

Trisomy

Explanation:

Since the maternal oocyte would have two copies of chromosome 21, the normal 21 and the derivative chromosome, the normal addition of the sperm would lead to a zygote with a total of three copies of chromosome 21, or trisomy 21. Polyploidy is a tempting choice, but polyploid is a state characterized by extra copies of multiple chromosomes, as might be expected from more than one sperm fertilizing an egg.

Example Question #76 : Cell Biology, Molecular Biology, And Genetics

Which of the following DNA changes is least likely to alter the translated amino acid sequence created by the sequence below?

5'-CGATGAACG-3'

Possible Answers:

5'-CGACGAACG-3'

5'-CGATAAACG-3'

5'-CGCTGAACT-3'

5'-CTATGAACG-3'

5'-CGATGGACG-3'

Correct answer:

5'-CGACGAACG-3'

Explanation:

First, you want to narrow down the answers to those with the fewest changes (1 base change). Next, you needed to select the single base change on the third base of the codon. A silent mutation (a DNA mutation that doesn't result in an amino acid change) is most likely to occur when the third base of a codon is changed.

Original sequence: 5'-CGA TGA ACG-3'

Unaltered mRNA transcript: 3'-GCU ACU UGC-5'

Note that the mRNA sequence is antiparallel, changing the orientation of the codons. mRNA is read in the 5'-to-3' direction by the ribosome, meaning that the third base in the codon will be to the left of the 3-base sequence. The given RNA sequence will be read as CGU UCA UCG.

Correct answer: 5'-CGA CGA ACG-3'

Altered mRNA transcript: 3'-GCU GCU UGC-5'

The sequence will be read as CGU UCG UCG.

The correct answer involves a mutation in the third nitrogenous base of the second codon. Both UCA (original) and UCG (altered) code for serine.

Example Question #77 : Cell Biology, Molecular Biology, And Genetics

Which of the following mutations will not change the amino acid sequence of the polypeptide created during translation?

Possible Answers:

Silent mutation

Insertion

Frameshift mutation

Misense mutation

Nonsense mutation

Correct answer:

Silent mutation

Explanation:

A silent mutation is a mutation that results in the inclusion of the same amino acid in the polypeptide product, despite a change in the DNA template sequence. Silent mutations generally involve the third nucleotide in the codon sequence, since this unit is often interchangeable due to the degeneracy of the coding sequence.

A missense mutation results in a different amino acid inclusion based on codon alteration. A nonsense mutation is the insertion of a new stop codon as a result of mutation, resulting in a truncated polypeptide and severely hindering functionality. An insertion or frameshift mutation can change the reading frame of the mRNA template, drastically changing the identity and primary sequence of the polypeptide product; these are the most dangerous type of mutation, as the product may be harmful as well as nonfunctional.

Learning Tools by Varsity Tutors