Chap 10 Problem Solutions

Which two experiments demonstrated that DNA is the genetic material?

A) Griffith and Watson-Crick

B) Avery-MacLeod-McCarty and Hershey-Chase

C) Meselson-Stahl and Hershey-Chase

D) Griffith and Franklin

Avery-MacLeod-McCarty and Hershey-Chase. Avery, MacLeod, and McCarty demonstrated that the transforming material was DNA; Hershey and Chase confirmed DNA as genetic material using bacteriophage T2.

How did Avery and colleagues prove the transforming principle was DNA?

A) Heating DNA destroyed transformation

B) DNase destroyed transforming activity, while proteases and RNase had no effect

C) 35S entered recipient bacterial cells

D) Transformation occurred only in the presence of RNase

DNase destroyed transforming activity, while proteases and RNase had no effect. Proteases and RNase degraded proteins and RNA without stopping transformation; only DNase (degrading DNA) eliminated activity.

In the Hershey-Chase experiment, what proved DNA is the genetic material in T2 bacteriophage?

A) Progeny phage released from 32P-labeled infections contained 32P

B) Progeny phage contained 35S in their coats

C) Bacteria absorbed 35S into their cytoplasm

D) Neither isotope entered the host cells

A) Progeny phage released from 32P-labeled infections contained 32P. 32P labels DNA (which entered host cells and passed to progeny), while 35S labels protein (which remained outside the host cells).

What are the three structural components of a nucleotide?

A) Ribose sugar, amino acid, and nitrogenous base

B) Phosphate group, five-carbon pentose sugar, and nitrogenous base

C) Purine base, pyrimidine base, and phosphate group

D) Glycerol, fatty acid, and phosphate group

Phosphate group, five-carbon pentose sugar, and nitrogenous base. Every DNA and RNA nucleotide is composed of a phosphate group, a pentose sugar, and a nitrogenous base.

Which pairing correctly describes the chemical differences between DNA and RNA?

A) DNA has ribose with 2'-OH; RNA has deoxyribose with 2'-H

B) DNA has deoxyribose (2'-H) and thymine; RNA has ribose (2'-OH) and uracil

C) DNA has uracil; RNA has thymine

D) DNA lacks a phosphate group at the 5' position

DNA has deoxyribose (2'-H) and thymine; RNA has ribose (2'-OH) and uracil. Deoxyribonucleotides have an -H at the 2' carbon; ribonucleotides have an -OH at the 2' carbon. DNA uses T, while RNA uses U.

How does a purine differ from a pyrimidine, and which bases belong to each?

A) Purines have a single ring (C, T, U); pyrimidines have a double ring (A, G)

B) Purines have a double ring (A, G); pyrimidines have a single ring (C, T, U)

C) Purines are only in DNA; pyrimidines are only in RNA

D) Purines contain sulfur; pyrimidines contain phosphorus

Purines have a double ring (A, G); pyrimidines have a single ring (C, T, U). Adenine and Guanine are double-ring purines. Cytosine, Thymine (DNA), and Uracil (RNA) are single-ring pyrimidines.

In canonical double-stranded DNA, how do bases pair and how many hydrogen bonds form?

A) A pairs with T (3 bonds); G pairs with C (2 bonds)

B) A pairs with G (2 bonds); T pairs with C (3 bonds)

C) A pairs with T (2 bonds); G pairs with C (3 bonds)

D) A pairs with C (2 bonds); G pairs with T (3 bonds)

A pairs with T (2 bonds); G pairs with C (3 bonds). Adenine forms 2 hydrogen bonds with thymine (A=T); Guanine forms 3 hydrogen bonds with cytosine (G≡C).

What is the structural role of phosphodiester bonds in a nucleic acid strand?

A) Form weak hydrogen bonds across complementary strands

B) Covalently link the 3' carbon of one sugar to the 5' carbon of the next sugar

C) Attach nitrogenous bases directly to the 1' carbon

D) Connect histone proteins to the DNA backbone

Covalently link the 3' carbon of one sugar to the 5' carbon of the next sugar. Phosphodiester bonds link adjacent nucleotides together, establishing the covalent sugar-phosphate backbone.

What are the functions of the 1', 3', and 5' carbons in a pentose sugar of DNA/RNA?

A) 1' binds base; 3' and 5' form phosphodiester backbone bonds

B) 1' binds phosphate; 5' binds base

C) 3' binds nitrogenous base; 1' and 5' bind phosphate

D) 5' determines whether the sugar is ribose or deoxyribose

1' binds base; 3' and 5' form phosphodiester backbone bonds. The 1' carbon forms a covalent bond to the nitrogenous base, while the 3' and 5' carbons participate in phosphodiester linkages.

What defines DNA strand polarity, and how are complementary strands oriented?

A) 5'-OH to 3'-phosphate; parallel

B) 5'-phosphate to 3'-OH; antiparallel

C) 3'-phosphate to 5'-OH; antiparallel

D) 1'-base to 5'-phosphate; parallel

5'-phosphate to 3'-OH; antiparallel. One end has a free 5'-phosphate, and the other has a free 3'-OH. The two strands run in opposite polar directions (antiparallel).

What are hairpins (stem-loops) and how do they form in single-stranded nucleic acids?

A) Triple helices formed by three identical DNA strands

B) Inverted complementary sequences on the same strand that fold back and base-pair

C) Supercoils stabilized by topoisomerase enzymes

D) Cross-links between non-homologous chromosomes

Inverted complementary sequences on the same strand that fold back and base-pair. Hairpins form when contiguous single-stranded regions contain inverted complementary repeats that anneal together.

If Griffith had injected mice with heat-killed IIIS (virulent) AND heat-killed IIR (non-virulent) bacteria, what would happen?

A) Mice contract pneumonia and die because IIIS is present

B) Mice survive because living recipient bacteria are required for transformation

C) Bacteria regenerate and kill the mouse

D) IIR turns virulent spontaneously without DNA

Mice survive because living recipient bacteria are required for transformation. Transformation requires live recipient cells (like living IIR) to take up and express the foreign DNA.

If bacteriophage T2 had an RNA genome instead of DNA, what would Hershey and Chase have observed?

A) 35S would enter the bacterial cells

B) 32P would still enter the bacteria and be passed to progeny phage

C) Neither 32P nor 35S would be detected in progeny

D) Phage would be unable to infect host cells

32P would still enter the bacteria and be passed to progeny phage. RNA contains phosphorus but no sulfur, so 32P would still label the viral RNA genome and pass into host cells and progeny.

What is the complementary strand to: 5'-ATTGCTACGG-3'?

A) 5'-TAACGATGCC-3'

B) 3'-TAACGATGCC-5'

C) 5'-ATTGCTACGG-3'

D) 3'-CGGTAGCAAT-5'

3'-TAACGATGCC-5'. Complementary base pairing requires antiparallel orientation: 5'-A-T-T-G-C-T-A-C-G-G-3' pairs to 3'-T-A-A-C-G-A-T-G-C-C-5' (or 5'-CCGTAGCAAT-3').

If a double-stranded DNA molecule contains 15% thymine, what are the percentages of the remaining bases? A) A = 15%, G = 35%, C = 35%

B) A = 35%, G = 15%, C = 35%

C) A = 15%, G = 70%, C = 0%

D) A = 35%, G = 35%, C = 15%

A = 15%, G = 35%, C = 35%. By Chargaff's rules: %A = %T = 15%. The remaining bases total 100% - 30% = 70%, which is split equally between G (35%) and C (35%).

In any canonical double-stranded DNA molecule, what is the expected ratio of purines to pyrimidines?

A) 75% purines, 25% pyrimidines

B) 60% purines, 40% pyrimidines

C) 50% purines, 50% pyrimidines

D) The ratio varies depending on GC content

50% purines, 50% pyrimidines. Because purines (A, G) pair strictly with pyrimidines (T, C) across the double helix, total purines must equal total pyrimidines (50% each).

Virus I base composition: 0% T, 12% C, 9% U, 12% G, 9% A. What type of genome is this?

A) Double-stranded DNA

B) Single-stranded DNA

C) Double-stranded RNA

D) Single-stranded RNA

Double-stranded RNA. Contains Uracil (RNA) and follows Chargaff parity (%A = %U = 9% and %G = %C = 12%), indicating double-strandedness.

Virus II base composition: 23% T, 16% C, 0% U, 16% G, 23% A. What type of genome is this?

A) Single-stranded RNA

B) Single-stranded DNA

C) Double-stranded RNA

D) Double-stranded DNA

Double-stranded DNA. Contains Thymine (DNA) and exhibits 1:1 complementary ratios (%A = %T = 23% and %G = %C = 16%).

Virus III base composition: 34% T, 42% C, 0% U, 18% G, 39% A. What type of genome is this?

A) Single-stranded DNA

B) Double-stranded DNA

C) Double-stranded RNA

D) Single-stranded RNA

Single-stranded DNA. Contains Thymine (DNA), but %A (39%) ≠ %T (34%) and %G (18%) ≠ %C (42%), ruling out a complementary double helix.

Virus IV base composition: 0% T, 24% C, 35% U, 27% G, 17% A. What type of genome is this?

A) Double-stranded DNA

B) Double-stranded RNA

C) Single-stranded RNA

D) Single-stranded DNA

Single-stranded RNA. Contains Uracil (RNA), but %A (17%) ≠ %U (35%), confirming a single-stranded genome.