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BCH5413 ACTUAL EXAM PAPER 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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BCH5413 ACTUAL EXAM PAPER 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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BCH5413 ACTUAL EXAM PAPER 2026
QUESTIONS WITH SOLUTIONS GRADED A+


◉ the melting temperature (Tm) of DNA increases with ----- GC content
and ----- salt concentration. Answer: higher/higher


◉ In the experiment that supports DNA as the genetic material, why
does the mouse live if it is injected with only the dead pathogenic
strain?. Answer: You need live bacteria to take up DNA from the dead
bacteria (and be transformed) for the mouse to die.


◉ The Hi-C conformation capture experiment can be used to investigate
the organization of chromatin within the nucleus of a cell under specific
cellular conditions. In this method, the purpose of the crosslinking step
is to.... Answer: To join two fragments of DNA that are interacting.


◉ In chromosome structure, the core histones function to form....
Answer: The nucleosome


◉ What type of mutation can be determined using the Western Blot
technique?. Answer: If a mutated protein has altered mobility compared
to the normal

, *The protein would be run on an acrylamide gel according to size and
charge, and then detected with an specific antibody. If a protein had
altered mobility it would be detected.


◉ Gel electrophoresis separates nucleic acids on the basis of.... Answer:
Size
*Larger sizes have more difficult traveling through the gel matrix an run
slower than smaller fragments.


◉ Which of the following statements regarding the stringency of a
Southern Blot is correct?. Answer: Decreasing the ionic strength
increases stringency because less ions in solution make it more difficult
for DNA fragments to anneal.


◉ A sample with higher Ct value would have ... starting material than a
sample with lower Ct value.. Answer: Less
*The higher the Ct value, the longer it takes a sample to cross the
threshold, meaning less RNA to start.


◉ In the development of knockout mice, it is critical to genotype the
mice to determine if the gene of interest is actually knocked out. Below
is a figure for the primer locations for triplex PCR. Which of the
following PCR results would represent true knockout mice?. Answer:
One band of relatively lower molecular weight
* The primer pair for the knockout gene produces a smaller fragment
than the endogenous gene. If both genes are replaced with PGK-NEO,

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