SBI4U- Molecular Genetics



SBI4U- Molecular Genetics

Name: __ANSWER______________________________

Evaluation Allocation

|Achievement Category |Marks Received |Achievement Level |

|Knowledge/Understanding (K/U) | 31 | |

|Thinking & Inquiry (I) | 25 | |

|Communication (C) | 18 | |

|Making Connections (MC) | 11 | |

Part A Multiple Choice (15 marks) (K/U) Using the test page provided, answer each of the following on the SCANTRON CARD provided. Make NO MARKS on this test page.

Part B Modified True/False (10 marks) (K/U) Indicate whether the sentence or statement is true or false. If false, change the italicized term or phrase to make the statement true in the space provided.

1. HIV virus is known to replicate without exonuclease activity. You would expect individual HIV viruses to be different. _________________________________________________________________________

2. DNA in the nucleus of a human eukaryotic cell is found on one circular chromosome. _________46 linear chromosomes_______________________________________________

3. You isolate DNA from a bacterium on the ocean floor of a deep sea vent. You measure the adenine content to be 31%. According to Chargaff’s rules, a correct estimate of the other bases would be: 31% Guanine, 38% Thymine, and 38% Cytosine. ___31% T, 19% G, 19% C___

4. There would not be a tRNA anticodon 3’ AUU 5’ to bond with the codon

5’ UAA 3’ on the mRNA _________________________________________________________

5. The mRNA transcript is likely to be most similar to the coding strand/anti-template found on the DNA. ___________________________________________________________________

6. In a healthy individual, a normal cellular response to a build up of DNA structural damage would be to start replicating the DNA rapidly and continuously.

___________apoptosis or suspend growth or stop replication________________

7. DNA rich in guanine/cytosine content would denature at a higher temperature than sample that is high in adenine/thymine content.

__________________________________________________________________________

8. Ribosomes move along the mRNA, accepting incoming tRNAs in at the P site. __________A site (except first tRNA but ribosome not moving then)_______________________________

1. The linkage connecting a phosphate and a deoxyribose sugar along the rails of the DNA ladder is called a glycosyl bond.

__________phosphodiester linkage________________________________________

9. The control of gene expression in prokaryotes is more clearly understood since structural and regulatory genes are found in clusters called promoters. ________operons_____________________________________________________

Part C Communication (C) (5 marks) Draw a small sketch of a DNA replication fork (with accompanying explanation) showing why DNA synthesis is different on the leading and lagging strands. Please name key components (eg. enzymes, DNA terminology) in your response.

Marks allocated for accuracy of replication fork with appropriate strand directions

Key enzymes demonstrating idea bolded above would be helicase and polymerase

Need to show that b/c replication is 5’-3’ and the DNA strands are antiparallel, as helicase opens up the helix, replication on one strand is continuous, while on the other strand the polymerase works away from the replication fork; therefore as helicase opens up more DNA, need more polymerases to make further Okazaki fragments

Part C Completion (C) (13 marks) Fill in the appropriate term/phrase to complete each of the following statements. All blanks carry a weight of 1 mark, except where indicated in brackets.

13. DNA is a double-stranded linear polymer made up of ___nucleotide_______________, each of which consists of a ____pentose sugar (deoxyribose)___ sugar, a _____phosphate_________________ and one of four __nitrogen_____ bases.

14. The sequence of __nitrogen bases______ in DNA determines the sequence of ___amino acids_____ in the polypeptide for which the gene codes.

15. Post-transcriptional processing occurs in the __nucleus_____ (pick location) of the cell. ___Intons______ are removed by _spliceosomes_____ allowing coding regions (called _exons___) to be fused back together. A ___modified guanine cap_____ is added to the 5’ end and a ___poly A tail__ made of many ___adenine____ nucleotides is added to the 3’ end.

Part D Short Answer (6 marks) (K/U) Answer the following in the space provided.

16. a) Explain what is meant by the “wobble hypothesis” and how this concept is useful in understanding the nature of the genetic code (ie. What codes for what). (2 marks)

1 tRNA can recognize more than 1 codon b/c of irregular base pairing at third position on mRNA; thus this can help explain why there might be more than 1 codon per aa.

b) How can what you mentioned in a) be a benefit for possible faults (mistakes) in other processes. (1 mark)

If a mutation occurs or mistake in transcription leading to an improper triplet, there is the possibility that the same aa might be placed in the polypeptide

16. Recent research has suggested about 80% of human genes are alternatively spliced. Explain what this itatlicized term means, and indicate the significance of this for genetic research. (3 marks)

Alternative splicing refers to the fact that when introns are cut out, the exons (coding regions) can be rearranged or even omitted in different ways producing different mRNA transcripts, thus affecting the resulting aa sequence produced in the polypeptides

This is significance b/c this helps to explain how one gene could correspond to more than 1 polypeptide refuting the previous thought of a 1:1 relationship

Part E Thinking and Inquiry (25 marks) (I) Answer the following in the space provided.

For Question 17 and 18, choose ONLY ONE of the following.

17. a) Imagine that you are a geneticist and are investigating a species of

bacteriophage about which little is known.

How might a technique from Hershey and Chase's experiment be used to

discover whether the nucleic acid in the bacteriophage is DNA or RNA? (2 marks)

Use radioactive thymine and uracil to distinguish between DNA or RNA. Label one test tube where you use only radioactive thymine; the other test tube is radioactive uracil

By seeing which one contains radioactivity inside the bacterium, can tell if it is DNA or RNA

b) In Griffith and Avery’s experiment with R and S pneumococcus was used. As an extension of their experiment, heat-treated S cells were combined with a polysaccharide digesting enzyme and with a protease (enzyme that breaks down protein). This solution was then combined with R cells and injected into a mouse. Predict whether or not the mouse will survive or not. As always, whether you choose living or dead, the mice were sacrificed and an autopsy was performed. Give a supportive reason of what one would find. (2 marks)

Mouse would die b/c the DNA is still intact; from these experiments, one learns that DNA is the transforming principle so as long as it is still intact it has the ability to transform R cells into S cells

OR

18. Suppose mammalian cells are cultured in a medium containing radioactive thymine. They grow and divide many times, until eventually every chromosome contains radioactive thymine. The cells are then REMOVED and allowed to replicate several more times in a culture medium containing normal thymine. Daughter chromosomes were tested with each successive generation to determine whether they carry the radioactive thymine.

a) Predict the radioactive status of the daughter chromosomes after one, two, and three rounds of division in the normal medium. (2 marks)

After one round – expect 2 hybrid DNA molecules each with a normal strand and a radioactive one

Two rounds – expect 2 hybrids, 2 normal DNAs

Three rounds – expect 2 hybrids, 6 normal DNAs

b) Explain how your predictions are consistent with what we know about the way DNA replicates itself. (2 marks)

DNA replicates semi-conservatively – means that there will always be an old strand and new DNA strand in the daughter DNAs produced

19.

a) A mutation occurs such that no functional enzyme can be made from the lacZ gene (B-galactosidase). Describe the possible ramifications of this mutation with respect to: a) the ability to turn on or off transcription, and b) the ability to use lactose from the surrounding environment for energy. (2 marks)

Does not affect switch b/c B galactosidase is not connected to the repressor/operator interaction

Bacterium will not be able to use lactose for energy since a functional B-gal enzyme cannot be made to break down the lactose (if present)

b) Two mutations of the lacI (repressor) gene have different effects on the repressor protein and its function in the operon. Mutant A and B represent these different examples; the last bacterium is normal.

|Lactose Availability |Lactose Absent |Lactose Present |

|Mutant A |Transcription occurs |Transcription occurs |

|Mutant B |No transcription |No transcription |

|Normal |No transcription |Transcription |

* Transcription infers that beta-galactosidase and permease are made

Provide a possible explanation for the different effects of these two mutations on the repressor protein molecule (lacI). (2 marks)

Mutant A – makes a repressor that changes shape and cannot bind to operator

Mutant B- makes a repressor that changes shape such that lactose can’t remove it from the operator

20. The following represents the DNA sequence close to where the beta-globulin gene (polypeptide from hemoglobin) begins on chromosome #11 of a normal patient and Maria’s cousin. The actual gene region covers over 1700 base pairs.

Normal_Patient_1 GAGCCAGGGCTGGGCATAAAAGTCAGGGCAGAGCCATCTATTGCTTACATTTGCTTCTGA

Maria’s Cousin GAGCCAGGGCTGGGCATAAAAGTCAGGGCAGAGCCATCTATTGCTTACATTTGCTTCTGA

************************************************************

Normal_Patient_1 CACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGAGGAG

Maria’s Cousin CACAACTGTGTTCACTAGCAACCTCAAACAGACACCATGGTGCACCTGACTCCTGTGGAG

******************************************************* ****

a) What is the name of the strand of DNA shown? What direction is it read (left to right)? (1 mark)

Coding strand/antitemplate 5’-3’ is universal way that geneticists store sequences

b) Use this information to find the start of the coding region of this gene. Explain your selection. (2 marks)

ATG is the start codon, representing the beginning of the first exon

c) The beginning of the coding region is not where transcription would begin. Highlight a sequence that you would hypothesize would be close to where transcription would start. Provide a rationale for your answer. (2 marks)

Any sequence resembling a TATA sequence- these sequences make it easier for opening the DNA double helix b/c less H bonds between A and T

d) Beginning from what you defined as the start of the coding region in b), write this sequence as mRNA. Show appropriate strand directions. (2 marks)

5’ AUG GUG CAC CUG ACU CCU GAG GAG 3’

e) Translate the sequence from the normal patient into amino acids. (2 marks)

Met- val – his – leu – thr – pro – glu - glu

f) How does this differ from the Maria’s cousin? (1 mark)

Maria’s cousin has valine in the place of glutamic acid

g) What type of mutation would you use to describe this situation? Circle an appropriate word from each column. (2 marks)

Column#1 – Nature of Mutation Column#2- Impact of Mutation

Substitution Silent

Insertion Frameshift

Deletion Nonsense

Missense

h) Locate the base pair underneath the star. Imagine if this base was deleted via a mutation in the family line from Normal Patient #1. What would the severity of this mutation be like? Provide a rationale for your answer, indicating the impact on the polypeptide and person. (3 marks)

The insertion would not only change the order of the triplets downstream of the insertion but this would create a premature stop codon in the next triplet. This would terminate translation earlier resulting in a severely malfunctional polypeptide made. This would dramatically influence the health of the person – in this case affect the ability to transport oxygen

i) Select a possible codon from the mRNA above. Show a possible tRNA anticodon (with attached amino acid) that would match up with a codon on the ribosome. (2 marks)

As shown in class on board

Part F Making Connections (11 marks) Answer the questions on another sheet of paper. NOTE: For question 23 and 24, you have a choice to do ONE of two questions.

21. In Europeans, it has been found that Type I diabetes is more likely to occur in people who have only low number repeats ( ................
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