a. Missense mutation that may interfere with glucocerebrosidase function and could be diagnostic of Gaucher disease

b. Silent mutation that should not interfere with glucocerebrosidase function and is not diagnostic of Gaucher disease

c. Nonsense mutation that produces a truncated, nonfunctional glucocerebrosidase polypeptide and is diagnostic of Gaucher disease

d. Suppressor mutation that interferes with glucocerebrosidase mRNA transcription and is diagnostic of Gaucher disease

e. Frameshift mutation that produces a nonfunctional glucocerebrosidase and is diagnostic of Gaucher disease

225. A 3-month-old Caucasian boy fails to grow and has very low muscle tone with strange “kinky” hair. A diagnosis of Menkes kinky hair syndrome (MIM*309400) is made by showing deficient copper in fibroblasts. A likely candidate gene encoding an ATP-dependent copper transporter protein is known, and antibodies to this protein are available. Experimental analysis of the patient’s fibroblast DNA is performed to see if a pathologic mutation can be found, thus confirming the copper transport gene as the cause of Menkes disease. The table below shows a partial DNA sequence from the transporter coding strand as amplified by PCR; the amplified sequence is from a region where a TATA box has been located.

Which of the following experimental results would be the most compatible with a pathologic mutation in the transporter gene

a. Usual band of usual intensity on Northern and Western blots

b. Absent bands on Northern and Western blots

c. Usual band plus a second band of lower size on Northern blot, usual band on Western blot

d. Usual band on Northern blot, band of unusual size and reduced intensity on Western blot

e. Two unusual bands on Northern blot, usual bands on Western blot

226. Lipoproteins transport lipids from intestine and liver, where they are absorbed and metabolized to peripheral tissues via the bloodstream. They contain core proteins (apo proteins) that bind various types of lipids, thus producing lipoprotein complexes of varying densities—very low density lipoproteins (VLDL), low-density lipoproteins (LDL), and high-density lipoproteins (HDL) after plasma lipoprotein electrophoresis. A disorder called abetalipoproteinemia (MIM*200100) was distinguished by its deficiency of the bands for very low density lipoproteins (VLDL) and low-density lipoproteins (LDL) after plasma electrophoresis. Affected patients have ataxia (wide gait, incoordination), retinopathy (retinal degeneration with blindness), and myopathy (muscle weakness). VLDL and LDL contain the same apoB core protein, their different densities produced by association with different lipids, and apoB became good candidate for mutation in the disorder. Initial characterization of the apoB gene together with its mRNA and protein products demonstrated identical mRNA sizes in all tissues. Western blotting using antibody specific for the amino-terminus of apoB protein showed a 100-kDa species in liver and a 48-kDa species in intestine, as did antibody specific for the C-terminus of the 48-kDa intestinal protein. Molecular study of a patient with abetalipoproteinemia showed normal mRNA sizes, but 100-kDa peptide species were identified in both liver and intestine using the amino- or carboxy-terminus antibody probes. Which of the following processes is most likely deficient in this patient with apobetalipoproteinemia?

a. RNA splicing

b. DNA amplification

c. Transcription initiation

d. Transcription factor phosphorylation

e. RNA editing

227. During the analyses described in Question 226, another patient with abetalipoproteinemia was found to have negligible amounts of apoB protein in either liver or intestine. Analysis of the patient’s DNA and RNA would most likely yield which of the following results?

a. Substitution of the last base of a codon near the middle of the first apoB exon

b. Insertion of two bases near the apoB amino-terminus

c. Transversion producing a nonsense mutation within the first apoB intron

d. Transition affecting the 30th base of the first apoB intron

e. Base substitution changing glycine for alanine near the apoB carboxyl-terminus

228. Part of the triplet genetic code involving mRNA codon triplets that start with U is shown below. Using the portion of the genetic code shown, which of the following mutations in the 3′ to 5′ DNA template segments corresponds to a nonsense mutation?

a. ACGACGACG to ACAAACACG

b. AGGAATATG to AGGAATATT

c. AGAATAACA to AAAATAACA

d. AAAATGAGC to AAAATAAGC

e. AACAACAAC to AACAAGAAC

229. The lactose operon is negatively controlled by the lactose repressor and positively controlled by which of the following?

a. Increased concentrations of glucose and cyclic AMP (cAMP)

b. Decreased concentrations of glucose and cAMP

c. Increased concentrations of glucose, decreased concentration of cAMP

d. Decreased concentrations of glucose, increased concentration of cAMP

e. Increased concentrations of glucose and adenosine triphosphate (ATP)

230. Which of the following regulators are said to act in “cis?”

a. The lac repressor and the lac operator

b. The lac operator and mammalian transcription factors

c. The lac operator and mammalian enhancers

d. Mammalian transcription factors and enhancers

e. The lac repressor and mammalian transcription factors

231. The proopiomelanocortin (POMC-MIM*176830) gene encodes several regulatory proteins that affect pituitary function. Children with severe brain defects such as holoprosencephaly (MIM*157170) often have abnormalities in the hypothalamic-pituitary axis. 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.

232. A 3-year-old Chinese boy has severe anemia with prominence of the forehead (frontal bossing) and cheeks. The red cell hemoglobin concentration is dramatically decreased, and it contains only β-globin chains with virtual deficiency of α-globin chains. Which of the following mechanisms is the most likely explanation?

a. A transcription factor regulating the α-globin gene is mutated.

b. A regulatory sequence element has been mutated adjacent to an α-globin gene.

c. A transcription factor regulating the β-globin gene is mutated.

d. A transcription factor regulating the α- and β-globin genes is deficient.

e. A deletion has occurred surrounding an α-globin gene.

233. A mutation that results in a valine replacement for glutamic acid at position 6 of the β chain of hemoglobin S hinders normal hemoglobin function and results in sickle cell anemia (MIM*602903) when the patient is homozygous for this mutation. This is an example of which of the following types of mutation?

a. Deletion

b. Frameshift

c. Insertion

d. Missense

e. Nonsense

Gene Expression and Regulation

Answers

195. The answer is b. (Murray, pp 396-398. Scriver, pp 3-45. Lewis, pp 188-194.) The “genetic code” uses 64 possible combinations of three-nucleotide “words” (codons) each specifying one of 20 amino acids or three stop signals that catalyze translational termination when present in the transcribed messenger RNA (answer b correct, answers a, c, and d incorrect). Single nucleotide changes in genes (point mutations) can thus change the codon in transcribed mRNA and produce single amino acid changes in the translated protein. Amino acid changes in proteins implied by altered enzyme activity or altered electrophoretic migration (eg, the abnormal Z form of AAT) can now be predicted by DNA sequencing of the encoding gene to demonstrate specific altered codons (eg, the GAG to AAG change in the AAT gene that changed the 342nd amino acid from lysine to glutamine. Mutations are best described by reference to amino acid position, since the linear correspondence of codons in DNA and of amino acids in protein domains is interrupted by the presence of introns in DNA (answer e is incorrect). Accumulation of abnormal AAT protein in liver or other cells can be detected by periodic acid Schiff (PAS) staining that recognized carbohydrate groups (also recognizing glycogen). The accumulation in liver to cause early disease is different from lung damage (emphysema) that occurs later due to lack of AAT protection.

196. The answer is c. (Murray, pp 396-398. Scriver, pp 3-45. Lewis, pp 188-194.) Insertion (correct answer c) or deletion (incorrect answer d) because no termination codon generated of nucleotides shifts the reading frame unless the change is a multiple of three (incorrect answer e). Frameshifts may create unintended stop codons as in correct answer c and are typically more severe than p. Point mutations with substitutions are named by their position in the protein, that is, P₂₁A (choice b). The protein change P₂₁A could also be denoted by the corresponding change in the DNA reading frame, that is, C₆₃A. Note that the single base substitution of C₆₅A for incorrect answer a does not result in an amino acid change. Deletions may be prefixed by the letter delta, as with ΔF₂₅ (incorrect answer e).

197. The answer is b. (Murray, pp 378-388. Scriver, pp 3-45. Lewis, pp 188-194.) Three RNA polymerases are responsible for RNA transcription in mammalian cells, RNA polymerase I for ribosomal RNA, RNA polymerase II for messenger RNA, and RNA polymerase III for transfer and small RNAs (5s, μRNAs). Mammalian RNA polymerase II is highly sensitive to the mushroom toxin α-amanitin, and thus synthesis of mRNA (correct answer b) rather than other RNA species (incorrect answers a, c-e) will be inhibited by α-amanitin (even experienced mushroom gatherers can confuse the toxic Amanita species with edible varieties). Inhibition of RNA polymerase II-catalyzed HnRNA/mRNA transcription seems to have the most dramatic effects in liver, perhaps, because of its active role in protein synthesis. The liver damage caused by α-amanitin ingestion cannot be treated, requiring regeneration in milder ingestions and transplant in severe ones.

198. The answer is c. (Murray, pp 389-394. Scriver, pp 3-45. Lewis, pp 215-218.) The size of the primary β-globin RNA transcript will include the 5′- and 3′-untranslated regions (50 plus 50 bp) and the exons encoding protein domains (450 amino acids requires 150 coding bp), producing a processed (mature) mRNA of 550 bp (correct answer c). Introns (900 bp) would be included in the 1450 bp primary or HnRNA (answer a incorrect). RNA processing removes the 150 and 750 bp introns, yielding a mature mRNA of 550 bp (excluding answers b, d, and e that incorrectly include an intron or exclude untranslated regions that remain in the mRNA). Thalassemias are caused by mutations that alter the ratio of α- and β-globin, with β-globin being reduced in β-thalassemia. The altered ratios reduce hemoglobin (α₂β₂) synthesis, leading to severe anemia and extramedullary hematopoiesis in liver in an attempt at compensation (causing liver enlargement or hepatomegaly). β-thalassemia requires mutations affecting both β-globin alleles to be severe, while α- thalassemia requires mutations affecting three of the four α-globin alleles to be severe (hemoglobin H disease).

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 polyA 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.

199. The answer is c. (Murray, pp 389-394. Scriver, 3-45. Lewis, pp 206-212.) The small size and alkaline lability of the factor suggests it is a small RNA (correct answer c) rather than mRNA (>100 to thousands of bp—incorrect answer b) or tRNA (75-90 bp—incorrect answer a). Some simple RNAs can self-catalyze their own splicing, but such mechanisms are less important during the complex processing of mammalian mRNAs (incorrect answer e). A transposon is a mobile DNA element that usually inhibits gene expression (incorrect answer d).

Small or mciroRNAs ranging from 21 to 30 nucleotides in length are now recognized in all eucaryotes and include more than 1000 species in humans. Micro RNAs may stimulate or interfere with gene transcription by binding to chromatin or DNA sequence elements. Regulation by microRNAs is complex, since many have double-stranded or hairpin regions that are cleaved and activated by specific nucleases such as Dicer, Drosha, or Pasha in flies. Silencing or interfering microRNAs are an important tool for genetic analysis in that they allow gene knockout to define their effects; they also are being examined as therapeutic agents.

200. The answer is c. (Murray, pp 389-394. Scriver, pp 3-45. Lewis, pp 206-212.) The splicing of messenger RNA is carried out by spliceo-somes—complexes of small ribonucleoprotein particles (snRNPs) and messenger RNA precursor. The presence of the GT to AT mutation in both PAH alleles, together with the larger (incompletely spliced) mRNA precursor, indicates a homozygote affected with PKU. Self-splicing by RNA catalysis (ribozymes) occurs in simpler organisms and some ribosomal RNA precursors, but not in spliceosome-mediated processing of mammalian mRNAs (incorrect answers a, b). Spliceosomes use snRNPs to recognize the 5′ splice site and the 3′ splice site followed by excision of the intronic RNA; RNA polymerase or helicase are not involved because it is the RNA within the spliceosome that catalyzes intron excision (incorrect answers d, e).

The indicated splicing mutation was the first molecular change demonstrated in PKU, a necessary step in diagnosis because PAH enzyme activity cannot be measured in blood leucocytes or in cultured fibroblasts. Although PKU caused by phenylalanine hydoxylase can be treated using phenylalanine-deficient formula, the diet is demanding in that most food protein is minimized, aspartame (a phenylalanine derivative) minimized, and tyrosine balance (becoming essential without dietary phenylalanine as its precursor) is critical. Thus, some parents do choose prenatal diagnosis, especially for severe forms of PKU reflecting defective biopterin cofactor binding.

201. The answer is d. (Murray, pp 389-394. Scriver, pp 3-45. Lewis, pp 206-212.) The primary transcripts of all eukaryotic mRNAs are capped at the 5′ ends as opposed to prokaryotic RNAs or eukaryotic tRNAs and rRNAs (incorrect answers b, c). Spliceosomes accomplish RNA splicing and excision of intronic sequences from mammalian mRNA (incorrect answer a), and poly(A) is added to the 3′ end (incorrect answer e). The cap is composed of 7-methylguanylate attached by a pyrophosphate linkage to the 5′ end. The cap protects the 5′ ends of mRNAs from nucleases and phosphatases and is essential for the recognition of eukaryotic mRNAs in the protein-synthesizing system.

202. The answer is d. (Murray, pp 389-394. Scriver, pp 3-45. Lewis, pp 243-248

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