Ap Bio Essay Biotech

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DNA QUESTION 1995: L. PETERSON/AP BIOLOGY

The diagram below shows a segment of DNA with a total length of 4,900 base pairs. The arrows indicate reaction sites for two restriction enzymes (enzyme X and enzyme Y).

(A) Explain how the principles of gel electrophoresis allow for the separation of DNA fragments.

(B) Describe the results you would expect from electrophoretic separation of fragments from the following treatments of the DNA segment above. Assume that the digestion occurred under appropriate conditions and went to completion. I. DNA digested with only enzyme X II. DNA digested with only enzyme Y

III. DNA digested with enzyme X and enzyme Y combined IV. Undigested DNA

(C) Explain both of the following:
(1). The mechanism of action of restriction enzymes
(2). The different results you would expect if a mutation occurred at the recognition site for enzyme Y.

PART A. Explain how the principles of gel electrophoresis allow for the separation of DNA fragments. (4 points max.)

Electricity..........Electrical potential (charge, field) moves fragments
Charge..............Negatively charged fragments move toward (+) anode
through gel (-) charge due to phosphate groups
Rate/size..........Smaller fragments move faster (farther)relative to larger
fragments. Describe logarithmic relationship
Calibraton.........DNAs of known molecular weights are used as markers/standards
Resolution........Depends on concentration of gel; is determined by pore size
Apparatus.........DNA is stained for visualization of bands/explains use of
wells, gel material, tracking dye, buffers.

PART B. Describe the results you would expect from electrophoretic separation of fragments from the following treatments of the DNA segment above. (4 points max.)

Treatment I........Describes 400, 1300, 1500, 1700,bp fragments –
or 4 bands _ or correct diagram with explanation.
Treatment II.......Describe 900, 4000 bp fragments – or 2 bands _
or correct diagram with explanation.
Treatment III......Describe 400, 500, 1200, 1300, 1500 bp fragments – or 5 bands – or correct diagram with explanation.
Treatment IV.....Describe 4900 bp fragment – or 1 band – or correct
diagram with explanation.

PART C1.
The mechanism of action of the restriction enzymes (4 point max for both C1 and C2; For a 10 must have at least one point from each section in part C)

Recogntion.................Binding of enzyme to target sequence / specific short bp
sequences of double stranded DNA are targeted / Recognizes specific targets 4-8 bp long / Site may be palindromic.
Cutting.........................Enzyme cuts at every target location / may cut frequently or rarely. Cuts but does not alter the sequence.

Alternate:One point may be given if instead of the above it is clear
that the student says that the enzyme cuts at specific point.

Detail point...................Fragment lengths correspond to leangths between cutting sites / May generate blunt or sticky ends.
Methylation or modification.
Breaks the phosphodiester bond / Describes mechanism in
living systems.
Restriction site may function as a genetic marker.
PART C2.
The different results if mutation occurred at the recognition site for enzyme Y.

Change in II...................Uncut / 1 band (looks like IV)

Change in III..................Like I / 4 bands

Alternate:One point may be given instead of the above, if it is clear
that the student says that Y sequence is no longer recognized and cut.

Detail Point...................Describes that RFLPs (markers) might correlate with
phenotypic variation.
Y site might become an X site.
Deletion/Insertion at Y site – changes fragment size
Silent...
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