Problem Set 4 Flashcards
(20 cards)
What is the difference between a missense and a nonsense mutation?
Missense changes the codon so that it codes for a different amino acid. Nonsense is
when codon is changed so it is not recognized by a tRNA and therefore is bound by a release
factor
What are the differences between neutral mutations and silent mutations?
Silent is when codon changes but still codes for the same amino acid. Neutral is
when a missense happens but the new amino acid is chemically similar to the original amino
acid…. Therefore the shape of the protein doesn’t change and it continues to function
In order to determine if exposure to radiation increased the rate of germinal mutations,
scientists studied the sex ratio of children produced by survivors of the Hiroshima nuclear
bomb. Why would it be expected that germinal mutations would affect the ratio of the sexes
of the survivors of this nuclear blast.
The fact that germinal mutations effect gametes, or sex cells, would cause
scientists to expect that germinal mutations would affect the ratio of the sexes of the
survivors of this nuclear blast. The sex cells are what are responsible for dividing and
allowing offspring to receive copies of genetic information. Germinal mutations would
involve mutations that could occur in x chromosomes or y chromosomes. In the case of a
female, if they received one mutated x chromosome and one normal x chromosome, then
the female would probably be fine. However, if the female received two mutated x’s then the
mutation would most likely be lethal causing less females to be born. If a guy received a
mutant x then it most likely would not develop properly. This would alter the ratio of the
sexes also. Overall, the germinational mutations could result in males not being able to be
produced or females not being able to be produced, which would obviously alter the ratio
of sexes in the offspring.
Many organisms are prone to carrying expanding nucleotide repeats. This refers to a
nucleotide sequence that is two, three, or more nucleotides long that is repeated over and over
again. The number of times that these sequences are repeated tend to change from one
generation to the next. Explain this phenomenon.
DNA replication could occur properly, and create a complimentary DNA strand to form a
new DNA molecule. DNA slippage could also occur, which would cause a “bubble” to along the
newly synthesized DNA strand. This would cause the newly synthesized strand to lose a little bit of
its genetic sequence. The next time that strand goes through DNA replication, there would be less
repeats included in the newly synthesized DNA molecule than there were in the original DNA
molecule. This process could occur again, in another DNA replication process, which would cause
even less of the sequence repeats. The change in the number of sequences could also increase if
the slippage occurs in the newly synthesized strand. The number of sequences would decrease if
the slippage occurred in the template strand.
list all single-base substitutions that would change a codon for Leu to a nonsense codon. For
each, indicate whether it would be a transition or transversion.
UUA could be changed to UAA which would produce a stop codon. This is a
transversion. It could also be changed to UGA to produce a stop codon. This is a transversion.
UUG also codes for leucine. This could be changed to UAG to produce a stop codon. This is a
transversion.
There are 4 other codons that would code for leucine, but none of them could be mutated with a
single base substitution to produce a stop codon.
The following nucleotide sequence is found on the template strand of DNA. Determine the
amino acids of the protein encoded by this sequence. Then determine the altered amino acid
sequence produced if the following mutations were to occur.
WT DNA 3’- TACTGGCCGTTAGTTGATATAACT -5’
a. Transition at nt 11
b. Transition at nt 13
c. One nt deletion at nt 7
d. T A transversion at nt 15
e. Addition of TGG after nt 6
f. Transition at nt 9
a. Transition at nt 11. The sequence would go from a TTA to a TCA. The codon
produced would become AGU. This would code for serine instead of asparagine, which
would be a missense mutation.
Met-Thr-Gly-Ser-Gln-Leu-Tyr
b. Transition at nt 13. The sequence GTT would become ATT, which would
become the codon UAA, which is a stop codon, so this would be a nonsense mutation.
Met-Thr-Gly-Asn
c. One nt deletion at nt 7
AUG/ACC/GCA/AUC/AAC/UAU/AUU/GA
Met-Thr-Ala-Ile-Asn-Tyr-Ile
d. T à A transversion at nt 15 This would cause the codon GTT to GTA. This would produce a codon that is CAU, which would produce the amino acid histidine
Met-Thr-Gly-Asn-His-Leu-Tyr
e. Addition of TGG after nt 6 This would insert the amino acid Threonine in between the already existing threonine and glycine.
Met-Thr-Thr-Gly-Asn-Gln-Leu-Tyr
f. Transition at nt 9 The sequence would become CCA, which would produce the codon GGU. This would code for the amino acid glycine, which is already coded for. This would be a silent mutation.
Met-Thr-Gly-Asn-Gln-Leu-Tyr
A protein has the following amino acid sequence:
Met Ser Pro Arg Leu Glu Gly
Determine the type of mutation that must occur to produce the following mutants.
a. Mutant 1: Met Ser Ser Arg Leu Glu Gly
b. Mutant 2: Met Ser Pro
c. Mutant 3: Met Ser Pro Asp Trp Arg Asp Lys
d. Mutant 4: Met Ser Pro Glu Gly
e. Mutant 5: Met Ser Pro Arg Leu Leu Glu Gly
a. Missense
b. Met Ser Pro - Nonsense
c. Insertion/Frameshift mutation
d. Inframe deletion of 6 nt (2 amino acids)
e. Inframe insertion
Why do insertions and deletions often have more drastic phenotypic effects than do base
substitutions?
Some base substitutions are allowable. For example, synonymous mutations occur because of base substitutions that produce a mutated codon that still codes for the same amino acid. Neutral mutations produce codons that code for different amino acids
that have similar properties, which usually cause little effect to the overall shape and function of the protein produced. Conversely, insertions and deletions are usually detrimental to the protein when they occur in the open reading frame. A single insertion could potentially change every codon within an open reading frame, which would obviously alter the primary structure (and most likely function) of the protein produced from the open reading frame.
How will the result of strand slippage on the newly synthesized strand differ from the result of
strand slippage on the template strand?
Slippage on the newly synthesized strand causes an insertion of nucleotides, because the DNA polymerase adds extra nucleotides to account for the ones in the
bubble. Slippage on the template strand causes a deletion of nucleotides, because the DNA polymerase does not copy the template nucleotides in the bubble.
What are three major causes of spontaneous mutations and how does each result in a mutation?
Chemical changes, transposon, and replication errors. Chemical changes such as depurination can result in spontaneous mutations. In depurination, adenines and guanines break apart, which leaves a gap in the DNA backbone. When the DNA polymerase tries to repair this, it guesses which nucleotide needs to be added, and will add an A most of the time. Sometimes this is okay, but other times this could cause a missense mutation if a new amino acid was coded when the A was added in. It could also cause a nonsense mutation. The other type of chemical change that can occur is considered deamination. This is when the amino group is removed from the base, which causes an improper base pair. This can change the chemical nature of the bases in some circumstances, and even cause cytosine to become uracil, and 5 methyl cytosine to become thymine, which would be a base substitution, potentially causing the protein produced to be nonfunctional. Transposons are DNA elements that can jump around which consist of around a couple hundred nucleotides. They can insert in an intron, intergenic space, or an ORF, which would cause a large frameshift to occur. If they inserted in a ORF, this would most likely produce a protein that is nonfunctional. Replication errors are the other major causes of mutation. DNA polymerase can run into sequences of repeated bases, which can cause slippage to occur. In slippage, a bubble can form DNA template strand which would lead to a deletion, because the nucleotides in the bubble wouldn’t get read by the RNA polymerase, causing fewer nucleotides to be added to the mRNA strand. This would be considered a deletion. If the bubble formed on the mRNA strand, then this would cause an insertion, because whichever bases that are in the bubble would not be a part of the normal mRNA strand that is complementary to the DNA template strand. If the bubble contained 2 nucleotides for example, and the mRNA strand was supposed to be 10 nucleotides long, the slippage that caused the bubble would result in an mRNA sequence that was 12 nucleotides long.
Three common causes of induced mutations include nitrous acid, alkylating agents, and hydroxylamine.
Explain how they result in mutations.
Nitrous acid is capable of causing the number of deamination events to occur. This is the process in which an amine group is removed from a nitrogenous
base. In the case of Cytosine, if an amino group were to be removed, this would cause the cytosine to become uracil, which is a point mutation. Alkylating agents are responsible for the addition of methyl and ethyl groups. For example, the addition of an ethyl group by ethyl/methyl sulfonate to guanine results in the formation of a base that is capable of pairing with thymine instead of cytosine. If you a base that no longer pairs with its normal complementary base, you could have a DNA strand that is an uneven width and the mutation could also result in different codons being produced in the mRNA strand that is transcribed from the DNA template strand. This could potentially cause missense, nonsense and other base substitution mutations. Hydroxylamine alters cytosines so that they pair with adenine. It adds a hydroxyl group to the base cytosine, this changes it to a chemically modified form. This type of mutation would cause similar effects to alkylating agents in which the width of the double stranded DNA molecule could
potentially be compromised. If the bases were transcribed to a mRNA strand, then the codon sequences could potentially code for different amino acids. Mutation like missense mutations, nonsense mutations, or any base substitution mutation could occur as a result. This is true for the alkylating agents and hydroxylamine.
Give a molecular explanation of dominance and recessiveness.
Dominant alleles, which are usually wild type alleles or gain of function mutants, can be displayed when an organism is homozygous or heterozygous dominant in regards to a specific gene. This is due to the fact that the dominant allele will produce a functional enzyme. If a heterozygous individual has a dominant allele, then the recessive allele, which is usually a loss of function mutant, producing a nonfunctional protein wouldn’t matter. The functional protein would still perform its role. If an organism is homozygous recessive, then it has two copies of the recessive alleles in which there are two nonfunctional proteins
produced. The protein wouldn’t carry out its role, resulting in a recessive phenotype being displayed in the individual.
Give a molecular explanation of dominance and recessiveness.
Dominant alleles, which are usually wild type alleles or gain of function mutants, can be displayed when an organism is homozygous or heterozygous dominant in regards to a specific gene. This is due to the fact that the dominant allele will produce a functional enzyme. If a heterozygous individual has a dominant allele, then the recessive allele, which is usually a loss of function mutant, producing a nonfunctional protein wouldn’t matter. The functional protein would still perform its role. If an organism is homozygous recessive, then it has two copies of the recessive alleles in which there are two nonfunctional proteins
produced. The protein wouldn’t carry out its role, resulting in a recessive phenotype being displayed in the individual.
The mutagen EMS converts guanine (G) to O-6-ethylguanine (G). O-6-ethylguanine (G) forms
base pairs with thymine (T) instead of cytosine (C). Suppose that exposure to EMS damages a
DNA molecule as shown below:
a. Diagram the steps required for the incorporated G* to induce a stably inherited
mutation. Your diagram should include all necessary rounds of replication.
- ---------A---------
- ----------T----------- → ----------T--------- stable mutation - -----------G*------------ → -----------G*---------- → --------G*--------
————C————- → ——–C——— ——–T———
-------G---------- wild type It took two rounds of mutation for a stable pair to be formed. TA formed instead of GC. b. Characterize the mutation induced by EMS as a transition, transversion, or frameshift.
The mutation would be a transition mutation, because guanine should pair with cytosine, pyrimidine, and instead it pairs with thymine, which is also a pyrimidine.
Can nonsense mutations be reversed by hydroxylamine? Why or why not?
No, because the only two possible that could be changed by hydroxylamine are UGA and UAG, because their complementary DNA strands would have Cytosine. If the mutation occurred that caused the cytosine to pair with Adenine thanks to the addition of a hydroxyl group, then thymine would replace cytosine in those places. This would result in the DNA sequences for both being ATT, which both code for another stop codon, UAA, meaning that it is impossible to reverse a nonsense mutation with a hydroxylamine.
The following nucleotide sequence is found in a short stretch of DNA:
5’–AG–3’
3’–TC–5’
a. Give all the mutant sequences that may result from spontaneous depurination occurring in this
stretch of DNA.
b. Give all the mutant sequences that may result from spontaneous deamination occurring in this
stretch of DNA.
Depurination results in dropping of a guanine base or an adenine base from a DNA strand. If the guanine was dropped from the sequence, hypothetically the DNA polymerase could add an A in which would result in the sequence AA in the top strand. It would also be possible for the first A to be dropped, and the DNA
polymerase could add a G in its place resulting in a GG sequence. If both were to be dropped from the sequence, the DNA polymerase could add a G in the first position and an A in the second resulting in a sequence that is GA.
B. Deamination events occur when an amine group is dropped from a nitrogenous base. This could result in cytosine being converted to uracil. If the C were converted to uracil via a deamination event then this would result in a sequence that reads TU 3’ to 5’.
An Ames test was carried out on a suspected mutagen, which we will call mutagen A. In each
case, one million bacteria were plated on media that lacked histidine. The following numbers
of colonies were observed to grow.
Trial Control With mutagen A.
1 7 43
2 9 19
3 4 27
4 5 33
5 7 44
You don’t need to conduct at-test now, but you can probably guess what the results would be.
Based upon the results, what would you conclude?
In conclusion, the Ames test would suggest that mutagen A is mutagenic, because the number of colonies increased in each trial compared to the control groups. The control groups serve to show the spontaneous mutation rate, and the trials that were treated with mutagen A show an increase in the spontaneous mutation rate.
Describe two different mechanisms of DNA repair, highlighting the type of damage that they are able to correct.
In prokaryotes, thymine dimers are caused by ultraviolet light. They cause thymine dimers
to form, which are where adjacent thymines in a single DNA strand form a covalent bond. They
are unable to form hydrogen bonds with their complementary bases on the other DNA strand. In
prokaryotes only, this can be corrected by photoreactivation. UV light also stimulates an enzyme
pholiase which breaks the covalent bonds between thymines, allowing them to form hydrogen
bonds with their complementary bases on the other strand.
Nucleotide excision repair also repairs thymine dimer formation, but in eukaryotes. It does so by the enzyme UVR endonuclease which recognizes the thymine dimers. It breaks apart a large chunk of the DNA strand around where the thymine dimers are, which creates a nick in the DNA strand. DNA polymerase then resynthesizes the missing nucleotides using the other strand as a
template. DNA ligase seals up the gaps.
During mismatch repair it is essential for the cell to be able to distinguish between the old template DNA strand and the newly synthesized DNA strand in order to make the appropriate correction of a mispairing created during DNA replication. How is this accomplished in E. coli?
This is accomplished with the help of adenine methylase. This enzyme adds methyl groups onto adenines that are found in the sequence GATC. Other complexes form at the mismatch repairs called mismatch repair complexes. It takes a relatively long time for the Adenine methylase to find all of these sequences. The template strand will already have methyl groups attached. The nontemplate strand will not have methyls attached yet, which is important for
mismatch repair, because this allows the cell to differentiate between the template strand and the
nontemplate strand. The proteins at the GATC sequences and the mismatches will interact with
each other, causing the DNA molecule to fold. The proteins in the repair complex will make cuts only on the nontemplate strand, which is identified by the lack of methyl groups on its adenines. The cuts cause the segment of DNA to be excised. A whole chunk is removed. The DNA template strand is still intact, therefore DNA polymerase can come around and use the template strand to synthesize a chunk of DNA to fill the hole that was left when segment of DNA that contained the mismatch was excised.
If a thymine dimer occurs within a DNA molecule, what are the steps that result in its removal?
It depends on whether the organism is prokaryotic or whether it is eukaryotic.
- –If the organism is prokaryotic, then the organism will use a process called photoreactivation to repair the thymine dimers. UV light stimulates photolyase which is an enzyme that will break the covalent bonds between the adjacent thymines which will allow for the hydrogen bonds to occur between the thymines and adenines across from each other.
- ——–If the organism is eukaryotic, then it will use nucleotide excision repair. This involves the use of an enzyme known as UVR endonuclease. This enzyme will detect the presence of thymine dimers because they form structural abnormalities in the DNA strand. The enzyme pulls the two strands of DNA apart, and will excise the part that contains the thymine dimers. The other strand of DNA still exists as a template strand and can serve as the template for DNA polymerase to replace the segment that was removed. Ligase will also be utilized to come around and fill in the nicks.
