Sbi4U – molecular genetics unit test

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Date:_________________________________

Name:______________________________________

SBI4U – MOLECULAR GENETICS UNIT TEST


Knowledge/Understanding

Communication

Thinking and Inquiry

Application

/19

/11

/14

/17



PART A: MULTIPLE CHOICE [10marks, K/U]

DF Answer all multiple choice questions on the scantron card provided. There is only ONE answer per question; select the best response for each question. Each question is worth one mark.
1) In eukaryotes, replication starts more or less simultaneously at multiple points of origin along a chromosome. What pattern of radioactivity would be found after briefly exposing to radioactive adenine a cell that is just beginning to replicate, and then quickly replacing the radioactive with nonradioactive adenine?

a.

many short sections of radioactive DNA

b.

one radioactive section of DNA starting at one particular end of the chromosomes

c.

many short radioactive strands of DNA in the leading strands only


d.

many short radioactive sections of DNA in the lagging strands only

e.

no radioactive sections in the DNA

2) A strand of DNA is composed of: A 30%, T 15%, G35%, and C 20%. What is the composition of the complimentary mRNA strand?



a.

A 30%, T 15%, G35%, and C 20%

b.

A 30%, U 15%, G35%, and C 20%.

c.

A 15%, T 30%, G20%, and C 35%.

d.

A 15%, U 30%, G20%, and C 35%.

e.

A 20%, T 35%, G15%, and C 30%.

3) Restriction endonuclease digestion of a DNA sequence yielded fragments of the following sizes:


1. 5.2 kb

2. 0.8 kb

3. 1.2 kb

4. 3.8 kb

5. 3.1 kb
After gel electrophoresis, what would be the order in which these fragments would be found—the last fragment listed being furthest from the negative pole.

a.


1,2,5,4,3

d.

3,2,4,5,1

b.

2,3,5,4,1

e.

1,4,5,3,2

c.

1,3,5,2,4




4) Line A below shows the peptide synthesized after transcription and translation of a piece of DNA. Line B shows the peptide synthesized after a mutation in this piece of DNA.


A. proline - arginine - aspartic acid - lysine - glycine

B. proline - glutamine - aspartic acid - lysine – glycine


The type of change in the DNA most likely to be responsible for the difference between the peptides is:


a.

A frame-shift mutation

d.

A base deletion

b.

A base insertion

e.

A silent mutation

c.

A base substitution




5) Which scientist(s)’ X-ray crystallography photograph was used to hypothesize the shape of DNA?

a.


Avery and McLeod

b.

Watson and Crick

c.

Meselson and Stahl

d.

Rosalind Franklin

e.

Frederick Griffith

6) The following are all steps in the production of a bacterium having recombinant DNA, which includes an inserted nonbacterial gene. They are in random order.



1. Gel electrophoresis of plasmid DNA from bacteria colonies that survived

2. Sticky ends are allowed to pair up

3. A restriction endonuclease is used to remove the gene to be inserted from its source and also to cut open a plasmid that includes a gene for antibiotic resistance

4. All bacterial colonies are treated with antibiotic

5. Treatment with ligase

6. Transformation (formation and growth of a new recombinant species)

The correct order of these steps is

A. 2,5,1,3,4,6 B. 4,3,6,2,5,1 C. 5,1,2,3,4,6



D. 3,2,5,6,4,1 E. 6,3,5,2,4,1

7) There are differences in the amino acid sequence of rabbit and frog hemoglobin polypeptides. If mRNA for rabbit hemoglobin is extracted from rabbit red blood cells, and is then placed in frog eggs, the cells will produce rabbit hemoglobin polypeptides. This shows that


a.

rabbit hemoglobin mRNA is the same as frog hemoglobin mRNA

b.

the genetic code and the machinery of translation are substantially the same in widely-different organisms

c.

the gene for haemoglobin is identical in all organisms

d.

the DNA for rabbit hemoglobin is reverse transcribed into DNA in the frog eggs

e.

frog ribosomes are incapable of binding to mammalian RNA

8) Post-transcriptional modifications include:

I. 5’ cap

II. Cleavage

III. Splicing of introns

IV. 3Poly A tail

V. Glycosylation




a.

I, II, V

b.

I, II, IV

c.

I, III, IV

d.

I, III, V

e.

II, IV, V

9) DNA acts as a template for transcription. Which of the following statements regarding the DNA of a gene being expressed is true?


a.

After unwinding, both of the DNA strands act as templates.

b.

After unwinding, only one of the DNA strands acts as a template.

c.

The two strands only act as a template when paired.

d.

In prokaryotes, the binding of RNA polymerase to unwound DNA occurs randomly on either of the two strands.

e.

The strand with the higher cytosine-guanine content acts as the template.

10) Karyotypes can be used to:

I. Map an individual’s chromosomes

II. Determine genetic mutations

III. Determine chromosomal abnormalities

IV. Directly treat abnormalities




a.

I, II, IV

b.

I, III

c.

I, II

d.

II, III

e.

I, II, III, IV

PART B: SHORT ANSWER [36 marks]

Answer the following questions on the papers provided. The number of marks is indicated in each bracket. You must show all work to get full marks for each question.



  1. Explain the effect(s) the following scenarios would have on DNA replication or translation. For each scenario, state whether DNA replication or translation would be able to proceed and explain your reasoning. [3 x 3 marks =9, T/I]




  1. Low amount of 7- methyl guanosine in the nucleus

A low amount of 7-methyl guanosine would result in a lower amount of mRNA receiving a 5’ cap (1 mark). This would mean that fewer mRNA would be able to leave the nucleus to participate in translation (1 mark). Protein synthesis would still occur at a much slower rate (1 mark).



  1. low amount of DNA polymerase I

DNA replication would be able to occur (1 mark), however, it would be at a much slower rate (1 mark). This is because DNA polymerase I is still present and functional which would allow DNA replication to continue (1 mark)


  1. lack of helicase

Without helicase, DNA replication could not occur (1 mark). DNA polymerase III as well as other enzymes needed for DNA replication wouldn’t be able to bind (1 mark) since the strands of DNA would still be intact with hydrogen bonds and not “unzipped”/separated (1 mark).


  1. Cancer cells have a large concentration of telomerase.

    1. Explain how this contributes to them being considered ‘immortal’. Be sure to explain the function

of telomerase in your answer. [4 marks, T/I]

Telomerase is a protein that adds telomeres to the ends of DNA strands (1 mark). As we age, the presence of telomerase diminishes and our telomeres, which are crucial for ensuring the entire DNA strand is replicated, shorten (1 mark). Cell death (or old age) occurs when there are not telomeres remaining and parts of our DNA aren’t replicated (1 mark). Since cancer cells have an abundance of telomerase, they can replicate without losing any genetic material. These cells would therefore not age and considered “immortal” (1 mark).

b. How could scientists use this knowledge to benefit humans? [2 marks, A]


Various answers accepted:


  • Could use this knowledge to find treatments for cancer (ex. Inhibiting telomerase to stop tumour growth)

  • Could find ways to elongate telomeres in humans. Therefore, longer cell life.

  • Could find ways to synthesize telomerase to lengthen cell life



  1. Use a graphic organizer of your choice to outline 4 differences between the leading and lagging strand of DNA replication [4 marks, C]




Leading Strand

Lagging Strand

  • Formed as a continuous strand




  • Template strand is oriented 3’  5’

  • Has one RNA primer at the start of the replication fork

  • DNA ligase used briefly to ligate primer (after it’s replaced with DNA) to the synthesized strand

  • Synthesized quickly (relative to the lagging strand) since it’s continuous

  • Formed in small fragments called Okazaki fragments

  • Template strand is oriented 5’  3’

  • Requires several RNA primers




  • DNA ligase used frequently to ligate Okazaki fragments as well as primers




  • Sythesized slowly (relative to leading strand) since it Okazaki fragments need to be ligated together

*Any four differences is fine ( 0.5 marks for each x 8) = 4 marks


  1. Orca is a new type of E. Coli bacterium. Suppa is a nutrient that Orca needs to survive. Suppa can be synthesized by four (4) enzymes (SupA, SupB, SubC, and SupD) in its cell or it can acquire it from its environment. It only synthesizes it’s own Suppa when it’s really desperate, because there’s none around! Draw and describe Orca’s operon controlling its synthesis of Suppa. [7 marks, C]



5’ (1/2 mark) (1/2 mark) (1/2 mark) 3’




Promotor

Operator




SupA

SupB

SupC

SupD






RNA polymerase (1/2 mark)


(1 mark)
Inactive Active

Repressor Repressor
When there’s an excess of Suppa, the Suppa will bind to the repressor (activating it) (1 mark) and it will bind to the operator, blocking transcription of SupA, SupB, SupC, and SupD (1 mark).

When there is a low amount of Suppa, the inactive repressor will remain inactive (1 mark), because there isn’t enough Suppa to bind to it. This will result in the repressor being unable to bind to the operator and the transcription of SupA, SupB, SupC, and SupD (1 mark).



  1. Thalassemia is a genetic disorder caused by a nonsense mutation, resulting in individuals who cannot produce normal hemoglobin. Hemoglobin is a protein found in red blood cells that binds to oxygen. Use your knowledge of mutations and protein structure to explain why individuals with Thalassemia need blood transfusions to live “normally.” (Be sure to explain what a nonsense mutation is) [4 marks, A]

A nonsense mutation occurs when a nucleotide is substituted resulting in a STOP codon (1 mark). A STOP codon would truncate the protein, resulting in an incomplete protein (1 mark). As a result, this protein that is supposed to be hemoglobin would not have the proper shape to carry on its role of binding to oxygen (1 mark). Individuals with Thalassemia would need blood transfusions to get a source of hemoglobin that can bind to oxygen or else they would suffer from severe oxygen deficiency (1 mark).


  1. PCR has revolutionalized biotechnology. Discuss three advantages and three disadvantages of PCR usage. [6 marks, K/U]

Any three for advantages and disadvantages is fine.

Advantages:

  1. It requires only a small DNA to work

  2. It provides relatively quick results

  3. Able to amplify DNA that is degraded or embedded in a medium

  4. Its simplicity enables PCR to be used in sequencing, cloning, DNA typing, etc.

Disadvantages:

  1. High sensitivity of technique means that it’s essential to eliminate sources of contamination during preparation.

  2. Amplification becomes problematic for large sequences
  3. Requires prior knowledge in order to construct specific primers for desired sequence


  4. DNA polymerases such as Taq lack 3’ to 5’ endonuclease activity; it lacks the ability to correct mis-incorporated nucleotides


PART C: PROBLEM SOLVING [15 marks]

Answer the following problems on the paper provided. You must show all your work for full marks.




  1. Hunting is banned in all national parks but not outside of them. During bear hunting season, conservation officers conduct routine checks. One day, the OPP receive a report from campers that they heard shots fired inside Algonquin National Park. The OPP transmits this information to the Natural Resources office. Based on the campers’ tip, check points are set up on the highway through Huntsville. All of the hunters that bring bears past the checkpoint have licenses, but in order to check out the camper’s story several hairs are pulled from each carcass for DNA checks.

The data below shows DNA containing two STR regions for the bear carcasses that were examined on the day after the campers’ complaint, and two known reference samples of bear sub-populations around Huntsville, including inside the park.


POSSIBLE POACHER’S DNA DATA

Yellow: GAC sequence

Red: TTA sequence

Reference #1: Genotype Indigenous to Park

TCAGGGACGACGACGACGACGACGACCCATTATCGGAGTTATTATTAGATCGATCCATTCGGATCGGATAT



7 repeats 3 repeats


Reference #2: Genotype Indigenous to Park

TCAGGGACGACGACGACGACGACGACGACCCATTATCGGAGGTCGAATTTATTATTATTATTATTATTATTA


8 repeats 8 repeats



(2 marks)


Genotype of Bear A shot by Hunter Suzy McDonough (1 mark)

TCAGGGACGACGACGACGACGACGACCCGATATCGCGGTTATTATTATTATTATTATTATTATTACGGATAT



7 repeats 9 repeats


Genotype of Bear B shot by Hunter Mandeep Pinky (1 mark)

TCAGGGACGACGACGACGACCCATTATCGGAGGTCGATTATTATTATTATTATTATTATTATTAGCGCGGATC



5 repeats 9 repeats


Genotype of Bear C shot by Hunter Ajda El-Zabet (1 mark)

TCAGGGACGACGACGACGACGACGACCCGATATCGCGGTTATTATTATTATTATTATTATTATTACGGATAT



7 repeats 9 repeats


Genotype of Bear D shot by Hunter Tomoko Satouchi (1 mark)

TCAGGGACGACGACGACGACGACGACGACCCATTATCGGAGGTCGATTATTATTATTATTATTATTATTA



8 repeats 8 repeats


Genotype of Bear E shot by Hunter Francis Kwapong (1 mark)

TCAGGGACGACGACGACGACGACGACGACCCATTATCGGAGGTCGATTATTATTATTAACGGATATGGCCG


8 repeats 4 repeats

Based on this data, which hunter hunted bears in Algonquin National Park? You must indicate why, for each hunter, they were not guilty, as well as identifying the perpetrator! [7 marks, A]
Since the number of repeats for both the GAC and TTA sequences in Bear D match reference #2, it suggests that Hunter Tomoko Satouchi shot the bear in the national park (1 mark)


  1. The following is a section of DNA removed from a cell nucleus:

5’ ATGAAATAATCAGTTAACAGCAGTTCCGATTTTTATACT 3’ __________________ strand

3’ TACTTTATTAGTCAATTGTCGTCAAGGCTAAAAATATGA 5’___________________ strand


  1. What does the Central Dogma state? [1 mark, K/U]

The Central Dogma states that the flow of genetic information is one way – from DNA to polypeptides. DNA sequences are transcribed into RNA, which are then translated into amino acids of a polypeptide chain.


  1. Label the strands above as the “sense” or “antisense” strand in the spaces provided above [1 mark, K/U]

0.5 mark for getting each right



  1. Using the chart below, transcribe ONLY the gene into mRNA and then translate the gene into its amino acid sequence. [ 4 marks, A]




The beginning of the gene is identified by the promoter, TATAAAA and ends at the terminator, TTATTT. (1 mark)
3’ TACTTTATTAGTCAATTGTCGTCAAGGCTAAAAATATGA 5

DNA: ATG GGA ACT GCT GTT AAC TGA (1 mark)


mRNA: UAC CCU UGA CGA CAA UUG ACU (1 mark)

tRNA: AUG GGA ACU GCU GUU AAC UGA (1 mark)

AA: Met Gly Thr Ala Val Asn Stp (1 mark)


  1. What would happen to the gene if the adenosine mutates to a thymine where the arrow indicates? [1 mark T/I] What type of mutation is this? [1 mark, K/U]


3’ TACTTTATTAGTCAATTGTCGTCAAGGCTAAAAATATGA 5


Nothing would happen to the gene since it lies outside of the coding sequence (1 mark). This is a silent mutation (1 mark).

Date:_________________________________

Name:______________________________________
SBI4U – MOLECULAR GENETICS UNIT TEST


Knowledge/Understanding

Communication

Thinking and Inquiry

Application

/19

/11

/14

/17


PART A: MULTIPLE CHOICE [ 10marks, K/U]

DF Answer all multiple choice questions on the scantron card provided. There is only ONE answer per question; select the best response for each question. Each question is worth one mark.

1) In eukaryotes, replication starts more or less simultaneously at multiple points of origin along a chromosome. What pattern of radioactivity would be found after briefly exposing to radioactive adenine a cell that is just beginning to replicate, and then quickly replacing the radioactive with nonradioactive adenine?

a.

many short sections of radioactive DNA

b.

one radioactive section of DNA starting at one particular end of the chromosomes

c.

many short radioactive strands of DNA in the leading strands only

d.

many short radioactive sections of DNA in the lagging strands only

e.

no radioactive sections in the DNA

2) A strand of DNA is composed of: A 30%, T 15%, G35%, and C 20%. What is the composition of the complimentary mRNA strand?



a.

A 30%, T 15%, G35%, and C 20%

b.

A 30%, U 15%, G35%, and C 20%.

c.

A 15%, T 30%, G20%, and C 35%.

d.

A 15%, U 30%, G20%, and C 35%.


e.

A 20%, T 35%, G15%, and C 30%.

3) Restriction endonuclease digestion of a DNA sequence yielded fragments of the following sizes:


1. 5.2 kb

2. 0.8 kb

3. 1.2 kb

4. 3.8 kb

5. 3.1 kb
After gel electrophoresis, what would be the order in which these fragments would be found—the last fragment listed being furthest from the negative pole.

a.

1,2,5,4,3

d.

3,2,4,5,1

b.

2,3,5,4,1

e.

1,4,5,3,2

c.

1,3,5,2,4




4) Line A below shows the peptide synthesized after transcription and translation of a piece of DNA. Line B shows the peptide synthesized after a mutation in this piece of DNA.


A. proline - arginine - aspartic acid - lysine - glycine

B. proline - glutamine - aspartic acid - lysine – glycine


The type of change in the DNA most likely to be responsible for the difference between the peptides is:

a.


A frame-shift mutation

d.

A base deletion

b.

A base insertion

e.

A silent mutation

c.

A base substitution




5) Which scientist(s)’ X-ray crystallography photograph was used to hypothesize the shape of DNA?



a.

Avery and McLeod

b.

Watson and Crick

c.

Meselson and Stahl

d.

Rosalind Franklin

e.

Frederick Griffith

6) The following are all steps in the production of a bacterium having recombinant DNA, which includes an inserted nonbacterial gene. They are in random order.



1. Gel electrophoresis of plasmid DNA from bacteria colonies that survived

2. Sticky ends are allowed to pair up

3. A restriction endonuclease is used to remove the gene to be inserted from its source and also to cut open a plasmid that includes a gene for antibiotic resistance

4. All bacterial colonies are treated with antibiotic


5. Treatment with ligase

6. Transformation (formation and growth of a new recombinant species)

The correct order of these steps is

A. 2,5,1,3,4,6 B. 4,3,6,2,5,1 C. 5,1,2,3,4,6

D. 3,2,5,6,4,1 E. 6,3,5,2,4,1
7) There are differences in the amino acid sequence of rabbit and frog haemoglobin polypeptides. If mRNA for rabbit haemoglobin is extracted from rabbit red blood cells, and is then placed in frog eggs, the cells will produce rabbit haemoglobin polypeptides. This shows that

a.

rabbit haemoglobin mRNA is the same as frog haemoglobin mRNA

b.

the genetic code and the machinery of translation are substantially the same in widely-different organisms

c.

the gene for haemoglobin is identical in all organisms

d.

the DNA for rabbit hemoglobin is reverse transcribed into DNA in the frog eggs

e.

frog ribosomes are incapable of binding to mammalian RNA

8) Post-transcriptional modifications include:

I. 5’ cap

II. Cleavage

III. Splicing of introns

IV. 3’ Poly A tail

V. Glycosylation


a.


I, II, V

b.

I, II, IV

c.

I, III, IV

d.

I, III, V

e.

II, IV, V

9) DNA acts as a template for transcription. Which of the following statements regarding the DNA of a gene being expressed is true?



a.

After unwinding, both of the DNA strands act as templates.

b.

After unwinding, only one of the DNA strands acts as a template.

c.

The two strands only act as a template when paired.

d.

In prokaryotes, the binding of RNA polymerase to unwound DNA occurs randomly on either of the two strands.

e.

The strand with the higher cytosine-guanine content acts as the template.

10) Karyotypes can be used to:

I. Map an individual’s chromosomes

II. Determine genetic mutations

III. Determine chromosomal abnormalities

IV. Directly treat abnormalities




a.

I, II, IV

b.

I, III


c.

I, II

d.

II, III

e.

I, II, III, IV



PART B: SHORT ANSWER [36 marks]

Answer the following questions on the papers provided. The number of marks is indicated in each bracket. You must show all work to get full marks for each question.




  1. Explain the effect(s) the following scenarios would have on DNA replication or translation. For each scenario, state whether DNA replication or translation would be able to proceed and explain your reasoning. [3 x 3 marks =9, T/I]




  1. Low amount of 7- methyl guanosine in the nucleus



  1. low amount of DNA polymerase I



  1. lack of helicase


  1. Cancer cells have a large concentration of telomerase.

    1. Explain how this contributes to them being considered ‘immortal’. Be sure to explain the function

of telomerase in your answer. [4 marks, T/I]

b. How could scientists use this knowledge to benefit humans? [2 marks, A]




  1. Use a graphic organizer of your choice to outline 4 differences between the leading and lagging strand of DNA replication [4 marks, C]


  1. Orca is a new type of E. Coli bacterium. Suppa is a nutrient that Orca needs to survive. Suppa can be synthesized by four (4) enzymes (SupA, SupB, SubC, and SupD) in its cell or it can acquire it from its environment. It only synthesizes it’s own Suppa when it’s really desperate, because there’s none around! Draw and describe Orca’s operon controlling its synthesis of Suppa. [7 marks, C]





  1. Thalassemia is a genetic disorder caused by a nonsense mutation, resulting in individuals who cannot produce normal hemoglobin. Hemoglobin is a protein found in red blood cells that binds to oxygen. Use your knowledge of mutations and protein structure to explain why individuals with Thalassemia need blood transfusions to live “normally.” (Be sure to explain what a nonsense mutation is) [4 marks, A]



  1. PCR has revolutionalized biotechnology. Discuss three advantages and three disadvantages of PCR usage. [6 marks, K/U]


PART C: PROBLEM SOLVING [15 marks]

Answer the following problems on the paper provided. You must show all your work for full marks.




  1. Hunting is banned in all national parks but not outside of them. During bear hunting season, conservation officers conduct routine checks. One day, the OPP receive a report from campers that they heard shots fired inside Algonquin National Park. The OPP transmits this information to the Natural Resources office. Based on the campers’ tip, check points are set up on the highway through Huntsville. All of the hunters that bring bears past the checkpoint have licenses, but in order to check out the camper’s story several hairs are pulled from each carcass for DNA checks.

The data below shows DNA containing two STR regions for the bear carcasses that were examined on the day after the campers’ complaint, and two known reference samples of bear sub-populations around Huntsville, including inside the park.


POSSIBLE POACHER’S DNA DATA

Reference #1: Genotype Indigenous to Park

TCAGGGACGACGACGACGACGACGACCCATTATCGGAGTTATTATTAGATCGATCCATTCGGATCGGATAT

Reference #2: Genotype Indigenous to Park

TCAGGGACGACGACGACGACGACGACGACCCATTATCGGAGGTCGAATTTATTATTATTATTATTATTATTA




Genotype of Bear A shot by Hunter Suzy McDonough TCAGGGACGACGACGACGACGACGACCCGATATCGCGGTTATTATTATTATTATTATTATTATTACGGATAT
Genotype of Bear B shot by Hunter Mandeep Pinky

TCAGGGACGACGACGACGACCCATTATCGGAGGTCGATTATTATTATTATTATTATTATTATTAGCGCGGATC


Genotype of Bear C shot by Hunter Ajda El-Zabet

TCAGGGACGACGACGACGACGACGACCCGATATCGCGGTTATTATTATTATTATTATTATTATTACGGATAT


Genotype of Bear D shot by Hunter Tomoko Satouchi

TCAGGGACGACGACGACGACGACGACGACCCATTATCGGAGGTCGATTATTATTATTATTATTATTATTA


Genotype of Bear E shot by Hunter Francis Kwapong

TCAGGGACGACGACGACGACGACGACGACCCATTATCGGAGGTCGATTATTATTATTAACGGATATGGCCG


Based on this data, which hunter hunted bears in Algonquin National Park? You must indicate why, for each hunter, they were not guilty, as well as identifying the perpetrator! [7 marks, A]

[7 marks, A]

  1. The following is a section of DNA removed from a cell nucleus:

5’ ATGAAATAATCAGTTAACAGCAGTTCCGATTTTTATACT 3’ __________________ strand



3’ TACTTTATTAGTCAATTGTCGTCAAGGCTAAAAATATGA 5’___________________ strand



  1. What does the Central Dogma state? [1 mark, K/U]


  1. Label the strands above as the “sense” or “antisense” strand in the spaces provided above [1 mark, K/U]



  1. Using the chart below, transcribe ONLY the gene into mRNA and then translate the gene into its amino acid sequence. [ 4 marks, A]





  1. What would happen to the gene if the adenosine mutates to a thymine where the arrow indicates? [1 mark T/I] What type of mutation is this? [1 mark, K/U]



3’ TACTTTATTAGTCAATTGTCGTCAAGGCTAAAAATATGA 5


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