Mini test 2 - Deep Thought q\'s PDF

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Mini Test 2 - Deep thought Questions The Klinefelter syndrome (disomy of the X chromosome in males) is a genetic disease caused by aneuploidy of the sex chromosomes in humans. Patients with the 47,XXY karyotype are male. Describe two genetic scenarios leading to patients with the Klinefelter syndrom...

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Mini Test 2 - Deep thought Questions 1. The Klinefelter syndrome (disomy of the X chromosome in males) is a genetic disease caused by aneuploidy of the sex chromosomes in humans. Patients with the 47,XXY karyotype are male. Describe two genetic scenarios leading to patients with the Klinefelter syndrome. During meiosis 2 the chromatids are unevenly separated and 3 chromosomes (XXY) are formed compared to the usual 2 which are formed, which causes aneuploidy of the sex chromosomes. Another scenario in which this could take place is when sister chromatids or homologous chromosome pairs do not split up, which can take place in meiosis 1/2 in which is known as ‘nondisjunction’. A female gamete that has an additional x chromosome, XX, could cross with a male gamete with a singular Y chromosome which would give the result of 47, XXY. This scenario would lead to aneuploid gametes being produced by mitosis as the gametes would be unequally distributed, n plus/minus one, instead of having 1n each, and they would have the wrong amount of chromosomes. This would be passed down as the resulting zygote would also become aneuploid, as the aneuploid gamete would combine another regular gamete in fertilisation. This zygote would have an irregular number of chromosomes, n+1 crossed with n = 2n+1. 2. A plant breeder breeds avocados, trying to obtain a plant with a unique combination of desirable fruit and pathogen resistance traits. After many years of breeding different wild populations of avocado, success is achieved: an avocado plant that produces fruits of excellent quality and has very strong resistance to a range of bacterial pathogens. To produce more plants like this one, should she reproduce this plant by self-fertilisation or produce more plants via asexual reproduction (using cuttings) and why? Fully explain your answer. If the breeder was to reproduce the plant through the process of self fertilisation, undesirable traits (such as lower fruit quality or low bacterial resistance) may appear in offspring because some of the alleles are recessive (don’t have a dominant allele). This would only become apparent in the phenotype of offspring. However, if the breeder produced plants via asexual reproduction the offspring would be genetically identical to the parent. This would provide her with offspring that produce excellent quality fruits with strong bacterial resistance. Through the use of cuttings, she is able to also to further select for desirable traits and ensure that undesirable traits do not become inherited by offspring.

3. Yeast genetics has been instrumental in deciphering biosynthesis pathways for many compounds. Explain why it might be easier and less time-consuming to use yeast (a haploid organism) to study the effects of mutations on protein function? A haploid organism only has one set of chromosomes, and therefore, no alleles. Consequently, there are no dominant genes for a particular characteristic, meaning that gene function is never masked by

a dominant gene from an allele counterpart. This allows the function of genes, and therefore the proteins they encode, to be analysed far more easily and precisely without having to work out if protein functions are a result of incomplete dominance, co-dominance, or total dominance of one allele over another. Therefore, any mutations on protein function would have a direct result in the organism’s phenotype, (since a homolog that carries a different version of the same gene does not exist in this scenario).

4. Even in the absence of genetic recombination, meiosis is an important source of genetic variation in sexually reproducing organisms. Use as an example a species with 2n=8, and an individual of that species that is a quadruple heterozygote for loci A, B, C, and D, with each loci on a different chromosome. Determine how many different types of gametes this individual can produce, and explain how segregation and independent assortment mechanisms produce genetic variation in the gametes produced by this individual. Use the following terms in your response: genes; alleles; homologous chromosomes; nonhomologous chromosomes. As an individual that has a diploid cell with 8 chromosomes, the individual can produce 2^4 genetically variant gametes. During meiosis I (1), all homologous chromosomes (Aa) are segregated, which means that the homologous chromosomes carrying heterozygous loci A, B, C and D are separated. In anaphase 2, this segregation is repeated with sister chromatids. Then the Non-homologous chromosomes are assorted independently during telophase 2, causing a random allocation of genes in each gamete. Gametes then carry a random assortment of A, B, C and D alleles. 5. The same chromosome can look very different depending on when in the cell cycle it is observed. Explain fully why this is so and also why the chromosome is less condensed during some parts of the cell cycle? How many DNA molecules does each chromosome contain at the beginning of mitosis? Use the following terms in your explanation: chromatin, histone, chromatid, interphase, metaphase, mitosis, chromosome, gene expression. DNA is approximately 1m long in every nuclei, therefore it needs to be condensed. It is wrapped around histones, and these specialised proteins create chromatin, a DNA protein complex. Chromosomes are then formed as a result of further compaction. Gene expression is stopped, as the condensing of chromatin renders the DNA inaccessible. Components of the chromosome are relaxed during duplication (S Phase) and transcription (interphase), which makes it less compacted and thus the chromosomes appearance is able to change. Two sister chromatids are formed from the duplication of chromosomes at the start of mitosis. Thus, both chromosomes each have 2 DNA molecules. Metaphase sees the condensing of each of these chromosomes, and during anaphase the 2 sister chromatids are pulled apart which forms 2 nuclei which both contain a copy of the genome. 6. A 3000 bp region of the human genome encodes two genes. One of the genes encodes a protein of 700 amino acids and the other gene encodes a protein of 310 amino acids. The mRNA sequences of the two genes do not contain any of the same nucleotide sequences. How is this possible? Fully explain your answer.

mRNA’s sequence is edited and altered after it has been translated from DNA. The maximum amount of amino acids which are able to be encoded for is 1000 due to the 3000bp length of the gene (amino acids are 3 bp). When adding the number of given amino acid values, 700 + 310 =1010, it can be seen that this exceeds the maximum amount by 10bp. However, this can be explained by the process of mRNA editing, as after the transcription of the 3000 base pair region, it then forms mRNA with 3000bp which is subject to editing that involves the insertion of new bases. Ten addition amino acids are formed as a result, explaining the 310 amino acids. Thus each of these genes don’t contain any identical nucleotide sequences as they each encode for different sections of the 3000 base pair region, one for 2100/3000bp and the other for 900/3000bp. 7. Which DNA mutation is more likely to have a detrimental effect on the protein it codes for: a 2 base pair deletion from the middle of the coding sequence of a gene, or a 12-base pair deletion from the same region of the gene? Explain why? If 2 bases pairs were to be deleted from the middle of the coding sequence of a gene, this would cause significant mutation in the cell as non-functional proteins could be expressed due to this frame shift mutation in the gene and every following codon would be effected. Whereas, a 12 base pair deletion would lead to the deletion of 4 codons, as 3 base pairs make up a codon, and thus the protein in question would only be missing 4 amino acids but would otherwise still be functional. However, if a stop codon is deleted, it could be missing more amino acids. b. Would the outcome be different if the deletions happened within the gene’s intron? Explain why? Proteins are unlikely to be affected if the 12 base pair deletion occurs in the intron of a gene as the initial transcription of mRNA is edited and introns are removed. Whereas all codons following the deletion would be affected if the 2 base pair deletion took place in the intron. The amino acids and protein that form would be different as the exons would be changed following the introns. Introns would also not be recognised and will be left in.

8. A mutant E. coli strain is found with a mutation affecting some of its tRNA(Cys). The wild type normally produces a tRNA that recognizes the codon 5’ UGC 3’, and is charged with the amino acid Cysteine (Cys) (its notation is tRNA(Cys)). The mutant tRNA is still charged with Cysteine, but the mutation is in its anticodon that now has the sequence 5’- UCA-3’. How will some of the proteins produced in these E. coli cells be different from the proteins produced in the wild type cells?

If the anticodon was in the five prime to three prime direction with a sequence of UCA, then it can be determined the subsequent codon that attaches would be in the three prime to five prime direction with the sequence AGU (UGA in the five prime-three prime direction). The translation to amino acids is then halted as UGA is one of the three stop codons. On the polypeptide chain, on the carboxyl tail, a water molecule (H20) is added and this stops translation. If the beforementioned anti-codon is charged with Cys (Cysteine), amino acids will continue to be translated (unless there is another stop codon soon after). The proteins produced in the E. coli cells will be different to the wild type cells as they will contain many more unessential amino acids. These longer proteins may be unable to perform the required function. 9. The Central Dogma of molecular biology states that the biological flow of information is from DNA to RNA to protein. Many examples of viral replication contradict aspects of the Central Dogma. Why have viruses evolved replication mechanisms that always relate to transcription as a means of replication of the viral genome rather than reversing the translation of RNA? When following the central dogma, RNA is transcribed from DNA, with the mRNA then being translated into a protein. In order to form proteins, viruses utilise the ribosomes of the host. As a result, this contradiction of the central dogma isn’t related to translation, as the translating of RNA to protein follows the central dogma. The ssRNA (+) genome gives viruses the ability to translate their RNA genome straight to protein in a similar fashion to if it was mRNA. The genome makes the template of -RNA to make +RNA by using RdRP. RNA intermediates are created with 2 stands and as a result, the transcription of mRNA form DNA doesn’t take place, thus viruses contradict the central dogma relating to transcription as a means of replication. 10. Researchers have identified mutations in the coding region of the lacI gene that lead to a change in the structure of the lacI repressor protein. In these mutants, the lacI protein’s allolactose binding pocket is now permanently in the “allolactose-bound” state in the absence of allolactose. Explain fully what consequence, if any, this will have on the production of lacZ protein in the mutant E. coli cells. ‘Negative control’ occurs when allolactose isn’t bound to wild-type lacl. RNA polymerase is blocked from synthesising mRNA as the protein binds to an operator sequence. Lac Z production is halted due to their being no allolactose for the lacl to bind to, due the absence of lactose in the cell. In an “Allolactose-bound” state, lac z production occurs despite the absence of lactose. The binding of CAP to the promotor, ‘positive control’, enables strong transcription. This only takes place when there is low levels of glucose which leads to high levels of cAMP.

11. A recent research paper demonstrated that cAMP bound to CAP acts more strongly on the lac operon than on the arabinose operon. Given what you know about catabolite regulation, and the lac operon, list the order of preference of utilisation of the three sugars; arabinose, lactose and glucose. The order of the preference of utilisation of the 3 sugars is: Glucose, lactose and then arabinose. The Lac operon uses glucose as its only source of sugar when glucose and lactose are both available. The glucose effect is the capability of glucose to inhibit the synthesis of certain enzymes. This is a valuable property which makes it the most preferred. Lactose is the second preference due to the ease of its metabolism and the low amount of energy required to break it down. Whereas Arabinose is the final preference as it requires much more modification to its structure and energy to break it down so It can be utilized by the lac operon.

12. a. What is meant by a repressible operon? Operators are DNA sequences which perform the role of binding to a corresponding regulatory protein. Repressible operons have their genes active despite the fact they are not bound to its specific corresponding regulatory protein. Their transcription is stopped when the repressor protein is bound to it. b. In the case of the trp operon, how is repression exerted? It encodes tryptophan synthesis, and the trp operon’s corresponding, inactive repressor protein is only activated when the binding of tryptophan to it occurs. This shows that tryptophan is a corepressor. The protein repressor is not active when there is little to no tryptophan, and this prompts the synthesis of tryptophan. However, the protein repressor becomes active again and represses the trp operon when there is a higher amount of tryptophan. c. If someone asked you to genetically engineer a strain of E. coli to produce large amounts of tryptophan, what would be the easiest way to do that? By modifying the repressor pathway, lots of tryptophan could be produced. The synthesis of the repressor protein would need to be halted by modifying the genome. Or the binding of the repressor protein to the tryptophan could be stopped by altering the repressor proteins shape or size. Another way could be by stopping the binding of the trp operon to the activated repressor protein by changing the trp operon.

13. Evolutionary change is more closely associated with changes in gene expression rather than with the evolution of new proteins with completely new functions. Describe how homeotic mutations illustrate this point. Homeotic mutations occur in the body and lead to the replacement of one body part by another. Gene expression is altered as the mutation is caused by a mutation in the hox transcription factor. A phenotypic change can occur due to the change in the hox transcription factor. Over a large period of time, new functioning/specialised proteins could evolve, but a significant number of mutations in the genome would need to take place to see any phenotypic change. Whereas phenotypic changes over time are the main contributor to evolutionary change, as mutations that effect gene expression lead to more significant and timely change in phenotypes compared to the evolution of new proteins. 14. In Labrador retrievers, a gene (B) determines the black or chocolate coat colour. The black colour (B) is dominant over chocolate (b). Your neighbour owns two Labradors from the same litter, both female, one black, one chocolate. He paid $10,000 to send his black Labrador female to breed with a champion black Labrador. This “champion” is owned by a dishonest character. Your neighbour is not a witness to the mating. Last weekend the neighbour’s dog gave birth to 12 puppies, 7 of which are chocolate colour. What are the most likely and the second most likely explanations for this? The neighbours dogs parents are unable to be homozygous black due to the fact that the neighbour has one black and one chocolate dog and as a result the neighbour’s dog has to be heterozygous (Bb). Hypothetically, if the other owner breeds the neighbours dog with a chocolate labrador (bb), the resulting genotype would be a 1:1 ratio of Bb(chocolate):bb(black). The offspring ratio of 7 chocolate to 5 black implies that the breeder most likely didn’t breed with his champion dog but rather with a chocolate male. This is due to the fact that 7:5 is close to the hypothetical 1:1 ratio between Bb & bb. Also, offspring’s colours can be different to what is expected as phenotype is based on the interaction of genotype and environment. A 3:1 ratio of black : chocolate, would occur when both the heterozygous black (Bb) champion male and female mate. The binomial distribution formula can be used to determine the chance of 7 chocolate and 5 black 12C 7 x 0.25^7 x 0.75^5= 0.0115 x 100=1.15% - which shows this is unlikely scenario. 15. Using Mendel’s lines of peas, describe how you would use a test cross experimental design to determine the genotype of a pea plant with purple flowers. Be very specific in describing your experimental design and how you would interpret your results based on the proportion of offspring phenotypes. Make sure to include all possible offspring phenotypes in your answer and what the proportion of phenotypes tells you about the purple plant genotype. We do not know dominant individuals genotype’s as they only express their phenotypes, they can be homozygous (BB) or heterozygous (Bb) for any trait. To find their genotype’s, test crosses are used which breed a homozygous recessive individual’s genotype with their unknown dominant phenotype. Breed the F1 generation (purple flowers) with a homozygous recessive pea plant (aa - white flowers) and take note of resulting phenotype of the F2 generation. If F1 is AA, all offspring have purple flowers. If F1 is Aa, then it is 1:1 of

purple flowers to white flowers. The dominant individual’s phenotype can be determined to be heterozygous if there is a recessive phenotype in the offspring.

16. The developmental origins of the germ line cells in plants and animals are quite different. Based on your understanding of the main differences, what are the relative odds (plants vs. animals) of a somatic mutation in a parent being heritable in a 10 year old apple tree vs. a 100 year old Galapagos tortoise? Explain your answer. Germ line cells undergo meiosis which produces gametes containing genetic information which is then inherited by offspring. Any somatic mutations that take place after the meiosis of the germline are not inherited. Germline cells in animals are found in organs and are designated in the early stages of development, in most animals, egg cells start in the foetus. For a somatic mutation to be heritable by a tortoise’s offspring, it would have to take place before this development. Whereas plants germ cells are designated much later and are developed via mitosis of adult somatic cells. Any mutations which occur in these somatic cells would be carried to the germ cells of the tree and it would be heritable to its offspring. As a result, somatic mutations are much more likely to be heritable in a 10 year old apple tree compared to a 100 year old tortoise. 17. The novel coronavirus, SARS-Cov2, which is responsible for the disease Covid-19 is a positive strand RNA ((+)ssRNA) virus. From what you learned in BIOL1020, suggest a model of how the SARS-Cov2 genome is replicated. The key enzyme for viral replication is unique to the virus and not found in the human host. What is the name of the enzyme, what is its enzymatic function, and how would you explain that a drug called Remdesivir (an adenosine analogue that causes premature termination of RNA synthesis) is currently the most promising drug against a range of RNA viruses, including Ebola ((-)ssRNA virus), MERS ( (+)ssRNA virus ) and SARS-Cov2 ((+)ssRNA virus)? 

Breaking Dogma - (probably want us to mention this somewhere within our answer)

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RNA dependent RNA polymerase (RdRp) is the key enzyme.

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Uses a positive ss RNA as a template to make a copy of itself. The copies can then be used as a template for more strands.

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RNA dependent RNA polymerase uses RNA to create more RNA by creating -ve ss RNA and using that acts as a template for more +ve sense RNA. (creates double stra...