Biology Life On Earth With Physiology 10th Edition By Audesirk – Test Bank

Chapter 11 DNA: The Molecule of Heredity

11.1 Multiple Choice Questions

1) The hereditary material present in all cells is
A) protein.
B) RNA.
C) DNA.
D) R-strain.
E) S-strain.
Answer: C
Diff: 1
Section: 11.1
Skill: Knowledge/Comprehension

2) What was the most significant conclusion of Griffith’s experiments with pneumonia in mice?
A) Mice exposed to the S-strain bacterium became resistant to the R-strain bacterium.
B) There is a substance present in dead bacteria that can cause a heritable change in living bacteria.
C) The genetic material was definitively proven to be DNA.
D) S-strain bacteria can cause pneumonia.
E) Heat destroys the hereditary material.
Answer: B
Diff: 2
Section: 11.1
Skill: Knowledge/Comprehension

3) What is the relationship among DNA, a gene, and a chromosome?
A) A chromosome contains hundreds of genes, which are composed of protein.
B) A chromosome contains hundreds of genes, which are composed of DNA.
C) A gene contains hundreds of chromosomes, which are composed of protein.
D) A gene is composed of DNA, but it has no relationship to a chromosome.
E) A gene contains hundreds of chromosomes, which are composed of DNA.
Answer: B
Diff: 2
Section: 11.1
Skill: Knowledge/Comprehension

4) In Griffith’s experiments, what happened when heat-killed S-strain bacteria were injected into a mouse along with live R-strain bacteria?
A) DNA from the live R-strain was taken up by the heat-killed S-strain, converting them to R-strain and killing the mouse.
B) DNA from the heat-killed S-strain was taken up by the live R-strain, converting them to S-strain and killing the mouse.
C) Proteins released from the heat-killed S-strain killed the mouse.
D) RNA from the heat-killed S-strain was translated into proteins that killed the mouse.
Answer: B
Diff: 2
Section: 11.1
Skill: Knowledge/Comprehension
5) When Griffith experimented with two types of Streptococcus pneumoniae, he found that if the
A) deadly strain was heat-killed before injection, the mice died.
B) nondeadly strain was mixed with the heat-killed, nondeadly strain before injection, the mice died.
C) nondeadly strain was mixed with the heat-killed, deadly strain before injection, the mice lived.
D) nondeadly strain was mixed with the heat-killed, deadly strain before injection, the mice died.
Answer: D
Diff: 2
Section: 11.1
Skill: Knowledge/Comprehension

6) DNA possesses
A) A, U, G, and C bases.
B) only C and T bases.
C) only A and G bases.
D) C, T, A, and G bases.
E) only U and T bases.
Answer: D
Diff: 2
Section: 11.2
Skill: Knowledge/Comprehension

7) If the quantities of the four bases in a DNA molecule are measured, we find that
A) A = C and G = T.
B) A = G and C = T.
C) T = A and C = G.
D) no two bases are equal in amount.
E) all bases are equal in amount.
Answer: C
Diff: 2
Section: 11.2
Skill: Knowledge/Comprehension

8) If the DNA of a certain organism has guanine as 30% of its bases, then what percentage of its bases are adenine?
A) 0%
B) 10%
C) 20%
D) 30%
E) 40%
Answer: C
Diff: 3
Section: 11.2
Skill: Application/Analysis
9) All of the following are found in DNA EXCEPT:
A) deoxyribose.
B) guanine.
C) a phosphate group.
D) a phospholipid group.
E) thymine.
Answer: D
Diff: 1
Section: 11.2
Skill: Knowledge/Comprehension

10) A DNA nucleotide is made up of
A) phosphate-deoxyribose-base.
B) phospholipid-deoxyribose-base.
C) phosphate-deoxyribose-phosphate-deoxyribose.
D) adenine-thymine-guanine-cytosine.
E) base-phosphate-glucose.
Answer: A
Diff: 2
Section: 11.2
Skill: Knowledge/Comprehension

11) The “rule” formulated by Chargaff states that
A) A = T and G = C in any molecule of DNA.
B) A = C and G = T in any molecule of DNA.
C) A = G and C = T in any molecule of DNA.
D) A = U and G = C in any molecule of RNA.
E) DNA and RNA are made up of the same four nitrogenous bases.
Answer: A
Diff: 1
Section: 11.2
Skill: Knowledge/Comprehension