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Integrated Topics in Genetics

QUESTION 1: A large population of hawkmoths lives in a valley in Canada. Genetic tests reveal the following frequencies of alleles for wing colour:

a) If moths continue to mate randomly, what would be the frequencies of moths with black, tan, and white wings in the next generation?

As per hardy Weinberg equilibrium,

Frequency of black wings = (0.5)2 + 2(0.5) (0.4) +2(0.5) (0.1) = 0.75

Frequency of tan wings = (0.4)2 + 2(0.4) (0.1) = 0.24

Frequency of white wings = (0.1)2 = 0.01

b) A small group of the moths flies to neighbouring Montana and starts a new population. After several generations, there is a large randomly-mating population of hawkmoths. Observations of the Montana moths reveal the following:

What are the allele frequencies for C, ct, and cw?

Since the tan wings is zero, then the frequencies of C, ct, and cw are 0.5, 0 and 0.5 respectively (C and Cw has equal frequencies)

c) The change in allele frequency in the Montana population as compared with the original Canadian population is an example of Genetic drift/founder effect.

QUESTION 2: Histone acetyltransferases add acetyl groups to lysines, which are positively charged amino acids. How might this affect the association of a nucleosome with DNA?

The association of a nucleosome with DNA reduces. Initially the positive nucleosome (as lysine has positive charge) attracts the negative DNA. Now the lysine does not have a positive charge and it has negative acetyl group attached to it after acetylation. This means nucleosome becomes negatively charged thereby repelling the DNA.

QUESTION 3: If ribonucleotides were depleted from a cell during S phase, how would the DNA synthesis be affected? Ignore energetic considerations.

Replication would cease because ribonucleotides are required to initiate DNA synthesis. This is because S phase is where the DNA synthesis starts and it requires ribonucleotides is used to initiate DNA synthesis from Okazaki fragments.

QUESTION 4. The following are melting temperatures for different double-stranded DNA molecules:

a) Using the letters (A-E) to symbolize the DNA molecules, arrange the above molecules from lower to higher content of G-C pairs.

High GC content has high melting point. Hence the order is as follows (Lower GC to higher GC) B<A<D<C<E.

b) Explain your answer in no more than 6 lines.

The reason for high melting point for high GC content is because GC bond is a triple bond compared to AT which is a double bond. This means more amount of energy is required to break a GC bond than an AT bond. This in turn means that the melting point for higher GC content will be higher than the samples with lower GC content.

QUESTION 5: A protein found in zebra fish normally has the following N-terminal amino acid sequence:


A mutant is altered so that the anticodon of a tRNA is changed from 5’-GAU-3’ to 5’-CAU-3’. Using the genetic code provided in the Appendix respond the following questions:

a) What would be the N-terminal amino acid sequence of this protein in the mutant?

The N terminal amino acid remains same as Met-Val-Ser-Ser-Pro-Met-Gly-Ala-Ala-Met-Ser. This is because the tRNA anticodon change from 5’-GAU-3’ to 5’-CAU-3’ does not have any corresponding effects as the mutated tRNA binds to methionine instead of Isoleucine. Since Isoleucine is not present in the given strand, the sequence remains same- Met-Val-Ser-Ser-Pro-Met-Gly-Ala-Ala-Met-Ser.

b) Would this mutation have an effect on the organism’s phenotype? Explain your reasoning.

No, this mutation does not effect on the organism’s phenotype. Change in the phenotype of a fish requires a large amount of changes and mutations in various base sequences. A change in the tRNA or mutation in it does not have any effects in the phenotype.

QUESTION 6: Klemke et al. (2001) discovered an interesting coding phenomenon in an exon within a neurologic hormone receptor gene in mammals, by which two protein entities (XLαs and ALEX) appear to be produced from the same exon. Below is the DNA sequence of the exon’s 5’-end derived from a rat. The lowercase letters represent the initial sequence for the XLαs gene, and the uppercase letters indicate the initial portion of the ALEX coding region. (For simplicity, and to correspond with the RNA coding dictionary, it is customary to represent the non-template strand of the DNA segment). gtcccaaccatgcccaccgatcttccgcctgcttctgaagATGCGGGCCCAG Using the genetic code provided to you in this document’s Appendix

a) Provide the mRNA sequence for each of the genes. Indicate the molecular orientation (3’ and 5’ ends) of the sequences.

3‘----gUcccaaccaUgcccaccgaUcUUccgccUgcUUcUgaag ---5’

b) Provide the amino acid sequence for each coding sequences. In the region of overlap, are the two amino acid sequences the same?

3’-----AUG CGG GCC CAG----5’: met arg ala gln

3’----gUc-cca-acc-aUg-ccc-acc-gaU-cUU-ccg-ccU-gcU-UcU-gaa- gAU ---5’: Val Pro thr met pro thr asp leu Pro pro ala ser glu -asp

Note: The free “g” superimpose with the AU of the AUG in the ALEX coding region thereby forming asp.

Also, in the region of overlap, the two amino acids are not same as the ALEX code overlap region has methionine whereas the XLαs has Aspartic acid.

c) Are there any advantages to having the same DNA sequence for two protein products? Discuss potential evolutionary advantages or disadvantages to having the same DNA sequence code for two protein products.)

Yes there is an advantage of having the same DNA sequence for two proteins. During production of recombinant proteins, the use of DNA fragment as insert into a vector has a limit. When a DNA sequence has the ability to code for two proteins, then one can produce two products using recombinant technology without a need for separate vectors. There is an evolutionary advantage of having the same DNA sequence for two protein products. The size of genome reduced and the efficiency of product synthesis increases to manifold. This is especially true for viruses that evolve very faster and have efficient genome with DNA encoding for multiple protein synthesis.

QUESTION 7. As a last-resort emergency response to severe DNA damage, Bacillus subtilis, like E. coli, activates an SOS response. The DNA damage activates the RecA protein, which results in the LexA protein cleaving itself. Intact LexA binds to a consensus sequence, 5’-CGAACNNNNGTTCG-3’, found in the promoter region of about 17 genes used to repair DNA damage. Cleaved LexA is unable to bind this consensus sequence sequence. Therefore, cleavage of LexA leads to the transcription of these genes.

Answer the following questions and concisely justify your answer.

a) Is LexA a positive or negative regulatory protein?

LexA is a negative regulator protein as it binds to the promoter region and stops the transcription process. When LexA is not present or cleaved, the transcription occurs. This means LexA is a negative regulator.

b) What role does the 5’-CGAACNNNNGTTCG-3’ sequence play?

The role of this sequence is to help bind LexA thereby not allowing the DNA to transcribe, which means, the other genes such as dinA, dinB, dinC and recA present after this sequence cannot be transcribed. This in turn means that the proteins required for correcting DNA strands cannot be made.

c) Tow allelic mutations, designated A and B, alter the 5’-CGAACNNNNGTTCG-3’ sequence of a gene so that LexA cannot bind the A sequence at all but it binds to the B sequence much more tightly than it does to the wild-type sequence. Suppose an SOS response is triggered in each mutant. What phenotype do your expect each to exhibit?

Since LexA cannot bind to the sequence, the promoter is always active and the DNA repair factors will be synthesized indefinitely. Irrespective of whether the SOS occurs, the DNA repair mechanism synthesizes the factors indefinitely, resulting an inflammation in the phenotype.

If the LexA is allowed to bind tightly to mutated 5’-CGAACNNNNGTTCG-3’ sequence, then, when the SOS response is triggered, the LexA will cleave slowly as the self cleaving feature of LexA is not affected by the tightness of LexA to the mutated sequence.

d) A plasmid bearing a mutant lexA gene is introduced into an otherwise normal strain of B. subtilis. The mutant gene makes a LexA protein that binds tightly to the 5’-CGAACNNNNGTTCG-3’ sequence but is unable to undergo self-cleavage. What phenotype do you expect this strain to have?

Since this strain cannot self cleave the LexA attached to the 5’-CGAACNNNNGTTCG-3’ sequence, the strain will exhibit severe DNA damages followed by death of the strain. This is because the LexA cannot be removed from the promoter sequence 5’-CGAACNNNNGTTCG-3’ which in turn means that he organism cannot synthesize the required proteins for repairing the DNA.

QUESTION 8: Cloned eukaryotic genes are not always able to be expressed in bacterial cells. What are some possible reasons for this?

The first reason for this is because the prokaryotes do not have a membrane bound nucleus. This means, the genes from eukaryotes cannot be expressed in the prokaryotes as the eukaryote gene requires the nucleus to be inside the membrane for better transcription of the gene followed by translation to produce proteins.

Post-translational modification is an important an essential phenomena observed in the eukaryotic genes. Post-translational modifications are rare with the prokaryote expression systems. This results in production of futile proteins or in other words, failure to express the desired proteins.

QUESTION 9: You are attempting to determine whether a plant-derived processed food sample (for example, a corn chip) contains material from a genetically modified organism (GMO). First, you crush the sample and attempt to extract DNA from it. Next, you perform PCR using two different sets of primers. One primer set will amplify a DNA sequence present in all plants. The second primer set will amplify a DNA sequence only found in GMO plants.

a) Why must you use both sets of primers for this experiment?

Both sets of primers are used to amplify the gene of interest present in the plants. For example, one primer is specially designed to bind to a unique sequence present in the normal plants and the other primer binds to the sequence not present in the plant but in a GMO component. This way, the primers will be able to amplify the gene of interest or the target gene and will show corresponding band in the PCR.

b) In addition to the test sample, you obtain a negative control sample (food material you are certain does not contain GMO material) and a positive control sample (food material you are certain does contain GMO material). You perform the DNA extraction on these three samples and then the PCRs with each of the two primer sets described above. Complete the following table with your expectations for (column 3) a test sample that does not contain genetically modified food and (column 4) a test sample that does contain GMO food.

A negative control does not have any GMO material. This means they give band for plant primers and no band for GMO primers. A positive control has GMO material with them and this means they give band for GMO primers also with the Plant based primers.


PCR reaction

Expect band? [non-GMO food]

Expect band? [GMO food]


Negative control – plant primers




Negative control – GMO primers




Test sample – plant primers




Test sample – GMO primers




Positive control – plant primers




Positive control – GMO primers



c) You obtain the results shown in the panel below. What conclusions can you draw from these reaction results? Does this test sample contain genetically modified components? Why or why not? [Lanes 1–6 are the same as listed above.]

The lane 2 shows the presence of a band when the GMO primers are added to a negative control. This means, the negative control is contaminated with other DNA. Therefore, it is impossible to clearly state whether the test sample 4 has GMO components or not (As the test sample 4 shows a band similar to the lane 6 bands- the positive control with GMO primers). This means, either the primers are contaminated or the sample is contaminated.

QUESTION 10: You are interested in a particular segment of rhinoceros DNA and would like to clone it into a cloning plasmid. Below is a restriction map of the region that includes the DNA of interest and the plasmid (E = EcoRI, H = HindIII, X = XbaI, S = SphI, N = NotI).

Which restriction enzymes would you choose to clone most easily the DNA of interest into the cloning vector?

The possibility of using E and H is rejected as these enzymes also cut the gene of interest. This leaves us with the only possibility of choosing N, X and S. It can be seen in the cloning vector that it does not have the site for the S restriction enzyme to cut. This leaves us with the only option of using E and H.

Remember, at the center of any academic work, lies clarity and evidence. Should you need further assistance, do look up to our Biology Assignment Help

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