BIOL 302 EXAM 6 Form

26 Jun BIOL 302 EXAM 6 Form

Question
Summer 2015 BIOL 302 EXAM 6 Form 0

Biology 302 – Molecular and General Genetics

Summer 2013 Exam 6

FORM 0

__________________________________________

Name, please print legibly

INSTRUCTIONS:

Fill in your name in the above blank.
On the Scantron sheet, fill in your name and student ID number.
On the Scantron sheet, fill in the exam form number shown above in box “K.”
Read the statements below, sign your name, and then tear off this cover sheet.
Turn in this cover sheet and Scantron sheet. You can keep your question sheets.
The answer Key will be posted in the glass case outside the elevators in the Biology building on the first floor.
Keep your answer sheet covered.
Note: The use of a cell phone during an exam in any manner is expressly prohibited and will result in a zero on the exam and a report filed to the UMBC ACC with no warning.

I pledge that I will neither seek help nor offer help to any other student while taking this exam. I will make every effort to assure that no other student can see my answers. I also state that I have no aids, such as crib sheets and that I know of none used by other students.

I understand that any confirmed case of cheating on an this exam will be dealt with through UMBC’s Academic Conduct Committee as described in the Policy for Academic Misconduct in Undergraduate Courses. Any and all instances of academic misconduct will be reported to the committee. Penalties include, but are not limited to, a grade of “F” in the class.

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Signature and Date

Summer 2013 BIOL 302 EXAM 6 Form 0

1. The DNA sequence below contains a palindromic sequence consisting of:

A G G A G C T T C T

of 4 bases

a. One palindromeT C C T C G A A G A

b. Two palindromes of 4 bases each

c. One palindrome of 6 bases

d. One palindrome of 4 bases and one palindrome of 6 bases

e. There is no palindromic sequence longer than three bases

Use the information below to answer the next two questions.

To the right is a map of the vector pUC19. Imagine you were to clone

(insert) a copy of the human insulin gene into this plasmid obtained by

performing RT-PCR on an mRNA sample taken from human

pancreatic beta cells. You then transform a strain of E. coli that is

ampS and lac- with the construct you have made.

MCS

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In this vector, what was the MCS used for?
Selection of transformants
Screening for conjugates
Insertion of DNA
Transformation of host cells
Production of pod people
When you succeed in putting the insulin gene into the E. coli cells, the resulting cells will be:
AmpS, lac-
AmpS, lac+
AmpR, lac-
AmpR, lac+
Why would E. coli need insulin?
A cDNA library:
Does not contain regulatory sequences
Requires the use of reverse transcriptase
Contains intron sequences
More than one of the above
Can be replaced with iTunes for many people

Summer 2013 BIOL 302 EXAM 6 Form 0

5. Which of the following will be incorporated into the DNA produced by PCR?

a. dNTPs, but not primers will be incorporated into the new DNA

b. Primers, but not dNTPs will be incorporated into the new DNA

c. Both dNTPs and primers will be incorporated into the new DNA

d. Neither dNTPs nor primers will be incorporated into the new DNA

e. I can still see the afterimage from the fireworks, why are you bothering me?

5’

A

C

B

D

6. Above is a melted fragment of DNA with a target you wish to amplify by PCR. Where would you design the primers to anneal?

a. A and B

b. B and C

c. B and D

d. A and D

e. None of the above

7. If you want to put DNA in a bacterium, and a plasmid won’t hold enough, you could consider using a:

a. Cosmid

b. BAC

c. YAC

d. Two of the above

e. All of the above

8. Chain terminators are:

a. Altered nucleotides

b. Used for PCR

c. Used for southern blotting

d. More than one of the above

9. Which type of sequencing includes the use of chain-terminators?

a. Sanger sequencing

b. Pyro-sequencing

c. Both Sanger sequencing and pyro-sequencing

d. Neither sanger sequencing nor pyro-sequencing

e. What does sequencing have to do with DNA?

Summer 2013 BIOL 302 EXAM 6 Form 0

In a southern blot, the probing is done with:
Antibodies
A digested vector
Single stranded DNA or RNA
Restriction enzymes
None of the above
A two channel gene array is performed comparing human male DNA, which is labeled green, and human female DNA, which is labeled red. An active Y-linked gene would be expected to fluoresce:
Red
Green
Yellow
Black

12. Three individuals underwent RFLP analysis to test for theS allele by digestion of some of their DNA with MstII, which cuts the gene approximate 200 bp downstream from the 5’ end. To the right is a southern blot showing where in the gel the DNA fragments of the three individuals is located. If a single probe that bound the first 20 bp of the gene (5’ end) was used, which bands would you see in the heterozygote?

a. The 0.2 kb only

b. The 1.1 kb only

A B C

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1.3 kb

1.1 kb

0.2 kb

c. The 0.2 kb and 1.1 kb bands

d. The 0.2 kb and 1.3 kb bands

e. The 1.1 kb and 1.3 kb bands

13. Exposing the larvae of normal fruit flies to sodium metaborate results in the development of greatly reduced eye size in a manner like that caused by the autosomal recessive mutation eyeless. This is a typical example of:

a. Parental Determination

b. Phenocopy

c. Genomic Imprinting

d. Simple Mendelian Inheritance

e. Anticipation

Summer 2013 BIOL 302 EXAM 6 Form 0

Occasionally, a random student is kidnapped from the commons

and a sample of the DNA is taken for testing before they are

released. When a particular 10 kb stretch of DNA from

four students is digested with BamHI and run on a gel.

10 kb

The gel is shown to the right.

9 kb

Bruce

Nicko

Steve

Dave

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14. Which student has the fewest BamHI sites? a. Bruce

5 kb

3 kb

2 kb

1 kb

b. Nicko

c. Steve

d. Dave

e. More than one of the above

15. Which student has the most BamHI sites?

a. Bruce

b. Nicko

c. Steve

d. Dave

e. More than one of the above

16. Spinocerebellar ataxia (SCA-1) shows differences in gene action dependent upon which parent the allele was inherited from. This is called:

a. Genomic Imprinting

b. Haplotyping

c. Anticipation

d. Parental Determination

e. Phenocopy

17. What makes Huntington’s a poor candidate for gene therapy?

a. The disease results from a single gene

b. It’s relatively easy to remove and replace the target cells

c. The mutation is recessive

d. The mutation is dominant

e. Two of the above are correct

18. An adult individual homozygous for HPFH will produce which type of hemoglobin?

a. HbA hemoglobin

b. HbS hemoglobin

c. HbGower hemoglobin

d. HbF hemoglobin

e. More than one of the above

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Summer 2013 BIOL 302 EXAM 6 Form 0

Below are maps of the wild type and mutant versions of the CFTR gene, which is responsible for cystic fibrosis, along with the binding location of a probe that can be used to investigate it. The cut sites by a restriction enzyme (labeled R) are also indicated. Also shown is a Southern Blot of parents and a fetus that was tested.

Father Fetus Mother

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WT CFTR

R

R

CFTR probe R

RR

Mutant CFTR

700 bp

300 bp

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What size band(s) would be visible in samples from individuals who are heterozygous for cystic fibrosis?
300 bp
700 bp
1000 bp and 700 bp
1000 bp and 300 bp
1000 bp, 700 bp, and 300 bp
Cystic fibrosis is a recessive disease. If we ignore the possibility of reduced penetrance and other such phenomena, will the fetus have the disease when born?
No
There is a 25% chance
There is a 50% chance
Yes
It is impossible to tell
What is the most likely nature of the CFTR probe when it is being used?
Double stranded DNA
Single stranded DNA
Double stranded RNA
Single stranded RNA
Surgical stainless steel

Summer 2013 BIOL 302 EXAM 6 Form 0

22. Huntington’s is a tri-base duplication caused disease like Fragile-X or SCA-1. With Huntington’s, the length of the duplication affects the disease such that a longer duplication results in:

a. Less severe symptoms

b. More severe sever symptoms

c. No disease occurrence

d. A later age of onset

e. An earlier age on onset

23. Like Huntington’s disease, the severity and age of onset of Myotonic Dystrophy, an autosomal dominant multi-system disease, results from a tribase tandem repeat. Children of individuals with the condition typically inherit longer repeats, though much more commonly and severely when inherited from their mother than their father. Which of the following phenomena does Myotonic Dystrophy demonstrate?

a. Gene amplification

b. Genomic imprinting

c. Phenocopy

d. Histone modification

e. None of the above

24. Which of the following pre-natal tests is performed by taking a sample of placental tissue, often by using a flexible catheter rather than a needle?

a. Amniocentesis

b. Fetal ultrasound

c. Free fetal DNA testing

d. Chorion villus sampling

e. Cell culturing

25. Prenatal testing can be done to identify a developing fetus for an identified gene. Which currently approved procedure can be done at the earliest point in the pregnancy?

a. Chorion villus sampling

b. Transgenic analysis

c. Amniocentesis

d. Tricorder scanning

e. More than one of the above

Summer 2013 BIOL 302 EXAM 6 Form 0

26. Which of the following sequencing methodologies use chain terminators?

a. Sanger sequencing

b. Ion torrent sequencing

c. Pyro sequencing

d. Illumina sequencing

e. More than one of the above use chain terminators

27. Early attempts by NIH to treat ADA with gene therapy were ultimately unsuccessful, while later attempts by the San Raffaele Telethon Institute for Gene Therapy in Italy were successful starting with Salsabil in 2001. Why did the Italian group have a more successful outcome?

a. Because ADA is a dominant disease

b. The Italian group used older patients who had been receiving drug therapy prior to the gene therapy

c. The NIH group used microinjection

d. They went after a different disease

e. The Italian group treat bone marrow cells