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Assume that the following is the entire sequence of a shortprotein which did not require mRNA splicing prior to itstranslation: NH2-Met-His-Leu-Tyr-Ile-His-Cys-Trp-Gln-Lys-Phe-COOH

a. Write out the sequence of a possible mRNA molecule encodingthis protein. You do not need to write out 5’-UTR or 3’-UTRsequences. Be sure to label 5’ / 3’ directionality.

b. Write out the sequences of both strands of thedouble-stranded DNA molecule corresponding to the mRNA and protein.You only need to write the portion of the double-stranded DNAmolecule which corresponds to the mRNA sequence in part (a). Labelthe template and non-template strands as well as 5’ / 3’directionality.

c. Identify three bases in the DNA molecule, each of which, ifmutated, will result in a nonsense mutation. Above each of thesenucleotides, write the base which will result in a nonsensemutation.

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Hubert Koch
Hubert KochLv2
28 Sep 2019
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Related questions

Can someone please check and help me with these? Especially II and III?

I. Below is a DNA sequence that is a template strand and contains an Open reading frame for a gene:

3’ TAA ATA CGT GAG GCA CTT CAA AGA CCG CAC ACT_ AAT TT 5’

5’ AUU UAU GCA CUC CGU GAA GUU UCU GGC GUG UGA _ UUA AA 3’

a. Transcribe the DNA template into a strand of mRNA by typing the mRNA sequence below the template strand. Label the 5’ and 3’ end of the mRNA 5’ AUU UAU GCA CUC CGU GAA GUU UCU GGC GUG UGA _ UUA AA 3’

b. Paste the mRNA below and do all 3 reading frames. Highlight all start codons green in all the reading frames. Highlight red the stop codons which are in frame with the start codons in all the 3 reading frames if they are present.

1) 5’ AUU UAU GCA CUC CGU GAA GUU UCU GGC GUG UGA UUA AA 3’

2) 5’ A UUU AUG CAC UCC GUG AAG UUU CUG GCG UGU GAU UAA A 3’

3) 5’ AU UUA UGC ACU CCG UGA AGU UUC UGG CGU GUG AUU AAA 3’

c) In which reading frame is the ORF found? The second one. Since it has AUG.

d) Paste the mRNA showing the most likely ORF below and translate the codons. Label the 5’ and 3’ end of the mRNA and the N- and C-termini of the peptide.

5’ A UUU AUG CAC UCC GUG AAG UUU CUG GCG UGU GAU UAA A 3’

MET HIS SER VAL LYS PHE LEU ARG CYS ASP STOP

N Terminus C Terminus

II. 3 pts. The base calling program made a mistake that you didn’t catch and deleted the bolded G in the original template strand above. Paste the template DNA below and delete the G. Transcribe (show the codons) and translate the resulting sequence. Label the 5’ and 3’ ends of the DNA and mRNA and the N and C-termini of the peptide. Comment on the resulting peptide sequence.

5’ AUUU AUG CAC UCC GUG AAG UUU CUG GCG UGU GAU UAA A ‘3

MET HIS SER VAL LYS PHE LEU ARG CYS ASP STOP

III. 3 pts. The base calling program made a mistake that you didn’t catch and added an extra G after the bolded G in the original template strand above. Paste the template DNA below and insert the extra G. Transcribe (show the codons) and translate the resulting sequence. Label the 5’ and 3’ ends of the DNA and mRNA and the N and C-termini of the peptide. Comment on the resulting peptide sequence.

5’ AUUU AUG CAC UCC GUG AAG UUU CUG GGG UGU GAU UAA A ‘3

MET HIS SER VAL LYS PHE LEU GLY CYS ASP STOP

QUESTION 1

A mutation caused by exposure to gamma rays is called a spontaneous mutation

a.

True

b.

False

c.

Sometimes, depending on the type of base change

d.

Gamma rays do not cause mutations

QUESTION 2

A recessive mutation affecting an essential biochemical pathway can be detected________

a.

when the organism is heterozygous for that mutation.

b.

when the organism is homozygous for that mutation.

c.

in either homozygous or heterozygous conditions.

d.

only by sequencing (recessive mutations never show a phenotype).

QUESTION 3

During DNA replication, the newly synthesized chain grows by adding

a.

nitrogenous bases

b.

the sugar-phosphate backbone

c.

nucleosides

d.

dNTPs

QUESTION 4

During replication, the DNA strand 5’- AAGTCTAGCCTAG -3’ will serve as a template for the polymerization of:

a.

5’- CTAGGCTAGACTT -3’

b.

5’ - TTCAGATCGGATC -3’

c.

5’ - GATCCGATCTGAA -3’

d.

3’ - AAGTCTAGCCTAG -5’

QUESTIOn 5

Given the following DNA sequences, select which statement(s) is(are) correct.

Sequence 1:

5’-TGGACGCTAA-3’

3’-ACCTGCGATT-5’

Sequence 2:

5’-AATCGCAGGT-3’

3’-TTAGCGTCCA-5’

Sequence 3:

5’-ACCTGCGATT-3’

3’-TGGACGCTAA-5’

a.

Sequences 1 and 2 are the same

b.

Sequences 1 and 3 are the same

c.

Sequences 2 and 3 are the same

d.

Sequences 1, 2 and 3 are all different

QUESTION 6

Repetitive sequences in the genome are hotspots for:

a.

Deamination

b.

Depurination

c.

Thymine dimer formation

d.

Replication errors

QUESTION 7

Select which statement(s) about the Ames test is(are) correct

a.

it has been designed to understand mutation repair systems in Salmonella typhimurium

b.

it uses mammalian liver extract

c.

it allows to study whether chemical compounds or their enzymatic breakdown products are mutagenic

d.

it is based on whether a chemical compound causes reversion mutations from his+ to his-

e.

two of the above are correct

f.

three of the above are correct

QUESTION 8

Select which statement(s) is(are) correct.

a.

All DNA strands have a direction, and it is specified by the carbons in the sugar backbone.

b.

All DNA strands have a direction, and it is specified by hydrogen bonds between nucleotides.

c.

In order to form proper base pairs in a double stranded DNA molecule, the two strands must run in opposite directions.

d.

A and C are correct.

e.

B and C are correct.

QUESTION 9

Studies of gene mutation frequencies have shown that:

a.

mutations are rare, and genomes are generally stable.

b.

mutation frequencies differ among organisms and also between genes, suggesting certain genes are more susceptible to mutation.

c.

mutation frequencies are consistent between organisms, and each region of DNA is equally susceptible to random mutations.

d.

Both A and B are correct.

e.

Both A and C are correct.

QUESTION 10

The compound 5-bromodeoxyuridine (BrdU) is a derivative of uracil, and if BrdU becomes incorporated during DNA replication, it pairs with adenine. This compound is best classified as which type of mutagen?

a.

base analog

b.

base inducer

c.

intercalating agent

d.

oxidative agent

e.

alkylating agent

QUESTION 11

The rate of mutation of the fruit fly is higher than the rate of mutation of Algae

True

False

QUESTION 12

Thymine dimers are most commonly caused by which of the following?

a.

X-rays

b.

Alkylating agents

c.

U.V. irradiation

d.

DNA intercalating agents

QUESTION 13

What chemical group is found in the 3’ end of a DNA strand?

a.

an alcohol

b.

a hydroxyl

c.

a methyl

d.

a phosphate

QUESTION 14

What chemical group is found in the 5’ end of a DNA strand?

a.

an alcohol

b.

a hydroxyl

c.

a methyl

d.

a phosphate

QUESTION 15

Which is the correct order of molecules binding to DNA during DNA replication?

a.

Helicase, SSB, primase, DNA pol III, DNA pol I, ligase

b.

SSB, DNA pol I, ligase, helicase, DNA pol I, primase

c.

primase, helicase, DNA pol III, ligase, DNA pol I, SSB

d.

SSB, helicase, primase, DNA pol III, DNA pol I, ligase

QUESTION 16

Which statement(s) is(are) correct about strand slippage?

a.

It can cause disorders such as Huntingon disease.

b.

It is a process that causes mutations altering the number of DNA repeats.

c.

It is a process that incorporates nucleotide base analogs and trinucleotide repeats.

d.

a and b are correct.

e.

b and c are correct.

f.

a, b and c are correct.

help needed asap.

1. In addition to identifying the genetic material, the experiments of Avery, MacLeod, and McCarty with different strains of Streptococcus pneumoniae demonstrated that
DNA is a double helix.
DNA may be taken up by bacterial cells and be active.
DNA is present in bacteriophage T2.
eukaryotic cells may be genetically transformed.
DNA binds to the cell wall of the bacterium.
2. In order to show that DNA in cell extracts is responsible for genetic transformation in Streptococcus pneumoniae, important corroborating evidence should indicate that _______ also destroy transforming activity.
temperatures high enough to kill cells
enzymes that hydrolyze proteins
enzymes that hydrolyze DNA
enzymes that hydrolyze polysaccharides
enzymes that hydrolyze RNA
3. Based on what you have learned about the experiments conducted by Griffith and Avery and colleagues with bacteria, which of the following would result in transformation of living R cells?
Pure RNA from S cells
Cell-membrane extracts from S cells
A mixture of pure RNA and protein from S cells
Cell-free extracts from S cells
Pure protein from S cells
4. A-T base pairs in a DNA double helix
form three hydrogen bonds with each other.
are more heat-stable than G-C base pairs.
are of a substantially different length than G-C base pairs.
are not accessible to DNA binding proteins.
are chemically distinct from G-C base pairs.
5. If 23 percent of the bases in a sample of double-stranded DNA are adenine, what percentage of the bases are uracil?
27
50
0
73
23
6. The uniform diameter of the DNA structure provides evidence for
antiparallel DNA strands.
a right-handed helix.
a double-stranded helix.
3′ end phosphorylation.
purine–pyrimidine pairing.
7. If a sequence of one strand of DNA is 5′-TGACTATC-3′, what is the complementary strand?
5′-ACTGATAG-3′
3′-TGACTATC-5′
3′-ACTGATAG-5′
3′-ACTGATAG-3′
5′-TGACTATC-3′
8. What structural aspect of the DNA facilitates dissociation of the two DNA strands for replication?
3′ end phosphorylation
Purine–pyrimidine pairing
Antiparallel DNA strands
Covalent bonds between sugars and phosphate groups
Hydrogen bonding between paired bases
9. If the Meselson–Stahl density gradient experiment had resulted in two bands of DNA molecules after only one round of replication, one containing only 15N and the second only 14N, this result would have indicated that replication was
unidirectional.
dispersive.
conservative.
bidirectional.

semiconservative.
10. The nucleoside analogue acyclovir, which is used to treat herpes simplex virus (HSV) infections, lacks a 3′ hydroxyl group (–OH). Predict what will happen if the host cell DNA polymerase incorporates a molecule of acyclovir into an elongating strand of HSV DNA.
The replication complex will no longer bind to origins of replication.
The DNA polymerase will only incorporate acyclovir molecules.
Acyclovir will bind single-strand binding proteins.
DNA polymerase will not be able to link a successive nucleotide.
Acyclovir will block the activity of the DNA helicase.
11. Which of the following does not demonstrate the stability of the DNA double helix?
Single-strand binding proteins are required to keep the strands apart.
Paired bases form hydrogen bonds.
High temperatures are required to denature it.
DNA synthesis requires energy.
DNA helicases require ATP.
12. What effect would a primase inhibitor have on DNA replication?
DNA polymerase would not be able to add bases to the DNA.
Primase would be blocked from joining the DNA replication complex.
Primase would not be able to provide primers for DNA polymerases.
DNA polymerase would not be able to use DNA as a template.
There would be no effect on DNA replication.
13. To replicate their DNA in a timely manner, most eukaryotic chromosomes
normally have uncoiled DNA.
have multiple origins of replication.
contain linear molecules of single-stranded DNA.
have two replication forks that move in the same direction.
are composed entirely of DNA.
14. Which statement about DNA replication is false?
Okazaki fragments are synthesized as parts of the leading strand.
Error rates for DNA replication are reduced by proofreading of the DNA polymerase.
The sliding clamp increases the rate of DNA synthesis.
Replication forks represent areas of active DNA synthesis on the chromosomes.
Ligases and polymerases function in the vicinity of replication forks.
15. In many eukaryotes, there are repetitive sequences called telomeres at the ends of chromosomes. After successive rounds of DNA replication, the _______ strand becomes shorter. In some cells, an enzyme called _______ repairs the shortened strand.
leading; telomerase
lagging; telomerase
lagging; DNA polymerase I
leading; DNA polymerase I
leading; replicase
16. A researcher studies normal human fibroblast cells. They can be maintained in culture but die off after about 30 cell generations. Unexpectedly, a colony of cells continues to survive and divide past 30 generations. Which scenario is most likely true for these cells?
DNA polymerase is constantly active.
Primase is able to extend the telomeres.
The mismatch repair system is extra efficient.
The telomerase gene is being expressed.
The cells do not need the ends of their chromosomes.
17. If DNA polymerase III introduces an incorrect nucleotide, what is the first corrective action made by the DNA repair system?
The replication complex excises the incorrect nucleotide.
The excision repair proteins remove the incorrect nucleotide.
The mismatch repair system scans for base-pairing mismatches.
There is no correction, because nucleotide errors by DNA polymerases are not repaired.
DNA ligase connects the correct nucleotide.
18. Choose the correct order of the following four events in the excision repair of DNA:
(1) Base-paired DNA is made complementary to the template.
(2) Damaged bases are recognized.
(3) DNA ligase seals the new strand to existing DNA.
(4) Part of a single strand is excised.
1, 2, 3, 4
2, 1, 3, 4
3, 4, 2, 1
2, 4, 1, 3
4, 2, 3, 1
19. Six complete cycles of PCR should result in a _______-fold increase in the amount of DNA.
64
32
12
128
6
20. When double-stranded DNA is heated to temperatures above 90°C, it denatures. Denaturation is a process that
breaks covalent bonds.
hydrolyzes DNA.
separates double-stranded DNA into two single strands.
causes new hydrogen bonds to form.
irreversibly destroys the double helical structure.

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