Concept explainers
Introduction:
The joining of two or more DNA molecules from different sources generates a recombinant DNA. The joining of two DNA (deoxyribonucleic acid) molecules is the main part of gene cloning. The DNA molecules are cut by the restriction enzymes, also called restriction endonucleases.
Answer to Problem 1TYK
Correct answer:
The sequence of
Explanation of Solution
Justification/ Explanation for the correct answer:
Option (a) is the sequence of nucleotides. The restriction enzymes help in recognition of restriction sites on the DNA fragments and cut them into different fragments. These restriction sites contain sequences of nucleotides that are recognized by enzymes and are palindromic in nature. There is a number of restriction enzymes identified for cutting DNA at specific restriction sites. Hence, option (a) is correct.
The explanation for the incorrect answers:
Option (b) is the length of DNA molecule. The activity of restriction enzymes depends on the presence of restriction sites on the DNA molecule. A long DNA molecule cannot be cut by an enzyme, if it does not contain particular recognition sites. So, it is an incorrect option.
Option (c) is whether it is closer to 5’end or 3´end of DNA molecule. The restriction enzymes recognize the palindromic sequence of DNA sequences, which can be read in both sides, from 5’ to 3´ and 3´ to 5´ ends. So, it is an incorrect option.
Option (d) is the number of copies of DNA molecule in a bacterial cell. A large number of copies of DNA cannot be cleaved by restriction enzymes without any restriction sites. So, it is an incorrect option.
Option (e) is the location of a start codon in a gene. Start codon only helps in the initiation of translation and protein synthesis. So, it is an incorrect option.
Hence, options (b), (c), (d) and (e) are incorrect.
Restriction sites are the sequence of DNA fragments which may vary in number and are recognized by a specific restriction endonuclease. The enzyme endonuclease cleaves the DNA fragment generating small fragments with sticky ends or blunt ends.
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Chapter 18 Solutions
Biology: The Dynamic Science (MindTap Course List)
- What is a restriction endonuclease? Select one: a. It is an enzyme that cleaves at a specific nucleotide sequence. b. It restricts the movement of the DNA outside the nucleus. c. It proofreads the DNA for accidental damages and corrects any errors. d. It is an enzyme that separates the DNA double helix.arrow_forwardDraw a figure showing annealing and extension of the oligos to form the DNA template.Label the following parts on your diagram: T7 Promoter, GG, Target Sequence, Tail Overlap, Tail. Explain the purpose of each labeled partarrow_forwardDiscuss the following statement: “from the nucleotide sequence of a cDNA clone, the complete amino acid sequence of a protein can be deduced by applying the genetic code. thus, protein biochemistry has become superfluous because there is nothing more that can be learned by studying the protein.”arrow_forward
- The restriction endonucleases used in recombinant DNA work: a. are synthesized by bacteria b. recognize sequences 14-16 bp long c. cut the DNA outside the recognition sequence d. all the above are truearrow_forwardA small section of bacterial DNA template (anti-sense) strand has the following nucleotide sequence: AAG TAT TAT GCA A mutation in the above sequence involved the insertion of a single base, leading to a shift in the reading frame of the gene, resulting in altered amino acids downstream from the insertion. Which of the following gene sequences exemplifies the mutation described above? a. AAG CTA TTA TGC A b. AG TAT TAT GCA c. AAG TATU TAT GCA d. AAG TAT TAGT GCAarrow_forwardThe base composition of the transcribable DNA template is: A = 10% G = 40% C = 30% T= 20% %3D %3D What is the base composition of the transcribed product? A. A = 20%; C = 40% B. G = 30%; C = 20% C. G = 10%; T = 20% %3D D. C = 10%; U = 40% O E. Cannot be determinedarrow_forward
- Define the following terms: a. processivity b. replisome c. exonuclease d. DNA ligase e. repliconarrow_forwardWhich of the following statements is TRUE concerning the synthesis of the leading and lagging strands of DNA in prokaryotic cells? a. O b. The leading strand is synthesized by one polymerase III continuously, and the lagging strand is synthesized by several molecules of DNA polymerase III. d. The leading and lagging strands are synthesized at the same time by the one DNA polymerase I. O c. The leading and lagging strands are synthesized at the same time by the one DNA polymerase III. The leading strand is synthesized by one polymerase III, and the lagging strand is synthesized by DNA polymerase I.arrow_forwardA genetic map provides a. the sequence of the DNA in a genome. b. the relative position of genes on chromosomes. c. the location of sites of restriction enzyme cleavage in a known sequence of DNA. d. the banding pattern of a chromosome.arrow_forward
- You obtained the sequence of the frog gene X you amplified in Question #16 through a process called automated sequencing. In automated sequencing, you are given a printout of the sense strand of your DNA. The first thing you need to do is use the correct reading frame. Having done this, the next thing to do is to write out the mRNA sequence using this sense strand reading frame. The last thing to do is to translate the sequence. a.The reading frame DNA sequence is: b.The mRNA sequence is: c.The polypeptide sequence is: A disease in frogs which causes their tongue to fall out of their mouths is killing the frog population in LA County. You obtain a dead frog and isolate its gene Xf. When you sequence this mutated gene, you find that the last ‘G’ at the end of the first line of this sequence has been deleted (i.e. the G at position 86). In order to determine how this mutation changes the resulting polypeptide, write the mutated polypeptide sequence in the space below. What kind of…arrow_forwardWhich of the following statements are NOT true? A. Replication is the process of making DNA and takes place in the nucleus of prokaryotic cells. B. Translation produces a polypeptide that may require additional processing to become a functional protein C. Transcription starts at the promoter of eukaryotic cells and scans until reaches the start codon. D. Splicing results in exons being put together and introns being removedarrow_forwarda. The reading frame DNA sequence is b. The mRNA sequence is c. The polypeptide sequence is a.The mutated polypeptide sequence is b.What kind of mutation was produced?arrow_forward
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