Please explain steps! Name Section Directions: Read the following. Highlight the
ID: 196035 • Letter: P
Question
Please explain steps!
Name Section Directions: Read the following. Highlight the restriction sites for each of the examples given. (If there are three nucleotides on either side of the dash it is a blunt cut. If there are less than three on either side of the dash it is cohesive cut.) Indicate the number of double stranded DNA generated after cutting with the enzymes. Apal: 5' GTT-AAC 3. 5' GGATGTTAACAATCTCTACGGGTTAACACCCTTGGGTTAACATCCGCGG 3' 3' CCTACAATTGTTAGAGATGCCCAATTGTGGGAACCCAATTGTAGGCGCC 3' Number of pieces of dsDNA 2. SspI: 5 AAT ATT 3' 5' GGATAATATTGTTAACAATCTCTACGGGTTAACACCCTTGGAATATTTTAA 3 CCTATTATAACAATTGTTAGAGATGCCCAATTGTGGGAACCTTATAAAATT 3" Number of pieces of dsDNA 3. PstI: 5' CTGCA G 3 5 ACGCTGCAGACGTATTATTATCCGCCGCTGCAGCCGTCATCA 3' 3' TGCGACGTCTGCATAATAATAGGCGGCGACGTOGGCAGTAGT 5' Number of pieces of dsDNA 4. Hindl I: 5' GTC GAC 3 5' ACGACGTAGTCGACTTATTATGTCGACCCGCCGCGTGTCGACCATCA 3 3 TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5 Number of pieces of dsDNA 5. EcoRI: 5 G AATTC 3' 5' ACGACCTATTAGAATTCTTATCCGCCGCCGGAATTCTCATCA 3 3' TGCTGCATAATCTTAAGAATAGGCGGCGGCCTTAAGAGTAGT 5' Number of pieces of dsDNA 5' ACGCCGGCCGTATTATCCGGATCCGCCGCCGGCTGTCCCGGATCA 3' 3' TGCGGCCGGCATAATAGGCCTAGGCGGCGGCCGACAGGGCCTAGT 5' Number of pieces of dsDNA 7. BamT: 5' CCTAG G3 5 ACGCCTAGGACGTATTATCCTAGGTATCCGCCGCCCTCATCA 3 3 TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5' Number of pieces of dsDNA 8. Hpal: 5' GTT AAC 3' and SspI: 5' AAT ATT 3 5' AGTTAACCGACAATATTGTATTATATCCGCCGCCGTCGTTAACATCA 3' 3' TCAATTGGCTGTTATAACATAATATAGGCGGCGGCAGCAATTCTAGT 5' Number of pieces of dsDNA 9. HindII: 5* GTC GAC 3' and HaeIIT: 5' CC GG 3' 5' ACGCTCGACACGTATTATTAGTCGACTCCGCCGCCGCCGGTCATCA 3 3 TGCCAGCTGTGCATAATAATCAGCTGAGGCGGCGGCGGCCAGTAGT 5 Nimber of pieces of dsDNA .HindI: 5 GTC GAC 3', HaeIII: 5 CC -G3 and BamI: 5 CCTAG -G3Explanation / Answer
AS PER CHEGG POLICY, ANSWERING QUESTIONS 1 TO 4 ALONE:
Our DNA is made up of nucleotides with 4 possible bases {2 Purines labelled Adenine (A) and Guanine (G), & 2 Pyrimidines labelled Cytosine (C) and Thymine (T)}. These 4 nucleotides are arranged sequentially across two strands that come together to form a ladder with the bases as its rungs. One strand goes from 5’ to 3’ direction and the other strand is anti-parallel to this going from the 3’ to 5’ direction.
Thus there end up being sequences in the middle that are palindromic, that is, read in the same sequence of bases on both strands when reading from the 5’ to 3’ direction. Example:
5’ – AGTTAACA – 3’
3’ – TCAATTGT – 5’
which is read as 5’ – GTTAAC – 3’ on both strands.
Certain enzymes called restriction enzymes were made by bacteria to counter genetic material insertions by viruses. These restriction enzymes treated these palindromic regions in the DNA as restriction sites and were able to cut the DNA there, leading to either a blunt cut:
5’ – AGTT ||| AACA – 3’
3’ – TCAA ||| TTGT – 5’
or a staggered/cohesive cut:
5’ – AGT ||| TAACA – 3’
3’ – TCAAT ||| TGT – 5’
If a linear stretch of double-stranded DNA (dsDNA) is cut at one restriction site, it will produce two end products. Similarly, ‘n’ number of such cuts will produce ‘n+1’ number of final fragments.
Proceeding with this logic we see that:
ANSWER 1.
Hpa1 looks for and bluntly cuts the sequence 5' -GTTAAC-3'. However we do not have two 5' to 3' strands in the first place as the lower strand is 3' to 3'. Hence no cuts will be produced and number of dsDNA = 1
ANSWER 2.
Ssp1 looks for and bluntly cuts the sequence 5' -AATATT-3'. However we do not have two 5' to 3' strands in the first place as the lower strand is 3' to 3'. Hence no cuts will be produced and number of dsDNA = 1
ANSWER 3.
Pst1 looks for and cohseively cuts the sequence 5' -CTGCA |||G-3'. In the given dsDNA this palindromic sequence occurs 2 times. Hence 2 cuts will be produced and number of dsDNA = 3
ANSWER 4.
Hind2 looks for and bluntly cuts the sequence 5' -GTC ||| GAC-3'. In the given dsDNA this palindromic sequence occurs 3 times. Hence 3 cuts will be produced and number of dsDNA = 4