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Door: serenacikalin23 • 11 maanden geleden

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Lewins Genes XI
Chapter 1
11. The basic building block of nucleic acids is the nucleotide. It is composed of three components. What are they?
Ans: They are a nitrogenous base, a sugar, and a phosphate.
Page 7
12. There are two types of nucleic acids in cells, DNA and RNA. List two differences between the structure of DNA and RNA.
Ans: 1) DNA has the pyrimidine base thymine, whereas RNA contains uracil. 2) DNA has 2′-
deoxyribose, whereas RNA has ribose. The sugar in RNA has an -OH group at the 2′ position of the pentose ring.
Pages 7, 8
13. If the mole % of A residues in double-stranded DNA is 29%, what would the mole % of G residues be?
Ans: The two strands of DNA are complementary with G residues on one strand paired with C
residues on the other strand, and with A residues paired with T residues. Therefore, if the mole
% of A is 29%, then the mole % of T is 29% and the mole % of G + C would be 42%. G must equal C, so as a result the mole % of G and of C would be 21%.
Page 10
14. The size of the smallest human chromosome is 4.7 × 107 bp. If the DNA is in the B-form helix, what would be the length in cm of the chromosome fully extended if not packaged by proteins?
Ans: 1.6 cm. In B-form DNA, the distance between adjacent nucleotides is 3.4 Å. Thus the length of the chromosome would be 4.7 × 107 × 3.4 Å = 16 × 107 Å. There are 108 Å per cm, so the length would be 1.6 cm.
Page 10
15. Meselson and Stahl used density labeling of DNA to show that DNA replication occurs via a semiconservative mechanism. In their experiment, they started with an organism grown in a heavy density label (15N). After two generations of growth in light medium (the more common 14N isotope), if the DNA is isolated and separated by density, how many bands would be observed and how would their density compare with the starting DNA?This study source was downloaded by 100000833532717 from CourseHero.com on 04-07-2022 14:08:45 GMT -05:00
https://www.coursehero.com/file/18331987/Chapter-1/ Lewins Genes XI
Ans: In each generation, the newly synthesized DNA is composed of one parental strand and one new daughter strand. After two generations, the DNA would yield two different density bands. Two of the daughters would contain one original heavy parental strand and one light daughter (hybrid density) strand, and two of the daughters would contain two light strands (light density).
Pages 12, 13
16. The genetic information of organisms is always in the form of nucleic acid. Give examples of cellular organisms or other genetic systems whose genomes are double-stranded DNA, single-stranded DNA, double-stranded RNA, or single-stranded RNA.
Ans: Living organisms (either prokaryotic or eukaryotic cells) always have genomes that are double-stranded DNA. There are, however, many viruses of both prokaryotes and eukaryotes whose genomes can be single-stranded DNA, or double- or single-stranded RNA, depending on the virus.
Pages 14, 15
17. The central dogma of molecular biology is that genetic information flows from DNA to RNA by transcription, and then to polypeptide by translation. Give an example of an exception to this generally true statement.
Ans: Retroviruses have genomes that are single-stranded RNA, and as part of their replication
cycle they convert the RNA into first a single-stranded DNA (using an enzyme called reverse transcriptase) and then a double-stranded DNA, in a process called reverse transcription. The double-stranded DNA copy of its genome integrates into the DNA of the host cell it infects and functions as any other gene, directing the synthesis of both RNA and polypeptide. In this example, genetic information is flowing from RNA to DNA.
Page 15
18. Is the enzyme reverse transcriptase an example of an RNA polymerase or a DNA polymerase?
Ans: Reverse transcriptases are RNA-dependent DNA polymerases that use deoxynucleotides as substrates.
Page 15
19. In the process of DNA replication, the two parental strands separate and serve as templates for the new daughter strands. During this process, are covalent bonds being broken?
Explain.This study source was downloaded by 100000833532717 from CourseHero.com on 04-07-2022 14:08:45 GMT -05:00
https://www.coursehero.com/file/18331987/Chapter-1/ Lewins Genes XI
Ans: No. The separation of the two parental strands involves breakage of the hydrogen (H) bonds (not covalent bonds) between the complementary base pairs. The covalent phosphodiester bonds between adjacent nucleotides on each strand are not broken.
Pages 10, 16
20. The separation of the double helical DNA in cells is mediated by enzymes called helicases. In the laboratory, however, what method is commonly used to denature (separate) the individual strands of a double helical DNA?
Ans: DNA may be denatured by heating the DNA above its T m, usually by boiling.
Page 16
21. A scientist at the San Diego Zoo is sent DNA samples to analyze. The samples were collected from two different bacterial species from different parts of the world. She determines that the G-C content of one of them is 32% and of the other is 62%. The labels have fallen off the tubes, however, so she does not know which sample is which. One of the organisms lived deep in the ocean at temperatures around 10ºC and the other was isolated from the Mojave desert, where temperatures can reach 45ºC. Which sample is likely to be which?
Ans: The bacteria from the Mojave desert is likely to have a G-C content of 62%, whereas the
ocean bacteria is likely to have a G-C content of 32%. The stability of the double-stranded DNA is greater with higher G-C content because G-C base pairs are more stable than A-T base pairs. Organisms that thrive at elevated temperatures have genomes with a higher fraction of G-C base pairs to stabilize the double helical structure of their DNA.
Page 16
22. Once a mutation has occurred in a particular gene, several types of events can occur to reverse the effects of the original mutation. Give three of these types.
Ans: 1) There may be a true reversion whereby the original base change is reversed, or the insertion of genetic material, such as a transposon, is deleted. 2) There may be a second site reversion whereby a second mutation in the same gene changes the amino acid sequence at a distinct site from the first mutation and allows the polypeptide to regain function. 3) There may be a suppressor mutation in another gene, altering a different polypeptide. This change bypasses the need for the original polypeptide or enables it to interact with the mutant form of
the original polypeptide.
Pages 19, 20
23. Why is the modified base 5-methylcytosine a hotspot for mutations?This study source was downloaded by 100000833532717 from CourseHero.com on 04-07-2022 14:08:45 GMT -05:00
https://www.coursehero.com/file/18331987/Chapter-1/

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