BIO 2133 Test#1 2021 key PDF

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Exam answers for midterm 1....

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BIO2133(Dr.Montpetit) TakeHomeTest#1(35Points) Due–February3rd830am‐midnight.  INSTRUCTIONS: 1. Please download a copy of the exam. You have the option to use the WORD or PDF version. 2. This exam is open book. Answer ALL questions as best you can with the information on hand. 3. This test must be completed individually. Should two or more students submit identical answers or answers that constitute plagiarism in any form, all students with those answers will face allegations of academic fraud. We are not using Respondus or any other lockdown software. 4. The test is designed to be written in a 90-minute block of time; however, you have up until midnight to submit your answers to the test. 5. You can answer questions in any way you choose/are able. This includes but is not limited to: a. Printing the exam, filling it in by hand, then scanning/photographing it. b. Answering each question by hand on a blank paper, then scanning/photographing it. *You do not need to recopy the questions to your sheet of paper but be sure to clearly indicate which question you are answering and if you must copy sequences/images/etc… to your sheet, do so very carefully to avoid making errors. c. Handwriting your answers digitally. (e.g. with an Apple pencil or stylus capable of writing in e-ink) d. Typing your responses directly into the file. 6. You may ask questions via email. They will be answered only during regular business hours (9am to 5pm). Only questions of a technical nature (i.e. – submission issues) will be answered in the final 30 minutes of the exam. Please do not expect quick answers given the size of the class. 7. Please name and save the digital file (as a PDF file- ONLY) containing your completed exam student#_BIO2133_surname.filetype (e.x - I would name my completed exam – 12345678_BIO2133_Montpetit.pdf). 8. You must submit your completed exam by midnight on Feb. 3rd . This is a hard deadline. a. Upload a single file (.docx or .pdf) of the entire completed exam by midnight on Feb. 3rd. b. You will receive an email confirmation of the receipt of your submission. c. If you experience technical difficulties resulting in you missing the submission deadline, please contact me before the deadline to let me know ([email protected]). d. Late submissions will not be assessed and considered an absence.

Helpful Hints and Tips:

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If completing the exam digitally, it is recommended that before beginning to write, that you name and save the document. I also recommended that you activate any automatic save function that your software may have.

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If scanning/photographing your exam to digitize a handwritten file, you may find it helpful to use a free app/program such as clear scanner (for android or iPhone) or pdf merge (web based) to compile

 multiple pages into a single file. Please try to create a PDF file prior to the exam, in order to test out your ability to successfully create a PDF file and to avoid any issues when trying to submit your exam.  Please use 12pt Times New Rom font if you type your answers. Write your answers in the space provided.

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 1. The figure below shown below is from a human individual. This karyotype was generated by collecting the chromosomes following duplication and condensation, but prior to entering mitosis. This particular individual is:  Heterozygote for the leptin gene: O/o  Heterozygote for the tyrosinase gene: C/c OR homozygous dominant (C/C)  Afflicted with an autosomal recessive form of the Ehlers-Danlos condition (Gene E) Also, suppose that the leptin and tyrosinase genes are said to be linked. The gene for Ehlers -Danlos is not linked to the leptin and tyrosinase genes. Using the karyotype diagram, provide a representation of the relative location of the various genes and alleles on the chromosomes and sister chromatids in such a way that is consistent with the scenario described above. In other words, use labels to show the locations of the genes and alleles in such a way that is consistent with the scenario described above. Choose the chromosomes of your liking to make the correct representations (3 marks)

e

O

Oo

o

C

Cc

c

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ee

e

 Location of genes on sister chromatids: 0.5 marks for each pair (O, C, and E) of correctly placed genes on sister chromatids – total 1.5 marks   

Please note that the location of the E, O and C genes can be located on any chromosome. Gene E is located on its own pair of homologous chromosomes. Genes O and C have to be on the same homologous chromosome. The relative location of all genes do not have to be localised in the exact position on the sister chromatids as represented in the answer key. They can be located anywhere on any arm (p or q arm) of the sister chromatids.

Correct genotypes:| 0.5 marks for each correct allele label on the homologous chromosomes – total 1.5 marks   .

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Dominant alleles for the O and C genes can be located on the same or different homologous chromosome. Must be heterozygote for O and C. Gene E is homozygote “ee”

 2. Priyanka, a 4th year undergraduate honours thesis student, recently identified a new allele causing cystic fibrosis. Genetic testing of multiple individuals afflicted with cystic fibrosis revealed that this new allele linked to cystic fibrosis is inherited in an autosomal dominant fashion and caused by a point mutation in the 3’ untranslated region that causes the mRNA to degrade at a very high rate. This is unlike the Δ504 mutation which causes an autosomal recessive form of cystic fibrosis. Priyanka’s study also demonstrated that 0.01% of the population is tested positive for this new allele.

Based on this information…. a. How many individuals in the population has the Cystic Fibrosis gene and why do some individuals have CF while others do not? Explain your reasoning. (2 marks)  

100% of individuals in this population has the gene (1 mark). CF is determined by genotype (autosomal recessive = cc, autosomal dominant = CC) (1 mark)

b. Every individual who tested positive for this new allele for CF are heterozygote (they have the new allele and a copy of the wild-type allele). Provide a genetic mechanism that could explain why this new allele causes CF in an autosomal dominant fashion. (2 marks)

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Possessing only 1 copy of the wild-type allele is haploinsufficient to achieve normal function. (1 mark). Insufficient amount of the protein to achieve a threshold level for normal function. (1 mark)

 3. You are studying a gene called COL5A1. You discover a deletion of 1 nucleotide in a variant of the COL5A1 gene. You are about to conduct experiments to compare the length of the wild-type COL5A1 protein and the mutant COL5A1 protein gene when your supervisor asks for your predictions. List 3 predictions and for each provide your reasoning (6 points)

Prediction 1: the wild-type protein will be longer than the protein coded by the mutant COL5A1. (1 mark) Reasoning – The deletion of 1 nucleotide causes a frameshift and an early stop codon thus making the mutant protein shorter. (1 mark)

Prediction 2: The protein coded by the mutant COL5A1 will be longer than the wild-type COL5A1 protein. (1 mark) Reasoning – The deletion of 1 nucleotide causes a frameshift and puts the stop codon out of frame thus making the mutant protein longer. (1 mark)

Prediction 3: The wild-type COL5A1 protein and the protein coded by the mutant COL5A1 allele will be the same length. (1 mark) Reasoning – the deletion is in an intron or in the promoter/5’ or 3’ untranslated region and thus would not affect the length of the protein. (1 mark) OR frameshift occurs at stop codon resulting in another stop codon in frame.

Prediction #4: Mutation in the AUG start codon of the mRNA which leads to no protein being produced. (1 mark) Reasoning – No start codon no protein. (1 mark)

Prediction #5: Deleting in core promotor element resulting in no mRNA being produced. Reasoning: Initiation of transcription does not occur as the promotor elements and TF factors, without this core promotor element, are not able to recruit the ARN pol complex and initiate transcription.

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 4. Joanne and Joan are about to have a baby. The baby is biologically Joanne’s. Ultrasounds and a karyotype have revealed that their child (to be named Joanna) is trisomic for chromosome 21. Joanne asks a genetic counsellor to explore the possibility that the chromosome nondisjunction occurred in her gamete or the biological father’s and in which stage of meiosis did it occur. To answer the question, a DNA test was conducted to measure the presence or absence of the chromosome marker STR (short tandem repeat) marker D21S11 in Joanne and Joanna (the baby). The STR marker D21S11 is located on chromosome #21 and 4 variants of the marker (A1, A2, A3 and A4 variants) can be found in the population. The DNA test revealed the following:  

Joanne – A2A4 Joanna (baby) – A1A2A3

Based on the test results, what is your conclusion and explain your reasoning. (4 marks)

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The results suggest that the baby inherited the A2 marker from Joanne. (1 mark) The A1 and A3 markers must come from the biological father (Joanne cannot contribute those markers). (1 mark)

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The nondisjunction must have resulted in the father during meiosis 1. (1 mark) The homologous chromosomes did not segregate into the 2 game cells. (1 mark)

 5. During translation, the tRNA molecule carrying the correct amino acid corresponding to its anticodon sequence must base-pair bind with the codon of the mRNA. What would happen in the case where tRNA carrying its correct amino acid incorrectly binds with the codon of the mRNA? Would this be considered a mutation? Explain your reasoning. (3 marks)

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The incorrect amino acid would be added to the growing peptide. (1 mark)

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No this would not be considered a mutation. A mutation is a change in base-pair in the DNA. (1 mark) Since this is not a change in base pair, this is not a mutation. (1 mark)

 6. Wild-type corn snakes are brown in colour, however, autosomal recessive mutations in the steps of colour determination can result in either orange, black or brown coloration or no colouration (white). Assume that the dominant alleles are haplosufficient. The biochemical pathway for colour determination is the following:

a. Assuming no confounding factors, which genotype will result in a phenotypically black corn snake? (use a clearly defined annotation system to indicate a genotype) (2 mark)

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AA or Aa (0.5marks) ; bb (0.5marks) ; CC or Cc (0.5marks) If all incorrect, award 0.5 (rubric is already setup this way) OR Not possible to have a phenotypically black corn snake (1 mark), because the membrane transporter only acts to transport the Brown pigment (1 mark).

b. Assuming that the transcription of enzyme A mRNA is completely inhibited by a transcription factor, what will be the snake’s phenotype? (1 mark)   

This depends on the answer provided in a). Orange (if they answered AA or Aa;bb; CC or Cc) or White (if they answered not possible to have a phenotypically black corn snake)

c. If the snake’s genotype is AaBBcc, what will its colour? (1 mark)

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No coloration or white

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7. Knockout mice are mice that have gone through genetic modification to inactivate certain genes that codes for an enzyme. One method of achieving knockout is through homologous recombination (you do not need to know how this is done for the purposes of this test). After breeding a homozygous wild-type mouse with a homozygous knockout mouse, you find the following results:

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a) Based on the results, what conclusions can you draw about this gene? (1 marks) - The knockout allele is a dominant allele OR causes haploinsufficiency in the heterozygote state. (1 mark) b) How would you annotate the alleles to represent the genotypes of the wild-type, heterozygotes, and homozygote mutants? (2 marks; all or none) Wild-types A+A+ or aa; Heterozygotes A-A+ or Aa ; homozygous KO: A-A- or AA

c) Assume there is a third allele of this gene. A heterozygote mouse with a copy of the wildtype allele and this third allele displays an enzyme activity that is lower than the homozygous wild-types but is above threshold for normal activity. What would be the order of dominance of the 3 alleles? (in other words which allele is the most dominant and so on) – rank the alleles with a 1 (most dominant), 2 or 3 (most recessive): (2 marks; all or none) Wild type:

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2

Knockout allele:

1

3rd allele:

3

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  8.

Consider the DNA molecule above, assuming no introns: (total 6 marks)

a. How many reading frames can you identify? ________6_________ (1 mark)

b. How many open reading frames can you identify? _______4_______ (1 mark)

c. Predict the amino acid sequence of each of the open reading frames you have identified: (4 marks – be aware that the number of marks does not necessarily reflect the number you need to identify).

1) 2) 3) 4)

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MET-ASP-ASP-SER MET-THR-HIS-ASN-LEU MET-THR-HIS MET-SER-HIS-LYS-LEU

OR M-D-D-S OR M-T-H-N-L OR M-T-H OR M-S-H-K-L

(1 mark) (1 mark) (1 mark) (1 mark)...