Lecture 9/14 - E2

Question 1

Common Temperature for Denaturation

Answer 1

94-98C

Question 2

Common Temperature for Annealing

Answer 2

37-60C

Question 3

Common Temperature for Extension

Answer 3

68-72C

Question 4

How do you calculate the total number of amplified products after each cycle?

Answer 4

2^n

n = the number of cycles

Question 5

How do you calculate the total number of ds DNA products when you begin with a single ds DNA molecule?

Answer 5

A(2^n - 2n)
A = the number of starting ds DNA template molecules
n = number of cycles

Question 6

Advantages of PCR over traditional approaches?

Answer 6

Faster and less expensive, can be optimized to detect low abundance mRNA splice variants, and it can be used for other purposes

Question 7

Limitations of PCR technology

Answer 7

Requires some precise knowledge of target DNA sequence in order to create the primers, size restrictions on amplicons, mutations may be introduced into the DNA by the polymerase during amplification

Question 8

Starting with genomic dsDNA isolated from a single somatic cell, how many fully dsDNA products are present after 4 cycles of PCR?
A) 0
B) 2
C) 4
D) 6
E) Cannot determine based on the information providded

Answer 8

E

Need to know how many molecules are present at the start of the reaction which depends upon the stage of the cell cycle

Question 9

Starting with genomic dsDNA from a single somatic cell at the G1 stage of the cell cycle, how many fully dsDNA products are present after 4 cycles of PCR?
A) 0
B) 4
C) 8
D) 16
E) 32

Answer 9

There are 2 tmplates because of mom and dad copies
A(2^n - 2n)
2(2^4 - 2(4)) = 2(16-8) = 16

Question 10

Haplosufficient

Answer 10

Where one copy of the WT allele is needed for the full WT functionality (Aa)

Question 11

Haploinsufficient

Answer 11

Where one copy of the fully functioning allele shows a mutant phenotype (Aa = mutant)

Question 12

Null/Amorphic Mutations

Answer 12

Complete loss of function mutation that produces no functional gene product
Completely inactivate gene function

Question 13

Weak/Leaky/Hypomorphic Mutations

Answer 13

Mutations that result in partial loss / reduced function

- Depends on the level of activity of the leaky mutant allele

Question 14

Possible defects associated with Null Alleles

Answer 14

Gene absent or inactivated
No transcription
Abberrant proessing of the mRNA transcript
No translation
Non-functional protein product

Question 15

Possible defects associated with Hypomorphic Alleles

Answer 15

No gene
Reduced rate of transcription
Abberant processing of the mRNA transcript
Reduced translation
Less-functional protein product

Question 16

Which types of DNA change in the yeast ACT1 gene is most likely to result in a hypomorphic mutation?
A) Deletion of all ACT1 protein coding sequence
B) Removal of all teanscriptional regulatory sequences
C) A codon change resulting in an amino acid substitution
D) ACT1 gene duplication
E) All equally likely to cause hypomorphic mutation

Answer 16

C) A codon change resulting in an amino acid substitution

Because it is the least mild change

Question 17

Hypermorphic Allele

Answer 17

Increased activity (more is not always better)

Question 18

Neomorphic Mutation

Answer 18

Mutation that results in an increased function

Question 19

Dominant Negative Mutation

Answer 19

Competes with the WT protein and blocks natural function
Produces poisonous gene products
“Spoiler Effect” - Homo dominant can be lethal and too toxic

Question 20

Which type of mutation is most likely to result in a haploinsuficient genotype?
A) A null allele
B) A weak hypomorphic allele
C) A wildtype allele
D) A gene duplication

Answer 20

A) A null allele