课程名称︰ 生物化学甲下 课程性质︰ 必修 课程教师︰ 余荣炽 开课学院： 生命科学院 开课系所︰ 生命科学系 考试日期（年月日）︰ 2010.5.18 考试时限（分钟）： 120分钟 是否需发放奖励金：是 （如未明确表示，则不予发放） 试题 : 单选题（以ABCDE作答；共25题，每题2分，答错不倒扣，共50分） C 1. Long-term maintenance of body weight is regulated by the hormone: (A) cortisol. (B) ghrelin. (C) leptin. (D) insulin. (E) testosterone. D 2. Which of the following descriptions is false? (A) Leptin is a small polypeptide that is produced in adipocytes. (B) The hormone leptin decreases appetite. (C) Insulin can act on insulin receptors in the hypothalamus to inhibit eating. (D) By altering synaptic transmissions from neurons in the arcuate nucleus, leptin suppresses the synthesis of thermogenin, and thus decreases energy expenditure. (E) α-Melanocyte hormone (α-MSH) is released by the neurons in the arcuate nucleus. It plays the function to inhibit appetite. B 3. The base of a nucleotide is joined covalently to the 1' carbon of the pentose in a(n): (A) amide linkage. (B) N-β-glycosyl bond. (C) ester linkage. (D) phosphodiester bond. (E) O-glycosidic bond. D 4. The deoxyribonucleotide polymer 5'-GTGATCAAG-3' could only form a stable double-stranded structure with: (A) 5'-GAACTAGTG-3'. (B) 5'-GTGATCAAG-3'. (C) 5'-CACTAGTTC-3'. (D) 5'-CTTGATCAC-3'. (E) 5'-TCCAGCTGT-3'. D 5. Which of the following descriptions is(are) false? (选择下方的ABCDE，非1 ~ 5) (1) When double-stranded DNA is heated to denature, the helical structure unwinds. (2) When double-stranded DNA is heated to denature, the viscosity of the solution decreases. (3) When a solution of DNA is heated slowly until the tm is reached, the DNA molecules are completely denatured at this temperature. (4) In denaturation of a nucleic acid duplex at neutral pH, the stability of a DNA-RNA hybrid duplex is greater than a DNA-DNA double helix with a comparable sequence. (5) When double-stranded DNA is heated to denature, the absorption of ultraviolet (260 nm) light decreases. (A) 3, 4, and 5. (B) 1 and 3. (C) 2 and 5. (D) 3 and 5. (E) 2, 3, and 5. C 6. In nucleotides and nucleic acids, syn and anti conformations relate to: (A) base stereoisomers. (B) rotation around the phosphodiester bond. (C) rotation around the sugar-base bond. (D) sugar pucker. (E) sugar stereoisomers. A 7. Which of the following descriptions is(are) correct? (选择下方的ABCDE，非1 ~ 5) (1) The difference between a ribonucleotide and a deoxyribonucleotide is that a ribonucleotide has an —H instead of an —OH at C-2. (2) The pentoses found in nucleic acids are always in the β-furanose forms. (3) In double-stranded DNA, sequences rich in G–C base pairs are denatured less readily than those rich in A–T pairs. (4) B form DNA is a left-handed helix; Z form DNA is a right-handed helix. (5) C-1 of the pentose in nucleic acids is joined to a phosphate group. (A) 2 and 3. (B) 1 and 3. (C) 2 and 5. (D) 3 only. (E) 2, 3, and 4 A 8. The phosphodiester bond that joins adjacent nucleotides in DNA: (A) associates ionically with metal ions, polyamines, and proteins. (B) is positively charged. (C) is susceptible to alkaline hydrolysis. (D) links C-2 of one base to C-3 of the next. (E) links C-3 of deoxyribose to N-1 of thymine or cytosine. C 9. Certain nucleotide bases in DNA molecules are enzymatically methylated. The methyl group donor in all known DNA methylation reactions is: (A) serine. (B) glycine. (C) S-adenosylmethionine. (D) N5,N10-methylenetetrahydrofolate. (E) N5-methyltetrahydrofolate. E 10. Several nucleotide bases undergo spontaneous loss of their exocyclic amino groups (deamination). Deamination of the nucleotide bases cytosine and 5-methylcytosine yields nucleotide bases: (A) thymine and uracil, respectively. (B) hypoxanthine and xanthine,respectively. (C) uracil only. (D) uracil and hypoxanthine, respectively. (E) uracil and thymine, respectively. E 11. Which of the following is a palindromic sequence? (A) TAATAA (B) CTCTCT (C) AAACCC ATTATT GAGAGA TTTGGG (D) CATACG (E) GCATGC GTATGC CGTACG E 12. Orotic aciduria is an inherited metabolic disease in which orotic acid (orotate) accumulates in the tissues, blood, and urine. The metabolic pathway in which the enzyme defect occurs is: (A) epinephrine synthesis. (B) purine breakdown. (C) purine synthesis. (D) pyrimidine breakdown. (E) pyrimidine synthesis. B 13. The Sanger sequencing procedure uses dideoxynucleoside triphosphate (ddNTP) analogs to interrupt DNA synthesis. When a ddNTP is inserted in place of a dNTP, the strand elongation is halted, because it: (A) lacks the 5'-phsophate group needed for the next step. (B) lacks the 3'-hydroxyl group needed for the next step. (C) lacks the 2'-hydroxyl group needed for the next step. (D) possesses the additional 2'-hydroxyl group, which interrupts DNA synthesis. (E) possesses the additional 3'-hydroxyl group, which interrupts DNA synthesis. B 14. Which of the following functional groups play the critical role in the reaction catalyzed by ribonucleotide reductase? (A) acetyl group. (B) thiol group. (C) hydorxyl group. (D) amino group. (E) carbonyl group. A 15. Deoxyribonucleotides are derived from the corresponding ribonucleotides by: (A) reduction at the 2’-carbon atom of the D-ribose. (B) reduction at the 3’-carbon atom of the D-ribose. (C) oxidation at the 2’-carbon atom of the D-ribose. (D) oxidation at the 3’-carbon atom of the D-ribose. (E) hydroxylation at the 2’-carbon atom of the D-ribose. B 16. The biosynthesis of deoxyribonucleotides from the corresponding ribonucleotides is catalyzed by: (A) adenosine phosphoribosyltransferase. (B) ribonucleotide reductase. (C) nucleoside kinase. (D) dihydrofolate reductase. (E) ribonucleotide oxidase. A 17. In the salvage pathways, the free purine bases react with to yield the corresponding nucleotides. (A) 5-phosphoribosyl 1-pyrophosphate. (B) adenosine 5'-phosphate. (C) guanosine 5'-phosphate. (D) ribose 5-phosphate. (E) ATP. D 18. The reaction, UMP + ATP → UDP + ADP, is catalyzed by a(n): (A) adenosine phosphoribosyltransferase. (B) adenylate kinase. (C) nucleoside diphosphate kinase. (D) nucleoside monophosphate kinase. (E) adenosine kinase. B 19. Free adenine bases are salvaged to make nucleotides. The reaction is catalyzed by: (A) adenylate kinase. (B) adenosine phosphoribosyltransferase. (C) hypoxanthine-guanine phosphoribosyltransferase. (D) adenosine deaminase. (E) DNA polymerase. E 20. The end product of purine degradation in humans is: (A) glutamate. (B) NH4+. (C) succinate. (D) urea. (E) uric acid. B 21. For a closed-circular DNA molecule of 10,000 base pairs in the fully relaxed form, the linking number (Lk) is about: (A) 10,000. (B) 950. (C) 100. (D) 9.5. (E) 2. D 22. Which of the following descriptions is(are) correct? (选择下方的ABCDE，非1 ~ 5) (1) The linking number (Lk) of a closed-circular DNA molecule can be changed only by breaking one or both strands. (2) The DNA of virtually every cell is underwound relative to B-form DNA. (3) Topoisomerases change the degree of supercoiling of a DNA molecule but not its linking number of DNA. (4) The linking number (Lk) of a closed-circular, double-stranded DNA molecule is changed by supercoiling without the breaking of any phosphodiester bonds. (5) Topoisomerases always change the linking number in increments of 1. (A) 1, 2, and 4. (B) 1 and 5. (C) 2 only. (D) 1 and 2. (E) 1 and 3. C 23. If the structure of a fully relaxed, closed-circular DNA molecule is changed so that the specific linking difference (σ) is –0.05, the number of: (A) bases is decreased by 5%. (B) bases is increased by 5%. (C) helical turns is decreased by 5%. (D) helical turns is increased by 5%. (E) helical turns is unchanged. E 24. Histones are very rich in amino acid(s): (A) cysteine. (B) glycine. (C) glutamate and glutamine. (D) asparte and asparagines. (E) arginine and lysine. D 25. Which of the following statements about type II restriction enzymes is(are) false? (1) Type II restriction enzymes cleave DNA only at recognition sequences specific to a given restriction enzyme. (2) Type II restriction enzymes cut both DNA strands at the same base pair. (3) They cleave and ligate DNA. (4) Some type II restriction enzymes make a staggered double-strand cut, leaving ends with a few nucleotides of single-stranded DNA protruding, while some restriction enzymes cut DNA to generate blunt ends. (5) They are part of a bacterial defense system in which foreign DNA is cleaved. (A) 1 and 2. (B) 3 and 5. (C) 3 only. (D) 2 and 3. (E) 2, 3, and 4. 问答题（共5 题，每题10 分，共50 分） 1. Why does lowering the ionic strength of a solution of double-stranded DNA permit the DNA to denature more readily (for example, to denature at a lower temperature than at a higher ionic strength)? Ans: Lower ionic strength reduces the screening of the negative charges on the phosphate groups by positive ions in the medium. The result is stronger charge-charge repulsion between the phosphate, which favors strand separation. 2. Diagram the biosynthetic pathways from the salvaging of hypoxanthine base to the formation of dATP. Use abbreviations (e.g., dATP), not complete structures. Name the enzyme of each reaction. (Note: The formation of AMP from IMP involve two steps, catalyzed by adenylosuccinate synthetase and adenylosuccinate lyase)（叙述主要的反应物及产物即可，不需要化学平衡式） Ans: Hypoxanthine + PRPP → IMP (hypoxanthine-guanine phosphoribosyltransferase) IMP + asparte → adenylosuccinate (adenylosuccinate synthetase) Adenylosuccinate → AMP + fumarate (adenylosuccinate lyase) AMP + ATP → ADP + ADP (adenylate kinase or nucleoside monophosphate kinase) ADP → dADP (ribonucleotide reductase) dADP + ATP → dATP + ADP (nucleoside diphosphate kinase) 3. Explain why inhibitors of dihydrofolate reductase be used to treat cancers and bacterial infections? Describe the complete reaction cycle that dihydrofolate reductase participates first, and then answer the question. Ans: Thymidylate synthase catalyzes the conversion of dUMP to dTMP, which is one of the building blocks of DNA. In conversion of dUMP to dTMP, an one-carbon unit at the hydroxymethyl(-CH2OH) oxidation level is transferred from N5,N10-methylenetetrahydrofolate to dUMP by thymidylate synthase, then reduced to a methyl group. The reduction occurs at the expense of oxidation of tetrahydrofolate to dihydrofolate, and dihydrofolate reductase is required to reduce the dihydrofolate to tetrahydrofolate. The activity of dihydrofolate reductase is required to complete the reaction cycle for the synthesis of dTMP from dUMP; thus inhibiters of dihydrofolate reductase could block the source of dTMP for DNA synthesis. As the cellular pools of nucleotides are quite small, perhaps 1% or less of the amounts required to synthesize the cell's DNA. Therefore, dividing cells must continue to synthesize nucleotides, and in some cases nucleotide synthesis may limit the rates of DNA replication and transcription. Thus inhibitors of dihydrofolate reductase, which block the source of dTMP for DNA synthesis, restrain the proliferation of fast-growing cells, such as cancer cells and bacteria, significantly, and thus are used to treat cancers and bacterial infections. 4. What is the essential difference between a genomic library and a cDNA library? Name one enzyme that is always used to make a cDNA library, but is generally not used to make a genomic DNA library. Describe its function briefly. Ans: A genomic library is produced when the complete genome of a particular organism is cleaved into thousands of fragments and the fragments are cloned by insertion into a cloning vector. cDNA library - first extracts mRNA from an organism or from specific cells and then prepares complementary DNAs (cDNAs) from the RNA catalyzed by the enzyme reverse transcriptase. The resulting double-stranded DNA fragments are then inserted into a vector and cloned, creating a population of clones called a cDNA library. A genomic library contains (in principle) all of the sequences present in the chromosome(s), including DNA sequences that are not transcribed. Because a cDNA library is made as a DNA copy of mRNA, it contains only those DNA sequences that are expressed in the cell. Reverse transcriptase is used to make first a single-stranded DNA complementary to mRNA, then a double-stranded DNA. 5. A DNA sequence that may be present as only a single copy in a large mammalian genome can be amplified and cloned using the polymerase chain reaction (PCR). Describe the steps and reaction components required in a PCR experiment. Illustrate the steps in just one round. Ans: DNA with the desired sequence is heated to convert it to single strands and cooled in the presence of an excess of oligonucleotide primers that flank the sequence to be amplified. A heat-stable DNA polymerase extends the primers, replicating the desired sequence. The reaction components required in a PCR experiment include: A DNA template containing the segment to be amplified. A pair of synthetic oligonucleotide primers, which specifically anneal to the 5’ and 3’ ends of the target DNA segment. The four deoxynucleoside triphosphates, dATP, dCTP, dGTP, and dTTP (dNTPs). A heat-stable DNA polymerase, such as the Taq polymerase. --

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