A. Incorrect! Garrod s experiment linked genes to enzymes. It is important to be familiar with the milestone experiments in genetics.

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1 Genetics - Problem Drill 11: DNA - The Chemical Basis of Genetics No. 1 of Which scientist first gave evidence that DNA is the genetic material? (A) Garrod, who postulated that Alcaptonuria, or black urine disease, was due to a defective enzyme. (B) Avery, MacLeod, and McCarty who repeated the transformation experiments of Griffith and chemically characterized the transforming principle. (C) Beadle and Tatum who used a mutational and biochemical analysis of the bread mold, Neurospora, to establish a direct link between genes and enzymes. (D) Meselson and Stahl who showed that DNA is replicated semiconservatively. (E) Watson and Crick who gave a model for the structure of DNA. Garrod s experiment linked genes to enzymes. B. Correct! Avery, MacLeod, and McCarty s transformation experiments are the classic ones to prove DNA is the genetic material. Beadle and Tatum proposed the one gene-one enzyme hypothesis, not proving DNA is the genetic material. Semi-conservative replication does not prove that DNA is the genetic material. The double helix model is the bases for modern molecular biology, but it by itself did not prove DNA is the genetic material. It is important to be familiar with the milestone experiments in genetics. (B)Avery, MacLeod, and McCarty who repeated the transformation experiments of Griffith and chemically characterized the transforming principle.

2 No. 2 of In this diagram of the process of DNA replication at a replication fork, the strand labeled B is the. (A) Primer (B) Template (C) Leading strand (D) Lagging strand (E) Okazaki fragment D and E are primers. A is a template. C. Correct! The leading strand is synthesized in the same direction as the fork. C is the lagging strand here. C is the Okazaki fragment. The key to solve this problem is to understand the mechanism of DNA replication. 1. On the lagging strand, primase reads the DNA and adds RNA sequences to act as a primer for POL III DNA polymerase. Multiple RNA primers are needed. 2. Pol III lengthens the primers, forming Okazaki fragment. 3. Pol I then removes the RNA and adds its nucleotides. 4. DNA ligase joins the fragments. (C)leading strand

3 No. 3 of For the DNA strand 5'-TACGATCATAT-3', the correct complementary DNA strand is. (A) 3'-TACGATCATAT-5' (B) 3'-ATGCTAGTATA-5' (C) 3 -AUGCUAGUAUA-5 (D) 3 -GCATATACGCG-5 (E) 3 -ATATACTAGCAT-5 This is the same sequence but reverse direction, not complementary. B. Correct! The complementary strand to a DNA is anti-parallel and follows the base-pairing principle. Therefore, the following two strands are complementary: 5'-TACGATCATAT-3' 3 -ATGCTAGTATA-5' This is the RNA strand, also complementary to the original DNA strand. This sequence comes out of nowhere. The direction is reversed. Key point: The complementary strand to a DNA is anti-parallel and follows the base-pairing principle. (B)3'-ATGCTAGTATA-5'

4 No. 4 of In the Meselson-Stahl DNA replication experiment, if the cells were first grown for many generations in N-15 containing media and then switched to N-14 containing media, what percent of the DNA had 1 light strand and 1 heavy strand after 2 generations of growth in N-14 growth media? (A) 0 (B) 25 (C) 50 (D) 75 (E) 100 Apparently, there will be some DNA with one light strand and 1 heavy strand. In the third generation, the ratio would be 25%. C. Correct! DNA replication follows a semi-conservative model; after one generation, the entire DNA will have one heavy strand and one light strand. Continue culturing in N-14 media, then the next generation would have half of all molecules containing light strand, half of all molecules containing heavy strand. The answer should be 50%. In this condition, 75% cannot be achieved. Obviously, not all DNA can have one light chain and one heavy chain. (C)50

5 No. 5 of Which of the following enzymes is NOT involved in DNA replication? (A) DNA ligase (B) DNA polymerase (C) RNA primase (D) RNA polymerase (E) Topoisomerase DNA ligase is required to ligate the Okazaki fragments together. DNA polymerase is required for DNA strand elongation. RNA primase is required for synthesis of RNA primers, which are required for initiating DNA synthesis. D. Correct! RNA polymerase is only required for RNA synthesis but is not required for DNA synthesis. Telomerase is also involved in DNA replication, only it is required for replicating the ends. It is important to understand how DNA is replicated and what enzymes are involved. (D)RNA polymerase

6 No. 6 of Which of the following statements about the Hershey-Chase experiment is true? (A) The experiments were done to study the possibility of protein and RNA as the genetic material. (B) They labeled both protein and DNA with radioactive 35 S to track them. (C) The ability of the phage to inject/transmit proteins and not DNA was the definitive result of these experiments. (D) The ability of the phage to inject/transmit DNA and not protein during infection was the definitive result of these experiments. (E) The bacteria were separated from the phage coats, and it was found that the bacteria contained the label 35 S. The experiments were done to study the possibility of DNA as the genetic material. They had to use different labels so they could identify which was which 35 S for protein and P 32 for DNA. The ability of the phage to inject/transmit proteins and not DNA was the definitive result of these experiments. D. Correct! The ability of the phage to inject/transmit DNA and not protein during infection was the definitive result of these experiments. The bacteria were separated from the phage coats, and it was found that the bacteria contained the label P 32. Hershey & Chase used the phage T2 to prove that DNA is the genetic material. In the first experiment, they radioactively labelled the phage DNA with P 32 and allowed the 2 phage to infect a bacterium. In the second experiment, the protein was labelled with 35 S isotope and, again, the phage infected the bacteria. The bacteria were again separated from the phage coats. It was found that there was no 35 S in the bacteria. This led to the conclusion that DNA, and not protein, was the genetic material. The bacteria were separated from the phage coats and it was found that the bacteria had P 32. (D)The ability of the phage to inject/transmit DNA and not protein during infection was the definitive result of these experiments.

7 No. 7 of What is additional evidence that DNA is the genetic material? (A) The fact that UV light mutagenesis has its best effect at 260nm, which matches with the absorption spectrum of DNA. (B) The fact that UV mutageneis has its best effect at 360nm, which matches with the absorption spectrum of DNA. (C) In transgenic animals treated with the same recombinant DNA, the DNA does not match the phenotype. (D) Mitochondria and chloroplasts both have DNA and both are known to carry some key proteins. (E) The fact that chloroplasts carry some key proteins unlike mitochondria. A. Correct! The fact that UV light mutagenesis has its best effect at 260nm, which matches with the absorption spectrum of DNA. The fact that UV light mutagenesis has its best effect at 260nm, which matches with the absorption spectrum of DNA. In transgenic animals treated with the same recombinant DNA, the DNA does match the phenotype. Mitochondria and chloroplasts both have DNA and both are known to carry some genetic information. Mitochondria and chloroplasts both have DNA and both are known to carry some genetic information. Mitochondria and chloroplast both have DNA and both are known to carry some genetic information. Mutagenesis study: UV has strong mutagenesis effect at 260 nm, which matches the absorption spectrum of DNA, but not that of protein. Recombinant DNA technology and transgenic organisms. Direct evidence: DNA matches the phenotype. In eukaryotes, the nucleus has genetic material, DNA and proteins. (A)The fact that UV light mutagenesis has its best effect at 260nm, which matches with the absorption spectrum of DNA.

8 No. 8 of Which of the flowing statements about the structure of DNA is correct? (A) DNA is made up of a phosphate base, a ribose, and a nitrogenous base. (B) One of the four DNA bases is Uracil. (C) Adenine and guanine are known as Purines. (D) Adenine always pairs with Guanine, according to Watson and Crick s base pairing rules. (E) Cytosine always pairs with Thymine. DNA is made up of a phosphate base, a deoxyribose, and a nitrogenous base. Uracil is an RNA base. C. Correct! Adenine and guanine are known as Purines. Adenine always pairs with Thymine, according to Watson and Crick s base pairing rules. Cytosine always pairs with Guanine. Four bases make up DNA: adenine (A), guanine (G), cytosine (C) and thymine (T). A only pairs with T, and G only pairs with C in DNA. DNA has a phosphate backbone. DNA nucleotides are linked at their phosphate groups in a phosphodiester bond. Hydrogen bonds are attractive forces between an electronegative atom and a hydrogen atom bonded to another electronegative atom. DNA nucleotides exhibit hydrogen bonding. Notice H is on the NH group and attracted to doubly bonded Oxygen and Nitrogen. (C)Adenine and guanine are known as Purines.

9 No. 9 of RNA. (A) Comes in 2 types: (1) rrna, and (2) trna. (B) Comes in 4 types: (1) mrna, (2) rrna, (3) lrna and (4) drna. (C) That carries information for the amino acid sequence of a protein is known as snrna. (D) Known as small nuclear RNA (snrna) is found in the nucleus and is important in RNA splicing. (E) Contains deoxyribose molecules, as opposed to DNA that contains ribose. There are 4 types of RNA: (1) mrna, (2) rrna, (3) snrna and (4) trna. There are 4 types of RNA: (1) mrna, (2) rrna, (3) snrna and (4) trna. RNA that carries information for the amino acid sequence of a protein is known as messenger RNA (mrna). D. Correct! RNA known as small nuclear RNA (snrna) is found in the nucleus and is important in RNA splicing. RNA contains ribose molecules, as opposed to DNA that contains deoxyribose. trna carries amino acids to the ribosome and transfers them to the growing polypeptide chain. mrna carries the information for the amino acid sequence of a polypeptide (protein) chain. rrna are pieces of RNA that form complexes with ribosomes (protein) to facilitate the transfer of amino acid from trna to the polypeptide chain. snrna (small nuclear RNA) are small RNA molecules found in the nucleus, which are important in RNA splicing. (D) Known as small nuclear RNA (snrna) is found in the nucleus and is important in RNA splicing.

10 No. 10 of Which of the following statements about telomeres is correct? (A) During DNA synthesis, the end of the DNA is not long enough to make the necessary Okazaki fragment; therefore, the 5ʹ end of each strand can t be completed. (B) During the DNA synthesis, the end of the DNA is not long enough for the DNA to be completed on the leading strand. (C) Telomerase is an enzyme that adds Okazaki repeat seguences to the 5ʹ end of the DNA strands. (D) Humans lose an estimated 100 bases from their telomeric DNA at each mitosis. Normal cells have no method or procedure to compensate for this. (E) Humans lose an estimated 100 bases from their telomeric DNA at each meiosis. Normal cells have no method or procedure to compensate for this. A. Correct! During DNA synthesis, the end of the DNA is not long enough to make the necessary Okazaki fragment; therefore, the 5ʹ end of each strand can t be completed. During DNA synthesis, the end of the DNA is not long enough to make the necessary Okazaki fragment; therefore, the 5ʹ end of each strand can t be completed. Telomerase is an enzyme that adds telomere repeat sequences to the 3ʹ end of the DNA strands. Telomerase compensates for this by adding repeat sequences, which can be used by DNA polymerase to complete the synthesis. Telomerase compensates for this by adding repeat sequences, which can be used by DNA polymerase to complete the synthesis. Replication of linear chromosome presents a special problem since the DNA polymerase can only synthesize a new strand in 5 to 3 direction. This works for the 3 to 5 strand on the chromosome because the polymerase can move from the origin of replication until it meets another replication bubble. Synthesis on the 5 to 3 strand is discontinuous. As the replication fork nears the end of the DNA, there is not enough templates to make Okazaki fragment. Therefore, the 5 end of each strand cannot be completed. This means that each daughter chromosome will have a shortened telomere. It is estimated that humans lose about 100 bases from their telomeric DNA at each mitosis. (A)During DNA synthesis, the end of the DNA is not long enough to make the necessary Okazaki fragment; therefore, the 5ʹ end of each strand can t be completed.