16.Bacterial cells protect their own DNA from restriction enzymes (endonucleases) by_______.  
(A) using DNA ligase to seal the bacterial DNA into a closed circle.
(B) adding methyI groups to adenines and cytosines.
(C) adding histones to protect the double-stranded DNA.
(D) forming "sticky ends" of bacterial DNA to prevent the enzyme (endonuclease) from attaching.
(E) inhibiting enzyme activity.

1. According to the fluid mosaic model of cell membranes, phospholipids_____. (A) have hydrophilic tails in the interior of the membrane. (B) frequently flip-flop from one side of the membrane to the other. (C) occur in an uninterrupted bilayer, with membrane proteins restricted to the surface of the membrane (D) can move laterally along the plane of the membrane. (E) have hydrophobic heads in the exterior of the membrane.

複選題2.The genetic code is essentially the same for all organisms. From this, what can be assumed? (A) The same codons in different organisms translate into the same amino acids. (B)A gene from an organism can theoretically be expressed by any other organism. (C) Different organisms have the same types of amino acids. (D) The sequence of amino acids could be predicted according to their DNA sequence. (E) The same genetic code could be translated into different amino acids.

3. Which of the following contradicts the one-gene, one-enzyme hypothesis? (A) Alkaptonuria results when individuals lack a single enzyme involved in the catalysis of homogentisic acid. (B) Sickle-cell anemia results in defective hemoglobin. (C) A mutation in a single gene can result in a defective protein. (D) A single antibody gene can code for different related proteins, depending on the alternative splicing. (E) No case contradicts the one-gene, one-enzyme hypothesis.

4.In 1928, English microbiologist Frederic Griffith demonstrated that DNA is the genetic material to determine the virulence of bacteria cells； what observations helped this conclusion? (A)' The heat-killed pathogenic bacteria + live non-pathogenic bacteria → pathogenic. (B) The heat-killed non-pathogenic bacteria+ live pathogenic bacteria → pathogenic. (C) The transformation of bacteria was carried by a heat-stable genetic material. (D) Pathogenicity reffects the action of the non-capsule gene. (E) Non-pathogenic bacteria could uptake the capsule gene from the bead pathogenic bacteria.

複選題5. What is (are) the mechanism of information transfer in eukaryotes? (A)Messenger RNA is transcribed from a single gene and transfers information from the DNA in the nucleus to the cytoplasm, where protein synthesis takes place. (B) Transfer RNA takes information from RNA directly to a ribosome, where protein synthesis takes place. (C) RNA from a single gene is synthesized and transferred to the cytoplasm, where it serves as a template for protein synthesis. (D) Proteins transfer information from the nucleus to the ribosome, where protein synthesis takes place. (E) A single gene of eukaryote only makes one mRNA.

複選題6. Which of the following occurs in prokaryotes but not in eukaryotes? (A) Post-transcriptional splicing (B) Translation in the absence of a ribosome (C) Concurrent transcription and translation (D) Gene regulation (E) Operon system

7. A ribozyme is_______. (A) a catalyst that uses RNA as a substrate (B) an enzyme that synthesizes RNA as part of the transcription process (C) an enzyme that catalyzes the association between the large and small ribosomal subunits (D) an RNA with catalytic activity (E) Group II intron could be one example of ribozyme.

8. A mutant bacterial cell has a defective aminoacyl-tRNA synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of the normal phenylalanine. The consequence of this for the cell will be that _______. (A) the positions of phenylalanine in proteins will still encode phenylalanine. (B) the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons. (C) proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU. (D) the ribosome will skip a codon every time a UUU is encountered. (E) No phenylalanine will be encoded.

9. A mutation that inactivates a regulatory gene of a repressible operon in an E. coli cell would result in_______.  (A) continuous transcription of the structural gene controlled by that regulator. (B) no binding of the repressor to the operator. (C) complete activation of transcription of the structural gene controlled by that regulator. (D) continuous translation of the mRNA because of alteration of its structure. (E) no gene will be affected.

10.The primary difference between enhancers and promoter-proximal elements is that enhancers_______.  (A) enhance transcription； promoter-proximal elements inhibit transcription. (B) are transcription factors； promoter-proximal elements are DNA sequences. (C) are at considerable distances from the promoter； promoter-proximal elements are close to the promoter. (D) are DNA sequences； promoter-proximal elements are also DNA sequences. (E) are regulated by RNA for their functions.

115年 - [無官方正解]115 國立臺灣大學_ 碩士班招生考試試題：基礎分子生物學#137679

115年 - [無官方正解]115 國立臺灣大學_ 碩士班招生考試試題：基礎分子生物學B#137647

114年 - 114 國立台灣大學_碩士班招生考試試題：基礎分子生物學#126024

113年 - 113 國立臺灣大學_碩士班招生考試：基礎分子生物學#118909

112年 - 112台大 基礎分子生物學.pdf#118452

111年 - 111 國立臺灣大學_碩士班招生考試試題：基礎分子生物學#120877

110年 - 110 國立臺灣大學_碩士班招生考試_部分系所:基礎分子生物學#101968

109年 - 109 國立臺灣大學_碩士班招生考試 :基礎分子生物學#125739

108年 - 108 國立臺灣大學_碩士班招生考試：基礎分子生物學#125746

107年 - 107 國立臺灣大學_碩士班招生考試：基礎分子生物學#125774