Do You Know USMLE Step
For those of you who love all things biology and living organisms especially anatomy and how stuff works in the human body then this is the quiz for you. If you are prepared to test your knowledge try it out.
Questions and Answers
- 1.
A sample of human DNA is subjected to increasing temperature until the major fraction exhibits optical density changes due to disruption of its helix (melting or denaturation). A smaller fraction is atypical in that it requires a much higher temperature for melting. This smaller, atypical fraction of DNA must contain a higher content of
- A.
Adenine plus cytosine
- B.
Cytosine plus guanine
- C.
Adenine plus thymine
- D.
Cytosine plus thymine
- E.
Adenine plus guanine
Correct Answer
B. Cytosine plus guanineExplanation
The melting temperature Tm of duplex DNA is the temperature at which half the base pairs are denatured. Adenine-thymine (A-T) base pairs have two hydrogen bonds, in contrast to cytosine-guanine (C-G) base pairs, which have three hydrogen bonds. Duplex DNA molecules rich in A-T base pairs have a much lower Tm than those rich in C-G base pairs. As DNA is heated, fractions with a higher A-T content melt or denature before those with a higher C-G content. Most mammals, including humans, have satellite DNA fractions that are highly repetitive and clustered in particular chromosome regions. Satellite DNAs are named for their altered density (satellite band) on centrifugation, caused by higher G-C content. Their function is unknown.Rate this question:
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- 2.
A newborn baby has a sibling with sickle cell anemia (141900) and is at risk for the disease. Which of the following is the appropriate diagnostic test for sickle cell anemia in this baby?
- A.
DNA amplification
- B.
Hemoglobin antibodies
- C.
DNA restriction
- D.
Red cell counting
- E.
DNA fingerprinting
Correct Answer
A. DNA amplificationExplanation
Sufficient DNA for analysis can be obtained by amplification of leukocyte DNA using the polymerase chain reaction (PCR). Short segments of DNA (oligonucleotide primers) are designed to be complementary to areas flanking the DNA region of interest—in this case, the portion of the beta -globin gene that may harbor the sickle cell anemia (141900) mutation. Some 20 to 30 cycles of cooling (to anneal the primers), synthesis (with heat-stable DNA polymerase), and heating (to melt the DNA and allow the next cycle) can amplify the targeted DNA segment over 1 million-fold. Hybridization in duplicate to allele-specific oligonucleotides (ASOs; one ASO for the hemoglobin A mutation, one ASO for the S mutation) can establish the diagnosis of normal (AA alleles), sickle trait (AS alleles), or sickle cell anemia (SS alleles). Newborns have fetal hemoglobin ( alpha- and gamma -globin) with little expression of hemoglobin A (alpha - and beta -globin genes) until 3 to 6 months of life, so testing for anemia or for abnormal hemoglobin with antibodies would not be helpful.
DNA polymorphisms (nucleotide sequence variations) occur approximately once per 200 to 500 base pairs (bp) of human DNA. If the sequence variation affects the recognition site for a restriction endonuclease, the altered segment sizes produced by endonuclease digestion allow detection of the sequence change [restriction fragment length polymorphism (RFLP)]. If the nucleotide change causing the RFLP is adjacent to (linked with) or coincident with a disease mutation, then one size variant of the RFLP may be diagnostic. However, mutations of known sequence (such as that for sickle cell anemia) are better detected by PCR and ASOs. The use of several highly variable RFLPs produces a pattern of restriction fragments that is highly distinctive for each individual (DNA fingerprinting) but not diagnostic for a particular disease.Rate this question:
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- 3.
. A farming couple in Northern Michigan consult their physician about severe skin rashes and ulcers noted over the past year. They also have lost many cattle over the past year, and claim that their cattle feed changed in consistency and smell about 1 year ago. Chemical analysis of the feed shows high concentrations of polychlorinated biphenyls, a fertilizer related to known carcinogens. The physician sends the chemical to a laboratory for carcinogen testing, which is performed initially and rapidly by
- A.
Inoculation of the chemical into nude mice
- B.
Incubation of mutant bacteria with the chemical to measure the rate of reverse or "back" mutations
- C.
Incubation with stimulated white blood cells to measure the impact on DNA replication
- D.
Computer modeling based on the structures of related carcinogens
- E.
Incubation with mammalian cell cultures to measure the rates of malignant transformation
Correct Answer
B. Incubation of mutant bacteria with the chemical to measure the rate of reverse or "back" mutationsExplanation
The Ames test is a rapid and relatively inexpensive bacterial assay for determining mutagenicity of potential toxic chemicals. Since many chemical carcinogens are mutagenic, it seems obvious that damage to DNA is a central event in carcinogenesis as well as mutagenesis. Dr. Bruce Ames developed a tester strain of Salmonella that has been modified not to grow in the absence of histidine because of a mutation in one of the genes for the biosynthesis of histidine. Toxic chemicals that are mutagens are placed in the center of the plate and result in reversions of the original mutations, so that histidine is synthesized and the mutated revertants multiply in histidine-free media. Since many carcinogens are converted to active forms by metabolism in the liver, preliminary incubation with liver homogenates may precede the bacterial assay. Essentially all chemicals known as carcinogens in humans cause mutagenesis in the Ames test. The other options—carcinogenicity screening in immunosuppressed (nude) mice, computer modeling, or incubation with mammalian cell cultures—may provide some information, but are less efficient and validated than the Ames test. Contamination of Michigan cattle feed with polychlorinated biphenyls (PCBs) did occur through an industrial mistake.Rate this question:
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- 4.
Patients with hereditary nonpolyposis colon cancer [HNPCC (114500)] have genes with microsatellite instability, that is, many regions containing abnormal, small loops of unpaired DNA. This is a result of a mutation affecting
- A.
Mismatch repair
- B.
Chain break repair
- C.
Base excision repair
- D.
Depurination repair
- E.
Nucleotide excision repair
Correct Answer
A. Mismatch repairExplanation
One of the most common types of inherited cancers is nonpolyposis colon cancer [HNPCC (114500)]. Most cases are associated with mutations of either of two genes that encode proteins critical in the surveillance of mismatches. Mismatches are due to copying errors leading to one- to five-base unmatched pieces of DNA. Two- to five-base-long unmatched bases form miniloops. Normally, specific proteins survey newly formed DNA between adenine methylated bases within a GATC sequence. Mismatches are removed and replaced. First, a GATC endonuclease nicks the faulty strand at a site complementary to GATC, then an exonuclease digests the strand from the GATC site beyond the mutation. Finally, the excised faulty DNA is replaced. In HNPCC, the unrecognized mismatches accumulate, leading to malignant growth of colon epithelium. The other forms of DNA repair are important for rectifying damage from ultraviolet light.Rate this question:
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- 5.
A culture of bacteria not resistant to tetracycline develops an infection from a virus that is derived from the lysis of tetracycline-resistant bacteria. Most of the bacterial progeny of the original culture is found to have become resistant to tetracycline. What phenomenon has occurred?
- A.
Conjugation
- B.
Colinearity
- C.
Recombination
- D.
Transformation
- E.
Transduction
Correct Answer
E. TransductionExplanation
The process of transduction involves the transfer of a portion of DNA from one bacterium to the chromosome of another bacterium by means of a viral infection. Conjugation is the transfer of a so-called male chromosomal DNA to the DNA of an acceptor, or female, bacterial cell. Colinearity defines the relationship between genes and proteins in that the sequence of amino acids in proteins is a result of the sequence of base triplets in template genes. Recombination is simply the exchange of sequences between two molecules of DNA. Transformation results when exogenous DNA fragments are incorporated into the chromosome of another organism, as in the transformation of pneumococcal bacteria that led Avery and McLeod to recognize the genetic significance of DNARate this question:
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- 6.
Which of the following molecules is found in a nucleoside?
- A.
A pyrophosphate group
- B.
A 1' base linked to a pentose sugar
- C.
A 5'-phosphate group linked to a pentose sugar
- D.
A 3'-phosphate group linked to a pentose sugar
- E.
A terminal triphosphate
Correct Answer
B. A 1' base linked to a pentose sugarExplanation
A nucleoside consists of a purine or pyrimidine base linked to a pentose sugar. The 1' carbon of the pentose is linked to the nitrogen of the base. In DNA, 2'-deoxyribose sugars are used; in RNA, ribose sugars are used. Nucleotides are phosphate esters of nucleosides with one to three phosphate groups, such as adenosine monophosphate (AMP), adenosine diphosphate (ADP), or adenosine triphosphate (ATP). The nitrogenous bases are adenine, thymine, guanine, and cytosine in DNA, with thymine replaced by uridine in RNA. Nucleotide polymers are chains of nucleotides with single phosphate groups, joined by bonds between the 3'-hydroxyl of the preceding pentose and the 5'-phosphate of the next pentose. Polymerization requires high-energy nucleotide triphosphate precursors that liberate pyrophosphate (broken down to phosphate) during joining. The polymerization reaction is given specificity by complementary RNA or DNA templates and rapidity by enzyme catalysts called polymerases.Rate this question:
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- 7.
Which of the following enzymes can be described as a DNA-dependent RNA polymerase?
- A.
DNA ligase
- B.
Primase
- C.
DNA polymerase III
- D.
DNA polymerase I
- E.
Reverse transcriptase
Correct Answer
B. PrimaseExplanation
Primase is a DNA-dependent RNA polymerase located in the primosome at the replication fork of DNA. Primase initiates DNA synthesis by synthesizing a 10-base RNA primer. The DNA-RNA helix formed binds DNA polymerase III, which synthesizes a DNA fragment (the Okazaki fragment) in a 5' to 3' direction. When the RNA primer of the previous Okazaki fragment is met, DNA polymerase I replaces III and digests the RNA primer, replacing it with appropriate DNA bases. When the RNA primer is completely removed, DNA ligase synthesizes the last phosphodiester bond, thereby sealing the space. What is left is a new lagging strand extended by the new Okazaki fragment with the 10-base RNA primer at its 5' end. Reverse transcriptase is a DNA polymerase that uses RNA as a template found in retroviruses as well as normal eukaryotic cells. Unlike DNA polymerase I and III, which proofread for errors during normal synthesis, reverse transcriptase has no proofreading capabilities. Hence, it has an exceedingly high error rate that contributes to the high rate of mutation in retroviruses like HIV.Rate this question:
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- 8.
. The DNA sequence M, shown below, is the sense strand from a coding region known to be a mutational "hot spot" for a gene. It encodes amino acids 21 to 25. Given the genetic and amino acid codes CCC = proline (P), GCC = alanine (A), TTC = phenylalanine (F), and TAG = stop codon, which of the following sequences is a frame-shift mutation that causes termination of the encoded protein? M 5'-CCC-CCT-AGG-TTC-AGG-3'
- A.
-CCA-CCT-AGG-TTC-AGG-
- B.
-GCC-CCT-AGG-TTC-AGG-
- C.
-CCA-CCC-TAG-GTT-CAG-
- D.
-CCC-CTA-GGT-TCA-GG—
- E.
-CCC-CCT-AGG-AGG——
Correct Answer
C. -CCA-CCC-TAG-GTT-CAG-Explanation
Insertion (choice c) or deletion (choice d) of nucleotides shifts the reading frame unless the change is a multiple of 3 (choice e). Frame shifts may create unintended stop codons as in choice c. Point mutations resulting in nucleotide or amino acid substitutions are conveniently named by their position in the protein, i.e., P21A (choice b). The protein change P21A could also be denoted by the corresponding change in the DNA reading frame, i.e., C63A. Deletions may be prefixed by the letter delta, as with F25 (choice e).Rate this question:
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- 9.
The hypothetical "stimulin" gene contains two exons that encode a protein of 100 amino acids. They are separated by an intron of 100 bp beginning after the codon for amino acid 10. Stimulin messenger RNA (mRNA) has 5' and 3' untranslated regions of 70 and 30 nucleotides, respectively. A complementary DNA (cDNA) made from mature stimulin RNA would have which of the following sizes?
- A.
500 bp
- B.
400 bp
- C.
300 bp
- D.
100 bp
- E.
70 bp
Correct Answer
B. 400 bpExplanation
Exons are the coding portions of genes and consist of trinucleotide codons that guide the placement of specific amino acids into protein. Introns are the noncoding portions of genes that may function in evolution to provide "shuffling" of exons to produce new proteins. The primary RNA transcript contains both exons and introns, but the latter are removed by RNA splicing. The 5' (upstream) and 3' (downstream) untranslated RNA regions remain in the mature RNA and are thought to regulate RNA transport or translation. A poly(A) tail is added to the primary transcript after transcription, which facilitates transport and processing from the nucleus. The discovery of introns complicated Mendel's idea of the gene as the smallest hereditary unit; a modern definition might be the colinear sequence of exons, introns, and adjacent regulatory sequences that accomplish protein expression. Using these principles, one can determine the size of the stimulin gene. It contains a coding region of 300 bp (100 amino acids × 3 bp per amino acid), plus 100 bp in the intron, plus 70 + 30 = 100 bp in the untranslated regions (total = 500 bp). The mature RNA contains the same number of bp except for the 100 bp in the intron (500 – 100 = 400 bp). Transcription begins at the start of the 5' untranslated region (70 bp) and the splice site occurs 30 bp (10 × 3) into the coding region at the beginning of the intron.Rate this question:
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- 10.
The hypothetical "stimulin" gene with two exons encoding a protein of 100 amino acids is found to have abnormal expression in cell culture. Which of the following mutations would produce a 500-bp stimulin mRNA and a 133–amino acid peptide that reacts with antibodies to stimulin protein?
- A.
Splice junction mutation preventing RNA splicing
- B.
Frame-shift mutation in codon #2
- C.
Silent point mutation in the third nucleotide of codon #50
- D.
Nonsense mutation at codon #2
- E.
Deletion of exon 1
Correct Answer
A. Splice junction mutation preventing RNA splicingExplanation
Splice junction mutations will theoretically produce a larger mRNA unless the mRNA is unstable; the larger protein may have abnormal function but retain peptide regions that react with antibody to the authentic protein. Nucleotide insertions or deletions other than multiples of 3 alter the reading frame of the code and scramble the amino acid sequence distal to the frame-shift mutation. Such altered mRNAs may be of increased or smaller size, depending on their stability, as may the translated protein, depending on the presence of stop codons within the shifted reading frame. Only the protein upstream of the frame-shift mutation retains immune cross-reactivity and normal function. Point mutations (nucleotide substitutions) may have substantial functional impact if the altered codon results in an amino acid substitution. If no amino acid substitution occurs, they are called silent mutations.Rate this question:
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- 11.
Template-directed RNA synthesis occurs in which of the following?
- A.
Point mutation
- B.
Triplet repeat expansion
- C.
Initiation of the polymerase chain reaction
- D.
Expression of oncogenes
- E.
Repair of thymine dimmers
Correct Answer
D. Expression of oncogenesExplanation
Oncogenes are cancer-producing genes. They are closely related to normal cellular genes and are often tyrosine kinases, growth factors, or receptors for growth factors. The expression of oncogenes leads to the translation and eventual transcription of the protein product of the oncogene. Thus, template-directed RNA synthesis but not DNA synthesis occurs during the expression of oncogenes. In contrast, template-directed DNA synthesis rather than RNA synthesis occurs during the repair of thymine dimers, the polymerase chain reaction, the functioning of the replication fork, and the growth of RNA tumor viruses. In the final stages of the repair of thymine dimers, once the dimer has been excised, DNA polymerase I enters the gap to carry out template-directed synthesis. In functioning of the replication fork, DNA polymerase III holoenzyme carries out synthesis of DNA during replication. Template-directed DNA synthesis is required for the growth of RNA tumor viruses (retroviruses). Once released into the host cytoplasm, retroviral RNA synthesizes both the positive and minus strands of DNA, using reverse transcriptase. This unique enzyme catalyzes the initial RNA-directed DNA synthesis, hydrolysis of RNA, and then DNA-directed DNA synthesis. The newly formed viral DNA duplex integrates into the host cell DNA prior to transcription. In this form, the retrovirus is inherited by daughter host cells. The polymerase chain reaction is a method of amplifying the amount of DNA in a sample or of enriching particular DNA sequences in a population of DNA molecules. In the polymerase chain reaction, oligonucleotides complementary to the ends of the desired DNA sequence are used as primer for multiple rounds of template-directed DNA synthesis.Rate this question:
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- 12.
Which of the following most correctly describes mammalian messenger RNAs?
- A.
They are usually transcribed from both DNA strands
- B.
They are normally double-stranded
- C.
Their content of uridine equals their content of adenine
- D.
They have an overall negative charge at neutral pH
- E.
Their ratio of ribose to purine bases equals 1
Correct Answer
D. They have an overall negative charge at neutral pHExplanation
At a physiologic pH of 7.4, mRNAs (like DNA) are polyanionic owing to the negatively charged phosphate hydroxyl groups. Mammalian mRNAs are synthesized from DNA as single-stranded linear molecules. Because they are not double-stranded, the concentrations of the different bases in mRNA are variable rather than exhibiting the A = T and G = C pattern of double-stranded DNA (A does not equal U in mRNA). The hybridization of RNA with its complementary template DNA is antiparallel. In both DNA and RNA, sugar units equal base units equal phosphate units. However, their bases consist of pyrimidines as well as purines.Rate this question:
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- 13.
The western blot uses what type of probe?
- A.
Antibody
- B.
MRNA
- C.
Products of polymerase chain reaction (PCR)
- D.
TRNA
- E.
Mutant and normal oligonucleotides
Correct Answer
A. AntibodyExplanation
In an expression library, cDNA clones are screened on the basis of their ability to direct bacterial synthesis of a foreign protein of interest. Radioactive antibodies specific to this protein can be used to identify the colonies of bacteria that contain the cDNA vector. As was the case for probing genomic libraries, bacteria grown on a master plate are blotted onto a nitrocellulose replica plate and then lysed. The released proteins may then be labeled with 125I antibodies. In contrast, northern blotting can be used to identify RNA molecules separated by gel electrophoresis. In northern blotting, RNA molecules separated by gel electrophoresis can be identified by hybridization with probe DNA following transfer to nitrocellulose. Mutant and wild-type oligonucleotides can be used as probes to analyze polymerase chain reaction products. Conversely, the products of polymerase chain reaction can be used to analyze cDNA libraries.Rate this question:
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- 14.
How many high-energy phosphate-bond equivalents are utilized in the process of activation of amino acids for protein synthesis?
- A.
Zero
- B.
One
- C.
Two
- D.
Three
- E.
Four
Correct Answer
C. TwoExplanation
ATP is required for the esterification of amino acids to their corresponding tRNAs. This reaction is catalyzed by the class of enzymes known as aminoacyl-tRNA synthetases. Each one of these enzymes is specific for one tRNA and its corresponding amino acid.
amino acid + tRNA + ATP aminoacyl-tRNA + AMP + PPi
As with most ATP hydrolysis reactions that release pyrophosphate, pyrophosphatase quickly hydrolyzes the product to Pi, which makes the reaction essentially irreversible. Since ATP is hydrolyzed to AMP and PPi during the reaction, by convention the equivalent of two high-energy phosphate bonds is utilized.Rate this question:
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- 15.
An immigrant family from rural Mexico brings their 3-month-old child to the emergency room because of whistling inspiration (stridor) and high fever. The child's physician is perplexed because the throat examination shows a gray membrane almost occluding the larynx. A senior physician recognizes diphtheria, now rare in immunized populations. The child is intubated, antitoxin is administered, and antibiotic therapy is initiated. Diphtheria toxin is often lethal in unimmunized persons because it
- A.
Inhibits initiation of protein synthesis by preventing the binding of GTP to the 40S ribosomal subunit
- B.
Binds to the signal recognition particle receptor on the cytoplasmic face of the endoplasmic reticulum receptor
- C.
Shuts off signal peptidase
- D.
Blocks elongation of proteins by inactivating elongation factor 2 (EF-2, or translocase)
- E.
Causes deletions of amino acid by speeding up the movement of peptidyl-tRNA from the A site to the P site
Correct Answer
D. Blocks elongation of proteins by inactivating elongation factor 2 (EF-2, or translocase)Explanation
The gene that produces the deadly toxin of Corynebacterium diphtheriae comes from a lysogenic phage that grows in the bacteria. Prior to immunization, diphtheria was the primary cause of death in children. The protein toxin produced by this bacterium inhibits protein synthesis by inactivating elongation factor 2 (EF-2, or translocase). Diphtheria toxin is a single protein composed of two portions (A and B). The B portion enables the A portion to translocate across a cell membrane into the cytoplasm. The A portion catalyzes the transfer of the adenosine diphosphate ribose unit of NAD1 to a nitrogen atom of the diphthamide ring of EF-2, thereby blocking translocation. Diphthamide is an unusual amino acid residue of EF-2.Rate this question:
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- 16.
Methionyl–transfer (t) RNA is used for initiation of protein synthesis by which of the following?
- A.
Chloroplast ribosomes
- B.
Eukaryotic mitochondrial ribosomes
- C.
Eukaryotic cytoplasmic ribosomes
- D.
Bacterial ribosomes
- E.
Bacterial cytoplasm
Correct Answer
C. Eukaryotic cytoplasmic ribosomesExplanation
Methionyl-tRNA is the special tRNA used in eukaryotes for initiation. Initiation of protein synthesis by bacterial, mitochondrial, and chloroplast ribosomes requires N-formylmethionyl-tRNA. The mitochondria of eukaryotic cells are similar to bacteria in the size of their ribosomal RNAs (23S and 16S) and their mechanisms for protein synthesis. Mitochondrial and prokaryotic ribosomes (including those of chloroplasts and bacteria) use formylmethionyl-tRNA for initiation of protein synthesis and are sensitive to inhibitors like streptomycin, tetracycline, and chloramphenicol that have little effect on eukaryotic cells. The latter drugs are useful as antibiotics in animals and humans since they inhibit bacteria but do not gain entry into mitochondria.Rate this question:
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- 17.
Which of the following is required for certain types of eukaryotic protein synthesis but not for prokaryotic protein synthesis?
- A.
Ribosomal RNA
- B.
Messenger RNA
- C.
Signal recognition particle
- D.
Peptidyl transferase
- E.
GTP
Correct Answer
C. Signal recognition particleExplanation
Signal recognition particles (SRPs) recognize the signal sequence on the N-terminal end of proteins destined for the lumen of the endoplasmic reticulum (ER). SRP binding arrests translation and an SRP receptor facilitates import of the SRP, ribosome, and nascent protein into the ER lumen. A signal peptidase removes the signal sequence from the protein, which may remain in the membrane or be routed for secretion. Common to both eukaryotic and prokaryotic protein synthesis is the requirement for ATP to activate amino acids. The activated aminoacyl-tRNAs then interact with ribosomes carrying mRNA. Peptidyl transferase catalyzes the formation of peptide bonds between the free amino group of activated aminoacyl-tRNA on the A site of the ribosome and the esterified carboxyl group of the peptidyl-rRNA on the P site; the liberated rRNA remains on the P site.Rate this question:
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- 18.
A major obstacle to gene therapy involves the difficulty of homologous gene replacement. Which of the following strategies addresses this issue?
- A.
A recombinant vector contains complementary DNA sequences that will facilitate site-specific recombination
- B.
A recombinant vector expresses antisense nucleotides that will hybridize with the targeted mRNA
- C.
A recombinant vector replaces inessential viral genes with a functional human gene
- D.
A recombinant vector transfects patient cells, which are returned to the patient
- E.
A recombinant vector contains DNA sequences that target its expressed protein to lysosomes
Correct Answer
A. A recombinant vector contains complementary DNA sequences that will facilitate site-specific recombinationExplanation
Challenges for gene therapy include the construction of recombinant viral genomes that can propagate the replacement gene (gene constructs or vectors), delivery of the altered gene to the appropriate tissues (gene targeting), and recombination at the appropriate locus so that replacement of the defective gene is achieved (site-specific recombination). The latter step positions DNA sequences in the vector so that the replacement gene pairs and recombines precisely with homologous DNA in the native gene. Ex vivo transfection (introduction of vector DNA into patient cells outside the body) is an ideal method for gene targeting if the engineered cells can repopulate the tissue/organ in question. Transfection of bone marrow stem cells with a functional adenosine deaminase gene, followed by bone marrow transplantation back to the patient, has been successful in restoring immunity to children with severe deficiency. Even when tissue targeting and precise gene replacement are feasible, mimicking the appropriate patterns of gene expression can be a substantial barrier to gene therapy. Injection of deficient enzymes into serum (enzyme therapy) has been successful in disorders such as Gaucher's disease [231000 (storage of lipids in the spleen and bone)], and takes advantage of cellular pathways that target enzymes to lysosomes.Rate this question:
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- 19.
A family in which several individuals have arthritis and detached retina is diagnosed with Stickler syndrome. The locus for Stickler syndrome has been mapped near that for type II collagen on chromosome 12, and mutations in the COL2A1 gene have been described in Stickler syndrome. The family became interested in molecular diagnosis to distinguish normal from mildly affected individuals. Which of the following results would be expected in an individual with a promoter mutation at one COL2A1 gene locus?
- A.
Western blotting detects no type II collagen chains
- B.
Southern blotting using intronic restriction sites yields normal restriction fragment sizes
- C.
Reverse transcriptase–polymerase chain reaction (RT-PCR) detects one-half normal amounts of COL2A1 mRNA in affected individuals
- D.
Fluorescent in situ hybridization (FISH) analysis using a COL2A1 probe detects signals on only one chromosome 12
- E.
DNA sequencing reveals a single nucleotide difference between homologous COL2A1 exons
Correct Answer
C. Reverse transcriptase–polymerase chain reaction (RT-PCR) detects one-half normal amounts of COL2A1 mRNA in affected individualsExplanation
After the locus responsible for a genetic disease is mapped to a particular chromosome region, "candidate" genes can be examined for molecular abnormalities in affected individuals. The connective tissue abnormalities in Stickler syndrome (108300) make the COL2A1 collagen locus an attractive candidate for disease mutations, prompting analysis of COL2A1 gene structure and expression. Western blotting detects gene alterations that interfere with protein expression, while use of the reverse transcriptase–polymerase chain reaction (RT-PCR) detects alterations in mRNA levels. Each analysis should detect one-half the respective amounts of COL2A1 protein or mRNA in the case of a promoter mutation that abolishes transcription of one COL2A1 allele. Southern blotting detects nucleotide changes that alter DNA restriction sites, but this is relatively insensitive unless large portions of the gene are deleted. Fluorescent in situ hybridization (FISH) analysis using DNA probes from the COL2A1 locus is a sensitive method for detecting deletions of the entire locus, and DNA sequencing of the entire gene provides the gold standard for detecting any alteration in the regulatory or coding sequences. Nucleotide sequence changes are still subject to interpretation, since they may represent polymorphisms that do not alter gene function. Population studies and/or in vitro studies of gene expression are often needed to discriminate DNA polymorphisms from mutations that disrupt gene function. For any autosomal locus, the interpretation of molecular analyses is complicated by the presence of two homologous copies of the gene.Rate this question:
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- 20.
Gyrate atrophy (258870) is a rare autosomal recessive genetic disorder caused by a deficiency of ornithine aminotransferase. Affected individuals experience progressive chorioretinal degeneration. The gene for ornithine aminotransferase has been cloned, its structure has been determined, and mutations in affected individuals have been extensively studied. Which of the mutations listed below best fits with test results showing normal Southern blots with probes from all ornithine aminotransferase exons but absent enzymatic activity?
- A.
Duplication of entire gene
- B.
Two-kb deletion in coding region of gene
- C.
Two-kb insertion in coding region of gene
- D.
Deletion of entire gene
- E.
Missense mutation
Correct Answer
E. Missense mutationExplanation
Missense mutations, which cause the substitution of one amino acid for another, may significantly alter the function of the resultant protein without altering the size of DNA restriction fragments detected by Southern blotting. In this case, northern blot results would most likely also be normal. Single-base changes may also result in nonsense mutations. Large insertions or deletions in the exon or coding regions of the gene alter the Southern blot pattern and usually ablate the activity of one gene copy. In the case of an autosomal locus like that for ornithine aminotransferase, the homologous allele remains active and gives 50% enzyme activity (heterozygote or carrier range with a normal phenotype). Similar effects on enzyme activity would be predicted from complete gene deletions at one locus, while a duplication might produce 150% or 50% of normal enzyme activity depending on the status of promoter sites.Rate this question:
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- 21.
. Hurler's syndrome (252800) is caused by a deficiency of L-iduronidase, an enzyme normally expressed in most human cell types. It was demonstrated by Neufeld that exogenous L-iduronidase could be taken up by deficient cells via a targeting signal that directed the enzyme to its normal lysosomal location. Which of the following therapeutic strategies would be the most realistic and efficient mode of therapy?
- A.
Germ-line gene therapy
- B.
Heterologous bone marrow transplant
- C.
Infection with a disabled adenovirus vector that carries the L-iduronidase gene
- D.
Injection with L-iduronidase purified from human liver
- E.
Autologous bone marrow transplant after transfection with a virus carrying the L-iduronidase gene
Correct Answer
B. Heterologous bone marrow transplantExplanation
All of the modes of therapy are theoretically possible, and enzyme therapy (i.e., injection of purified enzyme) has been successful in several lysosomal deficiencies, particularly those in which the central nervous system is not affected [i.e., Gaucher's disease (231000)]. Unfortunately, antibodies frequently develop to the injected enzyme and limit the term of successful enzyme delivery. Heterologous bone marrow transplant, preferably from a related donor, offers the most realistic and effective therapy since the graft provides a permanent source of enzyme. Bone marrow transplants do have a 10% mortality, however, and the enzyme diffuses poorly into the central nervous system. Somatic gene therapy (i.e., delivery of enzyme to somatic cells via viral vectors or transfected tissue) is now possible; however, targeting of the gene product to appropriate tissues and organelles is still a problem. Transfected autologous bone marrow transplant (i.e., marrow from the patient) has been used in a few cases of adenosine deaminase deficiency, an immune disorder affecting lymphocytes. Germ-line gene therapy requires the insertion of functional genes into gametes or blastomeres of early embryos prior to birth. The potential for embryonic damage, lack of knowledge regarding developmental gene control, and ethical controversies regarding selective breeding or embryo experimentation make germ-line therapy unrealistic at present.Rate this question:
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- 22.
Which of the following mutations is most likely to be lethal?
- A.
Substitution of adenine for cytosine
- B.
Substitution of cytosine for guanine
- C.
Substitution of methylcytosine for cytosine
- D.
Deletion of three nucleotides
- E.
Insertion of one nucleotide
Correct Answer
E. Insertion of one nucleotideExplanation
Insertion of one extra nucleotide causes a frame-shift mutation and mistranslation of all the mRNA transcribed from beyond that point in the DNA. All the other mutations cited in the question usually cause an error in the identity of only one amino acid (choice a or b), removal of one amino acid from the sequence (choice d), or no error at all in the amino acid sequence (choice c). There is a chance that the mutations in choices a or b will give a "nonsense," or chain-terminator, mutation, and this is about as likely to be lethal as is a frame shift.Rate this question:
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- 23.
. In the following partial sequence of mRNA, a mutation of the template DNA results in a change in codon 91 to UAA. What type of mutation is it? 88 89 90 91 92 93 94 GUC GAC CAG UAG GGC UAA CCG
- A.
Missense
- B.
Silent
- C.
Nonsense
- D.
Suppressor
- E.
Frame shift
Correct Answer
B. SilentExplanation
The replacement of the codon UAG with UAA would be a silent mutation since both codons are "stop" signals. Thus, transcription would cease when either triplet was reached. There are three termination codons in mRNA: UAG, UAA, and UGA. These are the only codons that do not specify an amino acid. A missense or a substitution mutation is the converting of a codon specifying one amino acid to another codon specifying a different amino acid. A nonsense mutation converts an amino acid codon to a termination codon. A suppression counteracts the effects of another mutation at another codon. The addition or deletion of nucleotides results in a frame-shift mutation.Rate this question:
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- 24.
Which one of the following causes a frame-shift mutation?
- A.
Transition
- B.
Transversion
- C.
Deletion
- D.
Substitution of purine for pyrimidine
- E.
Substitution of pyrimidine for purine
Correct Answer
C. DeletionExplanation
Point mutations that are frame-shift mutations put the normal reading frame out of register by one base pair. The insertion of an extra base pair or the deletion of one or more base pairs falls into this category. Transitions and transversions are not frame-shift mutations; they are substitutions of one base pair for another. Substitutions are the most common type of mutation. In transitions, a purine is replaced by a purine or a pyrimidine by a pyrimidine. In transversions, a purine is replaced by a pyrimidine or vice versa. It has been suggested that transitions occur spontaneously owing to the tautomeric changes in base-hydrogen-bond locations. Transversions can be caused by defective DNA polymerases.Rate this question:
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- 25.
Which of the following regulators are said to act in "cis"?
- A.
The lac repressor and mammalian transcription factors
- B.
The lac repressor and the lac operator
- C.
The lac operator and mammalian enhancers
- D.
The lac operator and mammalian transcription factors
- E.
Mammalian transcription factors and enhancers
Correct Answer
C. The lac operator and mammalian enhancersExplanation
Certain regulatory elements act on genes on the same chromosome ("cis"), while others can regulate genes on the opposite chromosome ("trans"). The terminology makes analogy to carbon-carbon double bonds where two modifying groups may both be above or below the bond (cis) or opposite it (trans). Cis regulatory elements like the lac operator and promoter or mammalian enhancers are usually DNA sequences (regulatory sequences) adjoining or within the regulated gene. Trans regulatory elements like the lac repressor protein or mammalian transcription factors are usually diffusible proteins (regulatory factors) that can interact with adjoining target genes or with target genes on other chromosomes. Classification of bacterial elements as cis or trans requires mating experiments where portions of a second chromosome are introduced by transfection (with bacteriophage) or conjugation (with other bacteria). The distinction between cis and trans is fundamental for understanding how regulators work.Rate this question:
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- 26.
Which of the structural domains of mammalian regulatory factors may be called intracellular receptors?
- A.
Response elements
- B.
Antirepressor domains
- C.
Transcription-activating domains
- D.
Ligand-binding domains
- E.
DNA-binding domains
Correct Answer
D. Ligand-binding domainsExplanation
Mammalian regulatory factors are much more diverse than those of bacteria, possessing several types of structural domains. Activators of transcription, such as steroid hormones, may enter the cell and bind to regulatory factors at specific sites called ligand-binding domains; these intracellular "receptors" are analogous to G protein–linked membrane receptors that extend into the extracellular space. Response elements are not regulatory factors but DNA sequences near the transcription site for certain types of genes (e.g., steroid-responsive and heat shock–responsive genes). Regulatory factors interact with specific DNA sequences through their DNA-binding domains, and with other regulatory factors through transcription-activating domains. Some regulatory factors have antirepressor domains that counteract the inhibitory effects of chromatin proteins (histones and nonhistones).Rate this question:
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- 27.
The proopiomelanocortin (POMC) gene encodes several regulatory proteins that affect pituitary function. In different brain regions, proteins encoded by this gene have different carboxy-terminal peptides. Which of the following best explains the regulatory mechanism?
- A.
POMC transcription is regulated by different factors in different brain regions
- B.
POMC translation elongation is regulated by different factors in different brain regions
- C.
POMC transcription has different enhancers in different brain regions
- D.
POMC protein undergoes different protein processing in different brain regions
- E.
POMC protein forms different allosteric complexes in different brain regions
Correct Answer
D. POMC protein undergoes different protein processing in different brain regionsExplanation
The POMC gene provides a mammalian example in which several proteins are derived from the same RNA transcript. Unlike the polycistronic mRNA of the bacterial lactose operon, mammalian cells generate several mRNAs or proteins from the same gene by variable protein processing or by alternative splicing. Variable protein processing preserves the peptide products of some gene regions but degrades those from others. Alternative splicing would often produce proteins composed of different exon combinations with the same terminal exon and carboxy-terminal peptide, but could remove the terminal exon in some proteins and produce different C-terminal peptides. Different transcription factors or enhancers in different brain regions could regulate the total amounts of POMC gene transcript but not the types of protein produced. Elongation of protein synthesis involves GTP cleavage but is not differentially regulated in mammalian tissues.Rate this question:
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- 28.
. Two boys with mental disability are found to have mutations in a gene on the X chromosome that has no homology with globin genes. Both are also noted to have deficiency of -globin synthesis with thalassemia. Which of the following is the best explanation for their phenotype?
- A.
The mutation disrupted an enhancer for an -globin pseudogene
- B.
The mutation disrupted an X-encoded transcription factor that regulates the -globin loci
- C.
There is a second mutation that disrupts an enhancer near the -globin gene
- D.
There is a DNA rearrangement that joins the mutated X chromosome gene with an -globin gene
- E.
There is a second mutation that disrupts the promoter of an -globin gene
Correct Answer
B. The mutation disrupted an X-encoded transcription factor that regulates the -globin lociExplanation
The boys have an X-linked recessive condition called thalassemia/mental retardation or ATR-X syndrome (309510). The X-encoded gene has an unknown function in brain as well as being a factor that regulates -globin gene transcription. In order to affect all four -globin genes, the X-encoded gene must produce a trans-acting factor; second mutations altering enhancers or promoters would be cis-acting and affect only one -globin gene. Pseudogenes are functionless gene copies, so altered expression would not influence -globin chain synthesisRate this question:
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- 29.
A middle-aged man presents with a markedly enlarged tonsil and recurrent infections with serum immunoglobulin deficiency. Chromosome analysis demonstrates a translocation between the immunuglobulin heavy chain locus on chromosome 14 and an unidentified gene on chromosome 8. Which of the following is the most likely cause of his phenotype?
- A.
The translocation has deleted constant chain exons on chromosome 14 and prevented heavy chain class switching
- B.
The translocation has deleted the interval containing diversity (D) and joining (J) regions
- C.
The translocation has activated a tumor-promoting gene on chromosome 8
- D.
The translocation has deleted the heavy chain constant chain C so that virgin B cells cannot produce IgM on their membranes
- E.
The translocation has deleted an immunoglobulin transcription factor gene on chromosome 8
Correct Answer
C. The translocation has activated a tumor-promoting gene on chromosome 8Explanation
This case is an example of Burkitt's lymphoma, which may affect the tonsils or other lymphoid tissues. The translocation places the myc oncogene on chromosome 8 downstream of the very active heavy chain locus on chromosome 14, activating myc gene expression in B cells and their derivatives. The translocation is likely an aberrant form of the normal DNA rearrangements that generate unique heavy chain genes in each B cell. The translocation joins one chromosome 8 to one chromosome 14, leaving their homologues unaffected. The cause for the phenotype must therefore be trans-acting, since cis-acting effects would pertain only to the translocated loci and not affect the homologous untranslocated loci. Activation of a tumor-promoting gene (oncogene) on chromosome 8 could produce an enlarged tonsil, while underactivity of immunoglobulin production due to one-half expression could decrease immune function but would not completely ablate the processes in choices a, b, d, and e. At the genetic level, trans-acting events are autosomal dominant in that one of the two homologous loci is abnormal and produces a phenotype. Mutations of cis-acting events must disrupt both homologous loci to produce phenotypes, making them autosomal recessive at the genetic level.Rate this question:
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- 30.
A 2-day-old neonate becomes lethargic and uninterested in breast-feeding. Physical examination reveals tachypnea (rapid breathing) with a normal heartbeat and breath sounds. Initial blood chemistry values include normal glucose, sodium, potassium, chloride, and bicarbonate (HCO3–) levels; initial blood gas values reveal a pH of 7.53, partial pressure of oxygen (PO2) normal at 103 mmHg, and partial pressure of carbon dioxide (PCO2) decreased at 27 mmHg. Which of the following is the most appropriate treatment?
- A.
Administer alkali to treat metabolic acidosis
- B.
Administer alkali to treat respiratory acidosis
- C.
Decrease the respiratory rate to treat metabolic acidosis
- D.
Decrease the respiratory rate to treat respiratory alkalosis
- E.
Administer acid to treat metabolic alkalosis
Correct Answer
D. Decrease the respiratory rate to treat respiratory alkalosisExplanation
Tachypnea in term infants may result from brain injuries or metabolic diseases that irritate the respiratory center. The increased respiratory rate removes ("blows off") carbon dioxide from the lung alveoli and lowers blood CO2, forcing a shift in the indicated equilibrium toward the left:
CO2 + H2O ==== HxCO3 ==== H+ + HCO3-
Carbonic acid (H2CO2) can be ignored because negligible amounts are present at physiologic pH, leaving the equilibrium:
CO2 + H2O ===== H+ + HCO3-
The leftward shift to replenish exhaled CO2 decreases the hydrogen ion (H+) concentration and increases the pH (-log10[H+]) to produce alkalosis (blood pH above the physiologic norm of 7.4). This respiratory alkalosis is best treated by diminishing the respiratory rate to elevate the blood [CO2], force the above equilibrium to the right, elevate the [H+], and decrease the pH. The newborn does not have acidosis, defined as a blood pH below 7.4, either from excess blood acids (metabolic acidosis) or from increased [CO2] (respiratory acidosis). The baby also does not have metabolic alkalosis, caused by loss of hydrogen ion from the kidney (e.g., with defective tubular filtration) or stomach (e.g., with severe vomiting).Rate this question:
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- 31.
. A newborn with tachypnea and cyanosis (bluish color) is found to have a blood pH of 7.1. A serum bicarbonate is measured as 12 mM, but the blood gas machine that would determine the partial pressures of oxygen (PO2) and carbon dioxide (PCO2) is broken. Recall the pKa of 6.1 for carbonic acid (reflecting the HCO3–/CO2 equilibrium in blood) and the fact that the blood CO2 concentration is equal to the PCO2 in mmHg (normal value = 40 mmHg) multiplied by 0.03. Which of the following is the most appropriate treatment?
- A.
Administer oxygen to improve tissue perfusion and decrease metabolic acidosis
- B.
Administer oxygen to decrease respiratory acidosis
- C.
Increase the respiratory rate to treat respiratory acidosis
- D.
Decrease the respiratory rate to treat respiratory acidosis
- E.
Administer medicines to decrease renal hydrogen ion excretion
Correct Answer
A. Administer oxygen to improve tissue perfusion and decrease metabolic acidosisExplanation
The equilibrium between an acid and its conjugate base is defined by the Henderson-Hasselbalch equation:
pH = pKa + log ( [base]/[acid] ) or
pH = 6.1 + log ( [HCO3-]/[CO2] )
in the case of carbonic acid. Note that CO2 is the effective acid and HCO3– the conjugate base for carbonic acid due to its complete dissociation in water.
Given a pH of 7.1 in the cyanotic newborn, then 7.1 – 6.1 = 1 = log (10) = log [HCO3–]/[CO2] = log [HCO3–]/0.03 × PCO2. Since the [HCO3–] is 12 mM, the PCO2 × 0.03 must be 1.2 mM and the PCO2 40 mmHg. This normal calculated value for PCO2 means that the baby must have metabolic acidosis, a common accompaniment of hypoxia (low PO2) that can be treated by providing oxygen or administering alkali to ameliorate the acidosis. If the baby had respiratory acidosis, the PCO2 would be elevated; this would be treated by increasing the respiratory rate to blow off CO2. Renal treatment of acidosis would require increasing acid excretion or alkali retention. The lungs compensate acidosis with increased breathing rates or tidal volumes to blow off CO2 and increase pH, the kidneys by retaining HCO3–. The lungs can compensate alkalosis somewhat by decreasing breathing rates or volumes to retain CO2 (and decrease oxygenation within limits), the kidneys by increasing excretion of HCO3–.Rate this question:
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- 32.
A diabetic teenager is found to have a pH of 7.1 and normal electrolyte levels (Na+ = 140 mM, K+ = 4 mM, Cl– = 103 mM) except for a bicarbonate of 11 mM (normal 22 to 28 mM). The urine tests positive for ketone bodies, mostly due to acetoacetic acid and acetoacetate (CH3C=OCH2COOH and CH3C=OCH2COO–), which have a pK of 4.8. In this case, it is assumed that acetoacetate is the only significant anion in the blood besides chloride, and that each acetoacetate anion binds and removes one sodium cation during excretion by the kidney. Given that the patient has a normal glomerular filtration rate of about 7 L of blood per hour without any retention of acetoacetate/acetoacetic acid, the rates of sodium, acetoacetate, and acetoacetic acid loss will be
- A.
10 mmol/h of each species
- B.
50 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretion
- C.
100 mmol/h of sodium and acetoacetic acid, virtually no acetoacetate excretion
- D.
200 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretion
- E.
300 mmol/h of each species
Correct Answer
D. 200 mmol/h of sodium and acetoacetate, virtually no acetoacetic acid excretionExplanation
The sum of the major cations in blood (Na+ plus K+) minus the sum of the major anions (HCO3– plus Cl–) is called the anion gap (filled by phosphate ions, negatively charged proteins, etc.). An anion gap over 20 suggests the presence of an abnormal anion, such as acetoacetate, which occurs in diabetics. The anion gap in this teenager is elevated at 30 (140 + 4) – (103 + 11). Assuming that all of the gap is made up by acetoacetate anion, then 7 L × 30 mmol/L = 210 mmol of acetoacetate and 210 mmol of sodium excreted per hour. Even with acidosis and a pH of 7.1, virtually all of the acetoacetic acid (pK 4.8) is present as acetoacetate and less than 1% is excreted as acetoacetic acid. To calculate the exact amount of acetoacetic acid present, the Henderson-Hasselbalch equation rearranges to (7.1 – 4.8) = 2.3 = log[base]/[acid]; [CH3C=OCH2COO–]/[CH3C=OCH2COOH] = antilog 2.3 = 102. Less than 0.3 mmol of acetoacetic acid is excreted for each liter of blood filtered through the kidney.Rate this question:
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- 33.
Which of the combinations of laboratory results below indicates compensated metabolic alkalosis?
- A.
Low PCO2, normal bicarbonate, high pH
- B.
Low PCO2, low bicarbonate, low pH
- C.
Normal PCO2, low bicarbonate, low pH
- D.
High PCO2, normal bicarbonate, low pH
- E.
High PCO2, high bicarbonate, high pH
Correct Answer
E. High PCO2, high bicarbonate, high pHExplanation
Pure metabolic acidosis (choice c) or pure metabolic alkalosis exhibits abnormal bicarbonate and normal lung function. Pure respiratory acidosis (choice d) or alkalosis (choice a) is associated with normal renal function (and normal blood acids) with a normal bicarbonate and abnormal PCO2. Thus choices b and e must involve compensation, since both the PCO2 and bicarbonate are abnormal. Choice e must represent compensated metabolic alkalosis since the PCO2 is high—if it were compensated respiratory acidosis with a high PCO2, the pH would be low.Rate this question:
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- 34.
. The pH of body fluids is stabilized by buffer systems. Which of the following compounds is the most effective buffer at physiologic pH?
- A.
Na2HPO4, pKa5 12.32
- B.
NH4OH, pKa5 9.24
- C.
NaH2PO4, pKa5 7.21
- D.
CH3CO2H, pKa5 4.74
- E.
Citric acid, pKa5 3.09
Correct Answer
C. NaH2PO4, pKa5 7.21Explanation
In any fluid, maximum buffering action is achieved by the acid whose pKa most nearly approximates the pH of the fluid. Physiologic pH is about 7.4, so that among those buffers listed in the question, NaH2PO4 is the most effective.Rate this question:
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- 35.
A child presents with severe vomiting, dehydration, and fever. Initial blood studies show acidosis with a low bicarbonate and an anion gap (the sum of sodium plus potassium minus chloride plus bicarbonate is 40 and larger than the normal 20 to 25). Preliminary results from the blood amino acid screen show two elevated amino acids, both with nonpolar side chains. A titration curve performed on one of the elevated species shows two ionizable groups with approximate pKs of 2 and 9.5. The most likely pair of elevated amino acids consists of
- A.
Aspartic acid and glutamine
- B.
Glutamic acid and threonine
- C.
Histidine and valine
- D.
Leucine and isoleucine
- E.
Glutamine and isoleucine
Correct Answer
D. Leucine and isoleucineExplanation
Leucine and isoleucine have nonpolar methyl groups as side chains. As for any amino acid, titration curves obtained by noting the change in pH over the range of 1 to 14 would show a pK of about 2 for the primary carboxyl group and about 9.5 for the primary amino group; there would be no additional pK for an ionizable side chain. Recall that the pK is the point of maximal buffering capacity when the amounts of charged and uncharged species are equal (see answer to question 104). Aspartic and glutamic acids (second carboxyl group), histidine (imino group), and glutamine (second amino group) all have ionizable side chains that would give an additional pK on the titration curve. The likely diagnosis here is maple syrup urine disease, which involves elevated isoleucine, leucine, and valine together with their ketoacid derivatives. The ketoacid derivatives cause the acidosis, and the fever suggests that the metabolic imbalance was worsened by an infection.Rate this question:
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- 36.
. Which of the hemoglobin designations below best describes the relationship of subunits in the quaternary structure of adult hemoglobin?
- A.
( alpha1- alpha2)( beta1-beta 2)
- B.
Alpha 1-alpha2- alpha 3-alpha 4
- C.
Beta-beta-beta-alpha
- D.
Beta1-beta2-beta3-alpha1
- E.
(alpha1-beta1)- (alpha2-beta2)
Correct Answer
E. (alpHa1-beta1)- (alpHa2-beta2)Explanation
Adult hemoglobin, or hemoglobin A, is composed of four polypeptide chains. Two of the chains are alpha chains and two are beta chains. The chains are held together by noncovalent interactions. The hemoglobin tetramer can best be represented as being composed of two dimers, each containing the two different polypeptides. Thus the designation ( alpha1- beta1) (Lpha 2- beta2), which refer to dimers 1 and 2, respectively, is the most correct way to refer to the quaternary structure of adult hemoglobin. Hydrophobic interactions are thought to be the main noncovalent interactions holding all four polypeptides together.Rate this question:
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- 37.
Blood is drawn from a child with severe anemia and the hemoglobin protein is degraded for peptide and amino acid analysis. Of the following results, which change in hemoglobin primary structure is most likely to correlate with the clinical phenotype of anemia?
- A.
Ile-leu-val to ile-ile-val
- B.
Leu-glu-ile to leu-val-ile
- C.
Gly-ile-gly to gly-val-gly
- D.
Gly-asp-gly to gly-glu-gly
- E.
Val-val-val to val-leu-val
Correct Answer
B. Leu-glu-ile to leu-val-ileExplanation
Primary protein structures denote the sequence of amino acids held together by peptide bonds (carboxyl groups joined to amino groups to form amide bonds). The types of amino acids then determine the secondary structure of peptide regions within the protein, sometimes forming spiral helices or flat pleated sheets. These regional peptide secondary structures then determine the overall three-dimensional tertiary structure of a protein, which is vital for its function. Amino acid substitutions that alter the charge of an amino acid side chain, like the change from glutamic acid (charged carboxyl group) to valine (nonpolar methyl groups) in choice b, are most likely to change the secondary and tertiary protein structure. A change in hemoglobin structure can cause instability, decreased mean cellular hemoglobin concentration (MCHC), and anemia. A change from glutamic acid to valine at position 6 in the beta -hemoglobin chain is the mutation responsible for sickle cell anemia.Rate this question:
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- 38.
Parents bring in their 2-week-old child fearful that he has ingested a poison. They had delayed disposing one of the child's diapers, and noted a black discoloration where the urine had collected. Later, they realized that all of the child's diapers would turn black if stored as waste for a day or so. Knowing that phenol groups can complex to form colors, which of the following amino acid pathways are implicated in this phenomenon?
- A.
The phenylalanine, tyrosine, and homogentisate pathway
- B.
The histidine pathway
- C.
The leucine, isoleucine, and valine pathway
- D.
The methionine and homocystine pathway
- E.
The arginine and citrulline pathway (urea cycle)
Correct Answer
A. The pHenylalanine, tyrosine, and homogentisate pathwayExplanation
Lack of the enzyme homogentisate oxidase causes the accumulation of homogentisic acid, a metabolite in the pathway of degradation of phenylalanine and tyrosine. Homogentisate, like tyrosine, contains a phenol group. It is excreted in the urine, where it oxidizes and is polymerized to a dark substance upon standing. Under normal conditions, phenylalanine is degraded to tyrosine, which is broken down through a series of steps to fumarate and acetoacetate. The dark pigment melanin is another end product of this pathway. Deficiency of homogentisate oxidase is called alkaptonuria (black urine), a mild disease discovered by Sir Archibald Garrod, the pioneer of biochemical genetics. Garrod's geneticist colleague, William Bateson, recognized that alkaptonuria, like nearly all enzyme deficiencies, exhibits autosomal recessive inheritance.Rate this question:
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- 39.
Certain amino acids are not part of the primary structure of proteins but are modified after translation. In scurvy, which amino acid that is normally part of collagen is not synthesized?
- A.
Hydroxytryptophan
- B.
Hydroxytyrosine
- C.
Hydroxyhistidine
- D.
Hydroxyalanine
- E.
Hydroxyproline
Correct Answer
E. HydroxyprolineExplanation
Hydroxyproline and hydroxylysine are not present in newly synthesized collagen. Proline and lysine residues are modified by hydroxylation in a reaction requiring the reducing agent ascorbic acid (vitamin C). The enzymes catalyzing the reactions are prolyl hydroxylase and lysyl hydroxylase. In scurvy, which results from a deficiency of vitamin C, insufficient hydroxylation of collagen causes abnormal collagen fibrils. The weakened collagen in teeth, bone, and blood vessels causes tooth loss, brittle bones with fractures, and bleeding tendencies with bruising and bleeding gums.Rate this question:
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- 40.
A newborn female has a large and distorted cranium, short and deformed limbs, and very blue scleras (whites of the eyes). Radiographs demonstrate multiple limb fractures and suggest a diagnosis of osteogenesis imperfecta (brittle bone disease). Analysis of type I collagen protein, a triple helix formed from two 1 and one 2 collagen chains, shows a 50% reduction in the amount of type I collagen in the baby's skin. DNA analysis demonstrates the presence of two normal 1 alleles and one normal 2 allele. Which of the following is the best explanation of these results?
- A.
. Deficiency of alpha1 collagen peptide synthesis
- B.
Inability of alpha 1 chains to incorporate into triple helix
- C.
Defective alpha1 chains that interrupt triple helix formation
- D.
Incorporation of defective alpha2 chains that cause instability and degradation of the triple helix
- E.
A missense mutation that alters the synthesis of alpha1 chains
Correct Answer
D. Incorporation of defective alpHa2 chains that cause instability and degradation of the triple helixExplanation
Collagen peptides assemble into helical tertiary structures that form quaternary triple helices. The triple helices in turn assemble end to end to form collagen fibrils that are essential for connective tissue strength. Over 15 types of collagen contribute to the connective tissue of various organs, including the contribution of type I collagen to eyes, bones, and skin. The fact that only one of two alpha 2 alleles is normal in this case implies that a mutant alpha2 allele could be responsible for the disease (even if the alpha 2 locus is on the X chromosome, since the baby is female with two X chromosomes). The mutant alpha 2 collagen peptide would be incorporated into half of the type I collagen triple helices, causing a 50% reduction in normal type I collagen. (A mutant alpha 1 collagen peptide would distort 75% of the molecules since two alpha 1 peptides go into each triple helix.) The ability of one abnormal collagen peptide allele to alter triple-helix structure with subsequent degradation is well documented and colorfully named protein suicide or, more properly, a dominant-negative mutation.Rate this question:
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- 41.
A child with tall stature, loose joints, and detached retinas is found to have a mutation in type II collagen. Recall that collagen consists of a repeating tripeptide motif where the first amino acid of each tripeptide is the same. Which of the following amino acids is the recurring amino acid most likely to be altered in mutations that distort collagen molecules?
- A.
Glycine
- B.
. Hydroxyproline
- C.
Hydroxylysine
- D.
Tyrosine
- E.
Tryptophan
Correct Answer
A. GlycineExplanation
The primary structure of collagen peptides consists of repeating tripeptides with a gly-X-Y motif, where gly is glycine and X and Y are any amino acid. The small CH2 group connecting the amino and carboxyl groups of glycine contrasts with the larger connecting groups and side chains of other amino acids. The small volume of glycine molecules is crucial for the helix secondary structure of collagen peptides. This in turn is necessary for their tertiary helical structure and their assembly into quaternary tripeptide, triple-helix structures. The most severe clinical phenotypes caused by amino acid substitutions in collagen peptides are those affecting glycine that prevent alpha helix formation. The child has a disorder called Stickler syndrome (108300) that exhibits autosomal dominant inheritance.Rate this question:
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- 42.
Which of the following techniques for purification of proteins can be made specific for a given protein?
- A.
Dialysis
- B.
Affinity chromatography
- C.
Gel filtration chromatography
- D.
Ion exchange chromatography
- E.
Electrophoresis
Correct Answer
B. Affinity chromatograpHyExplanation
Each of the techniques listed separates proteins from each other and from other biologic molecules based upon characteristics such as size, solubility, and charge. However, only affinity chromatography can use the high affinity of proteins for specific chemical groups or the specificity of immobilized antibodies for unique proteins. In affinity chromatography, a specific compound that binds to the desired protein—such as an antibody, a polypeptide receptor, or a substrate—is covalently bound to the column material. A mixture of proteins is added to the column under conditions ideal for binding the protein desired, and the column is then washed with buffer to remove unbound proteins. The protein is eluted either by adding a high concentration of the original binding material or by making the conditions unfavorable for binding (e.g., changing the pH). The other techniques are less specific than affinity binding for isolating proteins. Dialysis separates large proteins from small molecules. Ion exchange chromatography separates proteins with an overall charge of one sort from proteins with an opposite charge (e.g., negative from positive). Gel filtration chromatography separates on the basis of size. Electrophoresis separates proteins on the principle that net charge influences the rate of migration in an electric field.Rate this question:
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- 43.
. A solution of glutamic acid is titrated from pH 1.0 to 7.0 by the addition of 5 mL of a solution of 1 M NaOH. What is the approximate number of millimoles of amino acid in the sample (pKa1 = 2.19, pKa2 = 4.25, pKa3 = 9.67)?
- A.
1.5
- B.
3.0
- C.
6.0
- D.
12.0
- E.
18.0
Correct Answer
B. 3.0Explanation
To reach pH 7.0, approximately 100% of the -carboxyl group (pKa1 = 2.19) and 90% of the -carboxyl group (pKa2 = 4.25) of glutamic acid must be dissociated. At that pH, approximately twice the amount of NaOH as glutamic acid molecules has been utilized to titrate the two carboxyl groups. Since each milliliter of a 1 M NaOH solution contains 1 mmol of OH– ion, about 3 mmol of the amino acid is present.Rate this question:
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- 44.
Which one of the following proteolytic enzymes is activated by acid hydrolysis of the proenzyme form?
- A.
Trypsin
- B.
Chymotrypsin
- C.
Elastase
- D.
Pepsin
- E.
Carboxypeptidase
Correct Answer
D. PepsinExplanation
Pepsin is secreted in a proenzyme form in the stomach. Unlike the majority of proenzymes, it is not activated by protease hydrolysis. Instead, spontaneous acid hydrolysis at pH 2 or lower converts pepsinogen to pepsin. Hydrochloric acid secreted by the stomach lining creates the acid environment. All the enzymes secreted by the pancreas are activated at the same time upon entrance into the duodenum. This is accomplished by trypsin hydrolysis of the inactive proenzymes trypsinogen, chymotrypsinogen, procarboxypeptidase, and proelastase. Primer amounts of trypsin are derived from trypsinogen by the action of enteropeptidase secreted by the cells of the duodenum.Rate this question:
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- 45.
Which one of the following structures may be classified as a hydrophobic amino acid at pH 7.0?
- A.
Isoleucine
- B.
Arginine
- C.
Aspartic acid
- D.
Lysine
- E.
Threonine
Correct Answer
A. IsoleucineExplanation
The carbon next to a carboxyl (C=O) group may be designated as the alpha carbon, with subsequent carbons as beta , gamma , delta , etc. alpha -amino acids contain an amino group on their carbon, as distinguished from compounds like gamma-aminobutyric acid, in which the amino group is two carbons down (gamma -carbon). In alpha-amino acids the amino acid, carboxylic acid, and the side chain or R group are all bound to the central alpha-carbon, which is thus asymmetric (except when R is hydrogen, as for glycine). Amino acids are classified as acidic, neutral hydrophobic, neutral hydrophilic, or basic, depending on the charge or partial charge on the R group at pH 7.0. Hydrophobic (water-hating) groups are carbon-hydrogen chains like those of leucine, isoleucine, glycine, or valine. Basic R groups, such as those of lysine and arginine, carry a positive charge at physiologic pH owing to protonated amide groups, while acidic R groups, such as glutamic acid, carry a negative charge owing to ionized carboxyl groups. Threonine with its hydroxyl side chain is neutral at physiologic pH. Neutral hydrophilic side chains have uncharged but polar or partially charged groups.Rate this question:
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- 46.
Which of the following amino acids is ionizable in proteins?
- A.
Leucine
- B.
Histidine
- C.
Valine
- D.
Alanine
- E.
Glycine
Correct Answer
B. HistidineExplanation
Except for terminal amino acids, all alpha -amino groups and all alpha-carboxyl groups are utilized in peptide bonds. Thus only amino acids with side chains may be considered. Of these, 7 of the 20 common amino acids have easily ionizable side chains. These are the basic amino acids lysine, arginine, and histidine; the acidic amino acids aspartate and glutamate; and tyrosine and cysteine. Leucine, valine, and alanine have hydrocarbon side chains.Rate this question:
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- 47.
The oxygen carrier of muscle is the globular protein myoglobin. Which one of the following amino acids is highly likely to be localized within the interior of the molecule?
- A.
Arginine
- B.
Aspartic acid
- C.
Glutamic acid
- D.
Valine
- E.
Lysine
Correct Answer
D. ValineExplanation
The structure of myoglobin is illustrative of most water-soluble proteins. Globular proteins tend to fold into compact configurations with nonpolar cores. The interior of myoglobin is composed almost exclusively of nonpolar, hydrophobic amino acids like valine, leucine, phenylalanine, and methionine. In contrast, polar hydrophilic residues such as arginine, aspartic acid, glutamic acid, and lysine are found mostly on the surface of the water-soluble protein.Rate this question:
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- 48.
A child stops making developmental progress at age 2 years and develops coarse facial features with thick mucous drainage. Skeletal deformities appear over the next year, and the child regresses to a vegetative state by age 10 years. The child's urine tests positive for glycosaminoglycans that include which of the following molecules?
- A.
Collagen
- B.
gamma-aminobutyric acid
- C.
Heparan sulfate
- D.
Glycogen
- E.
Fibrillin
Correct Answer
C. Heparan sulfateExplanation
Glycosaminoglycans (mucopolysaccharides) are polysaccharide chains that may be bound to proteins as proteoglycans. Each proteoglycan is a complex molecule with a core protein that is covalently bound to glycosaminoglycans—repeating units of disaccharides. The amino sugars forming the disaccharides contain negatively charged sulfate or carboxylate groups. The primary glycosaminoglycans found in mammals are hyaluronic acid, heparin, heparan sulfate, chondroitin sulfate, and keratan sulfate. Inborn errors of glycosaminoglycan degradation cause neurodegeneration and physical stigmata described by the outmoded term "gargoylism." Glycogen is a polysaccharide of glucose used for energy storage, and has no sulfate groups. Collagen and fibrillin are important proteins in connective tissue. gamma-aminobutyric acid is a gamma-amino acid involved in neurotransmission.Rate this question:
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- 49.
Under normal conditions in blood, which of the following amino acid residues of albumin is neutral?
- A.
Arginine
- B.
Aspartate
- C.
Glutamine
- D.
Glutamate
- E.
Histidine
Correct Answer
C. GlutamineExplanation
In blood and other solutions at physiologic pH (approximately 7.0), only terminal carboxyl groups, terminal amino groups, and ionizable side chains of amino acid residues in proteins have charges. The basic amino acids lysine, arginine, and histidine have positive charges (protonated amines). The acidic amino acids aspartate and glutamate have negative charges (ionized carboxyls). Glutamine possesses an uncharged but hydrophilic side chain.Rate this question:
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- 50.
Which of the substances below is primarily found in tendons?
- A.
Collagen
- B.
Troponin
- C.
Fibrillin
- D.
Fibrin
- E.
Fibronectin
Correct Answer
A. CollagenExplanation
Collagens are insoluble proteins that have great tensile strength. They are the main fibers composing the connective tissue elements of skin, bone, teeth, tendons, and cartilage. Collagen is composed of tropocollagen, a triple-stranded helical rod rich in glycine, proline, and hydroxyproline residues. Troponin is found in muscle, fibrillin in heart valves, blood vessels, and ligaments [it is defective in Marfan's syndrome (154700)]. Fibrin is a component of blood clots and fibronectin is a component of extracellular matrix.Rate this question:
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