9. The restriction endonuclease AluI leaves blunt ends on DNA fragments after digestion. These blunt ends

a) form hydrogen bonds with the blunt ends of other fragments produced by AluI
b) are fully base paired
c) have only one unpaired base
d) always include the base adenine
e) require a polymerase enzyme to bond them to the other blunt ends
MY ANSWER: d

10. You digest a piece of DNA that is 49000 base pairslong, with a restriction endonuclease having a recognition site four base pairs long. Approxiamately, how many fragments would you expect to be produced?
a) 20
b)98
c)192
d) 256
e)331
MY ANSWER: e
11. Consider the ends of the DNA fragments shown below . They have been produced by digestion of a single sequence of DNA usinga number of restriction endonucleases
1) 5’A 3’
3’TTCGA5’
2) 5’G 3’
3’CAGCT5’
3) 5’AATTC3’
3’ G5
4) 5’TCGAC
3’ G5’3’
5) 5’GGG3’
3’CCC 5’

Which of these ends are capable of annealing and being joined by DNA ligase?
a) 1 and 2
b) 1 and 5
c) 2 and 4
d) 5 and 3
e) 3 and 4
MY ANSWER: d

12. By mistake, DNA fragments were loaded onto a gel using a buffer with a PH so low that all the phosphate groups of the DNA nucleotides were protonated (i.e…had H+ ions added to them.) Subsequent electrophoresis failed to separate the fragments. This was because
a) the voltage difference across the gel was inadequate
b) the double- stranded fragments became single-standed and all migrated very swiftly to the positive pole
c) the fragments joined together into one huge molecule
d) the fragments had very little charge, and thus were not attracted to the positive pole
e) the fragments became circular

MY ANSWER: b

13. Restriction endonuclease digestion of a DNA sequence yielded fragments to the following size
1) 5.2kb
2)0.8kb
3)1.2kb
4)3.8kb
5)3.1kb
After gel electrophoresis, what would be the order in which these fragments would be found- --the last fragment listed being the furthest from the negative pole.
a) 1,2,5,4,3
b) 2,3,4,5,1
c) 1,3,5,2,4
d) 3,2,4,5,1
e) 1,4,5,3,2
MYANSWER: e

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9. The correct answer is b) are fully base paired. Blunt ends are fully base paired with no unpaired bases.

10. The correct answer is e) 331. To calculate the number of fragments, divide the length of the DNA (49000 base pairs) by the length of the recognition site (4 base pairs) and subtract 1. (49000 / 4) - 1 = 12250 - 1 = 12249 fragments.

11. The correct answer is d) 5 and 3. The ends of fragments 5 and 3 are complementary and can anneal and be joined by DNA ligase.

12. The correct answer is b) the double-stranded fragments became single-stranded and all migrated very swiftly to the positive pole. Protonation of the phosphate groups would result in a lower negative charge on the DNA fragments and hinder their separation through gel electrophoresis.

13. The correct answer is e) 1,4,5,3,2. The fragments are listed in order of decreasing size, with the smallest fragment (0.8kb) being closest to the negative pole and the largest fragment (5.2kb) being the furthest from the negative pole.

9. The correct answer is b) are fully base paired. Blunt ends are created when a restriction endonuclease cuts DNA without any overhangs or unpaired bases. This means that the DNA fragments produced will have ends that are fully base paired.

10. To determine the number of fragments produced, divide the length of the DNA by the length of the recognition site and subtract 1. In this case, the DNA is 49000 base pairs long and the recognition site is 4 base pairs long. So, (49000 / 4) - 1 = 12250 - 1 = 12249. Therefore, approximately 12249 fragments would be expected. None of the given options match the correct answer, so I will mark it as incorrect.

11. In order for DNA fragments to anneal and be joined by DNA ligase, the ends need to have complementary base pairs. Looking at the given options, only option d) 5 and 3 has fragments with complementary sequences (5'GGG3' and 3'CCC5'). Therefore, option d) is the correct answer.

12. The correct answer is b) the double-stranded fragments became single-stranded and all migrated very swiftly to the positive pole. The low pH of the buffer would protonate the phosphate groups of the DNA nucleotides, causing the DNA to become single-stranded. Single-stranded DNA migrates faster in gel electrophoresis and would not separate properly.

13. To determine the order of the fragments, arrange them in ascending order based on their sizes. The correct order is 2, 3, 4, 5, 1. Therefore, the correct answer is b).