Download Biology 321 Spring 2011 Answers to Assignment Set #5

Biology 321 Spring 2011 Answers to Assignment Set #5
Answers to questions in Griffiths 10th edition
Chapter 7
8.
NOTE: primes used in the lab are short polymers of DNA (15-30 bases long)
Chapter 8
19.
21.
Chapter 9
12
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1
14
18
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2
Chapter 16
9.
10.
13
21.
d. A nonsense mutation causes premature termination of translation. A frameshift mutation shifts
the reading frame and often results in a truncated polypeptide due to the presence of a stop codon in
the new reading frame.
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3
22.
31.
Answers to Biology 205 review questions:
1. peptide bonds
2. tertiary
3. isoleucine & phenylalanine
3B 4 & 5
4. The tertiary structure of a globular protein is determined in part by the hydrophobic effect –
nonpolar amino acid side chains are sequestered in the interior of a protein. Polar side chains will
typically be on the surface of the protein (interacting with water). This mutation has resulted in the
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substitution of an amino acid with a polar side chain for an aa with a non-polar side chain and
might very well affect the stability and/or foramtion of the proper tertiary structure.
5. & 6. single-stranded DNA 5’ carbon is at the top of the chain – with 3 phosphates; 3’ carbon
is at the bottom with the free hydroxy
7.
a.
8. a
9. c
10. c
11. b & c
12. c
13. a
14A a. A
b. no anticodon will H-bond here since the 5’ UAA 3’codon is a stop codon and
will terminate polypeptide synthesis
c. mRNA is written and translated 5’ 3’
d. protein is synthesized from N-terminal to C-terminal
14B: e.
15: a. nucleoside = nucleotide without a phosphate
b. 1. carbonyl 2. pyrimidine 3. hydroxyl 4. modified deoxyribose
16. DNA consists of two polymers; adenine base-pairs with thymine and guanine with cytosine
17. the top pyrimidine
18. Watson drew guanine in its rare tautomeric form which (due to its pattern of H-bond donor and
acceptors) H-bonds with thymine not cytosine
19. 5’ GUU 3’ anticodon will base-pair with a 5’ AAC 3’ codon which specifies asparagine (ASN)
20. promoter
21. a&b. see problem 20 c. specific sequences which are typically high in A’s and T’s (TATA
box)
22.
5 translation begins at the first AUG and ends with the first inframe stop codon: UAA
23. DNA is the permanent archive of the genetic database; RNAs typically have a short half-life.
Errors made during DNA synthesis in a multicellular organism (that remain unrepaired) are passed
onto the next generation of cells (if somatic) or progeny (if in the germline)
24.
The codon is 5’ UAC 3’ and will specify tyrosine
25. See text description of mRNA, tRNA and rRNA 26. b 27. d
28. See lecture notes and text 29. T T F
30. a. ase means the protein has catalytic activity b. transcript in the standard context means an
RNA molecule that is generated from a DNA template C. the retrovirus infection cycle involves
generating DNA molecules using RNA as a template – the reverse of what usually happens in the
cell during transcription
31. Answer is a. The mRNA reads 5’ CGUAAUCAUG…. etc. (antiparallel to the template).
Translation starts with the first AUG and ends with the first in-frame stop codon: UAG
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Additional Problems
 Problem 1 The mutation rate per gamete in Drosophila represents mutations that have
accumulated after a series of mitoses that gave rise to the germline in the embryo and to the cells
that eventually under go meiosis. So more DNA replication events are represented by the per
gamete rate.
 Problem 2 The 3’-5’ exonuclease activity associated with DNA polymerases corrects most
mistakes that occur during the DNA chain elonation process. DNA mismatch repair systems correct
errors that remain post-DNA synthesis. RNA polymerases do not have proofreading capabilities
and there are no RNA repair systems. Why does this make sense in the context of the different
roles of DNA and RNA in the cell?
 Problem 3
Answer: c
 Problem 4
a. See lecture notes for labelling of the carbons.
b. dideoxy means that there is no hydroxyl group on either the 2’ or 3’ carbon.
c. This compound must be converted into the triphosphate form (added to the 5’ carbon) before it
can serve as a substrate for viral DNA polymerase
d. The TP version of dideoxyI can be added to the 3’ end of a DNA chain, but DNA replication
terminates because there is no 3’ hydroxyl -- which is required for addition of the next nucleotide.
 Problem 5A
a. After a couple of rounds of replication: CG to TA transition
b. Thymine replaces uracil in DNA so that the cell can recognize the cytosine deamination product
(uracil) as abnormal and initiate a repair process. If uracil were used as adenine’s pairing partner in
DNA, there would be a problem differentiating “normal” uracils that should be in the DNA from
uracils that are a cytosine decay product. NOTE: I think I remember reading that a G:T repair
system was just discovered that changes the T to a C in a GT mismatch. This is not a postreplication mismatch repair system which could change the G to an A OR the T to C depending on
which strand was the parental.
 Problem 5B
 Problem 6
a. thymine b. a GC to AT transition will result
Answer is c.
A
G
C
G*
C
C
T
A
rep
G
C
___ ____ = apurinic site = the phosphodiester backbone (phosphate –ribosephosphate) is still intact but the ribose doesn’t have a base attached to it
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 Problem 7
a. Some spontaneous mutations are the result of mistakes made during DNA
replication. Furthermore, many (all of the examples discusssed in class) spontaneous
and induced mutations require rounds of replication to establish the error in both
strands of DNA. Therefore the more a cell divides, the greater the risk of mutation.
Paternal gametes undergo many more rounds of replication than maternal gametes.
Since the germline cells divide continually in a male, in an older father, the germ
cells will have undergone more round of replication.
b. NOTE: the mutation (GGU  AGU) is designated in codon jargon, which, by
convention, is in RNA language. BUT, we are not talking about a transcription error
-- which obviously does not meet the definition of a mutation.
diagram on the next page
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 Problem 8
See answers to a revised version of this problem: Winter 2010 Take-home quiz (#4)
Problem #2.
 Problem 9
Answer is c.
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 Problem 10
MUTANT #1
a. nonsense
b. tyr
stop
tyr
UAU --› UAA
or
UAC --› UAA
--› UAG
--› UAG
each case is a transversion
c. a base modifying agent might, depending on which bases it chemically
modified and whether an apurinic/apyrimidinic base resulted.
MUTANT #2
a. frameshift
b. trp
UGG =› one of the G's is deleted
c. intercalating agent
MUTANT #3
a. missense
b. tyr
UAU --› CAU or UAC --› CAC both are transition mutations
c. base analog or base modifying agent
d. mutant #3
 Problem 11
(i) e (b & c)
Mutant 1 could result from (i) UAu/c --> UAG or UAA transversion , (ii) from a
deletion of the 3rd base in the tyr codon which would generate a UAA stop codon or
(iii) from an insertion of a G or an A after the UA of the tyr codon resulting in a UAA
or a UAG stop codon
(ii) e (a & b) (iii) a
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 Problem 12
a. FH380: missense
FH 763: frameshift
FH 683: nonsense
b. UAU/C tyrosine  UGU/G cysteine A to G transition
c. mRNA 5’….GUG UAC CAA….
DNA
5’ ….GTG TAC CAA….3’
3’ ….CAC ATG GTT… 5’
bottom strand serves as the template for mRNA synthesis
5’ ---A--- 3’
3’ ---T--- 5’

----G ---  --- G------ T* ------ C ---
T* = thymine in rare tautomeric form
 = round of DNA replication (drawing follows one parental strand only)
d. I’m going to testify for the defense for the following reasons
1. The father and mother need to be assessed. It is likely that one or the other is
heterozygous for the same mutation their son has. Do either have a history of heart
attacks? Have their genotypes been determined?
2. Even if the son results from a new mutation it couldn’t have been caused by the
father’s exposure to hydroxylamine. The son’s mutation is a AT to GC transition.
Hydroxylamine cannot cause this mutation because it only modifies cytosines and
would not act at an AT base pair.
3. Defense probably pays more (just kidding)
e. Note ARG is specified by 6 different codons.
• The mutation could be CGn  UAG or AGa/g  UAG
• The first alternative CGn  UAG requires transition mutations in the first two
base pairs of the codon (assuming that n=G). Neither aflatoxin or proflavin will
accomplish this.
The second alternative AGa/g  UAG is a possibility. Inserting a T before this
codon would result in a UAG and, as an intercalating agent, proflavin could cause
this type of mutation. Since both parents are likely to be homozygous wild-type, I’ll
go with the prosecutors on this one.
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 Problem 18
a. Neutral Sequence variation does not affect the Darwinian fitness of the organism.
b. Yes. The alternative sequence variation exceeds 1%
c. These sequence variations are not disease causing. You would want to focus your screen on the
mutations in Table 2.
d. It is in the third codon position so likely to be a same sense or silent mutation
e. It is in the first codon position so likely toe be a neutral missense
 Problem 19
Answer not available just yet. Will post it on FRiday
 Problem 20
a. Review definition of polymorphism in earlier lecture
b. silent or same sense mutation
c. Neutral missense mutation (note legend at bottom of table that indicates that all people
genotyped were healthy non-NIDDM)
d. Examination of a normal control group is important because some sequence variations will be
associated with disease and others will have no obvious effect on the encoded protein or on the
organismal phenotype. In order to understand how the gene product functions and to assess the
phenotypic implications of a particular genotype, it is critical that researchers be able to distinguish
between these two types of sequence variations.
e. Intron sequences are non-coding – not translated into protein
 Problem 21
GAU/C (asp)  CAU/C
transversion
GC  CG
Start here: Have G* pair with G in first round of replication, etc
5’ GAT 3’
3’ CTA 5’
hhhhhhhhhhhhhhhhhhhh
 Problem 22
Answer will not be available
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