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AP Bio Ch. 16 Study Guide Test Your Knowledge

Multiple Choice
Identify the choice that best completes the statement or answers the question.
 

 1. 
One of the reasons most scientists believed proteins were the carriers of genetic information was that
a.
proteins are more heat stable than nucleic acids
b.
the protein content of duplicating cells always doubles prior to division
c.
proteins are much more complex and heterozygous than nucleic acids
d.
early experimental evidence pointed to proteins as the hereditary material
e.
proteins are found in DNA
 

 2. 
Transformation involves
a.
the uptake of external genetic material. often from one bacterial strain to another
b.
the creation of a strand of RNA from a DNA molecule
c.
the infection of bacterial cells by phage
d.
the type of semiconservative replication shown by DNA
e.
the replication of DNA along the lagging strand
 

 3. 
The DNA of an organism has thymine as 20% of its bases. What percentage of its bases would be guanine?
a.
20%
b.
30%
c.
40%
d.
60%
e.
80%
 

 4. 
In his work with pnemonia-causing bacteria, Griffith found that
a.
DNA was the transforming agent
b.
the pathogenic and harmless strands mated
c.
heat-killed harmless cells could cause pnemoniawhen mixed with heat-killed pathogenic cells
d.
a substance was transferred to harmless cells to transform them into pathogenic cells
e.
A T2 phage transformed harmless cells to pathogenic cells
 

 5. 
T2 phage is grown in E.coli with radioactive phosphorous and then allowed to infect other E. coli cells. The culture is blended to separate the viral coats from the bacterial cellls and then centrifuged. Which of the following statements best describes the expected results of such an experiment?
a.
Both viral and DNA molecules are labeled; radioactivity is found in the liquid above the pellet
b.
Viral DNA is labeled; radioactivity is found in the pellet
c.
Viral proteins are labeled; radioactivity is found in the eliquid but not the pellet
d.
Both viral and bacterial proteins are labeled; radioactivity is present in both the liquid and the pellet
e.
The virus destroyed the bacteria; no pellet is formed
 

 6. 
Watson and Crick concluded that each base could not pair with itself because
a.
there would not be room for the helix to make a full turn every 3.4 nm
b.
the uniform width of 2 nm would not permit two purines or two pyramidines to pair together
c.
the bases could not be stacked .34 nm apart
d.
identical bases could not hydrogen-bond together
e.
they would be on antiparallel strands
 

 7. 
In their classic experiment, Meselson and Stahl
a.
provided evidence for the semiconservative model of DNA replication
b.
were able to seperate phage protein coats from E. coli by using a blender
c.
found that DNA labeled with 35N was of intermediate density
d.
grew E. coli on labeled phosphorous and sulfer
e.
found that DNA composition was species specific
 

 8. 
The joining of nucleotides in the polymerization of DNA requires energy from
a.
DNA polymerase
b.
the hydrolysis of the terminal phosphate group of ATP
c.
RNA nucleotides
d.
the phosphate groups of the sugar-phosphate backbone
e.
the hydrolysis of the pyrophosphates removed from nucleoside triphosphates
 

 9. 
The continuous elongation of a new DNA strand along one of the template strands of DNA
a.
requires the action of DNA ligase as well as polymerase
b.
occurs because DNA ligase can only elongate in 5’ --> 3’ direction
c.
occurs on the leading strand
d.
occurs on the lagging strand
e.
a, b, and c are correct
 

 10. 
Which of the following statements about DNA polymerase is incorrect?
a.
It forms the bonds between complimentary base pairs
b.
It is able to proofread and correct errors in base pairing
c.
It is unable to initiate synthesis; it requires an RNA primer
d.
It only works in the 5’ --> 3’ direction
e.
It is found in eukaryotes and prokaryotes
 

 11. 
Thymine dimers - covalent links between adjacent thymines bases in DNA - may be induced by UV light. When these dimers occur, they are repaired by
a.
excision enzymes (nucleases)
b.
DNA polymerase
c.
ligase
d.
primase
e.
a, b, and c are all needed
 

 12. 
How does DNA synthesis along the lagging strand differ from that on the leading strand?
a.
Nucleotides are added to the 5’ end instead of the 3’ end
b.
Ligase is the enzyme that polymerizes DNA on the lagging strand
c.
An RNA primer is needed on the lagging strand but not on the leading strand
d.
Okazaki fragments, which each grow 5’ --> 3’, must be joined along the lagging strand
e.
Helicase synthesizes Okazaki fragments, which are then joined by ligase
 

 13. 
Which of the following enzymes or proteins is paired with an incorrect or inaccurate function?
a.
Helicase - unwinds and seperates parental double helix
b.
Tolemerase - adds telomere repititions to ends of chromosomes
c.
Single strand binding protein - holds strands of unwound DNA apart and straight
d.
Nuclease - cuts out (excises) damaged DNA strands
e.
Primase - forms DNA primer to start replication
 

 14. 
Which letter indicates the 5’ end of this single DNA strand?
a.
A
b.
B
c.
C
d.
D
e.
E
 

 15. 
At which letter would the next nucleotide be added?
a.
A
b.
B
c.
C
d.
D
e.
E
 

 16. 
Which letter indicates a phosphodiester bond formed by DNA polymerase? the 5’
a.
A
b.
B
c.
C
d.
D
e.
E
 

 17. 
The base sequence of the DNA strand made from this template would be (from top to bottom)
a.
ATC
c.
TAC
e.
ATG
b.
CGA
d.
UAC
 

 18. 
Which of the following statements about telomeres is incorrect?
a.
They are-ever shortening tips of chromosomes that may signal cells to stop dividing at maturity
b.
They are highly repetitive sequences at the tips of chromosomes that protect the lagging strand during replication
c.
They are repetitive sequences of nucleotides at the centromere region of a chromosome
d.
They are enzymes, in germ cells that allow these cells to undergo repeated divisions
e.
Both a and b are correct
 

 19. 
You are trying to test your hypothesis that DNA replication is conservative - that is, that the parental strands seperate, newly made complimentary join together to make new DNA molecules, and the parental strands then rejoin. You take a sample of E. coli grown in a medium containing only heavy nitrogen (15N) and transfer it to a medium containing light nitrogen (14N). After allowing time for only one DNA replication, you centrifuge a sample and compare the density band(s) formed to the bands formed from bacteria grown on either normal 14N or 15N medium. Which band location would support your hypothesis of conservative DNA replication?
a.
A
b.
B
c.
C
d.
D
e.
E
 

 20. 
Given the experimental procedure explained in question #19, which centrifuge tube (obtained after one DNA replication) would represent the band distribution indicating that DNA replication is semiconservative?
a.
A
b.
B
c.
C
d.
D
e.
E
 

 21. 
If the following structures were put in order from smallest to largest, which structure would be in the middle of the size range?
a.
looped domain
b.
histone
c.
nucleosome
d.
30-nm fiber
e.
metaphase chromosome
 

 22. 
Biologists have learned from the technique of of “painting chromosomes” with different colored molecular tags that
a.
DNA replication proceeds at different rates on different chromosomes
b.
the two homologs of a pair of chromosomes differ enough that they stain different colors
c.
chromosome packing occurs only as the cell prepares for mitosis or meiosis
d.
heterochromatin is concentrated at the tips and centromeres of chromosomes
e.
in the interphase nucleus, each chromosome appears to occupy a secific area
 



 
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