Question

Question 2 (50 points) The template strand of a buman DNA and a plasmid are given below. The strands are read from left to right. Both DNAs are to be digested by the four base pair cutter restriction enzymes shown in the table Bacteria DNA: 3-ACAGACCATGGAGGCTCCCATGGTTAGCCTGAAGCTTCGA- 3'--CTTIGTGTOCCACCATGGCAGGTACCGAACCTCAACTTTCCTC.-5, Human DNA: Restriction Enzyme Name Ai Bi Ci Di Ei Sequence and type of eut GATC- 3 CTAG 5 T CGA-3 5- G TAC-3 CAT GS 3'--TC GA-S a Number each DNA from left to right with the first base mumbered1" and determine the b. Which restriction enzymels) can be used in cloning the human DNA into the bacteria? cutting sites of the restriction enzymes on both the human and bacterla DNA. Explain? c. What will be the total number of fragmeats in both the human and bacteria DNA d. What kind of ends will be produced after the cut, a cohesive or blunt end? c. How many nucleotides are in each fragment? (lahel each fragment acconding to number of nucleotides contained in the fragment) Describe what will happen if the the fragmems from both DNAs are mixed with the uncut DNAs of both species and f. restriction enzyme in (b) is used to cut the DNAs and all ugarose gel electrophonesis is conducted on the mixed fragmens Show the bands that E Assuming the fragments of the human DNA is to be cloned into the bacteria, describe h. L Determine if the bacteria cells will express any poten from the human DNA r yes, will forn on the agarose gel. how this could be achieved will be formed? cerc o be uansformed widh he recombinant DNA, which kind of library what kind of amino acid chain will be produced (don't worry about promoters, enhancers and transcription factors in the recombinant plasmid, just consider only the fragments cloned into the plasmid and assume thai any genes can be transcribed from the fragment)

i need from a to i please step by step

Answer 1

a.

Bacterial DNA - Ci, Di, Ei

Ci – 33, 34, 35, 36 nucleotides – 3’ AGC –T5’

Di – 7, 8, 9, 10 nucleotides – 3’ CAT ---G5’

Ei – 37, 38, 39, 40 nucleotides – 3’ TC – GA5’

Human DNA – Di

Di - 14, 15, 16, 17 nucleotides – 3’ CAT ---G5’

b.

Ei is the the restriction enzyme with cutting edge of 3’ TC – GA 5’ and 5’ AG --- CT 3’ can be used for cloning human DNA into bacterial DNA. As the ends of human DNA have CT at 3’ end and TC at the 5’ end, this restriction enzyme can be used for cut at the GA and AG positions of the bacterial DNA to ligate with the human DNA.

c.

There will be four fragments in bacterial DNA after restriction cuts and two fragments in human DNA.

d.

Cohesive ends by Ci and Di and blunt ends by Ei

e.

Number of nucleotides in every fragment of bacterial DNA are

3’ ACAGACCAT----GGAGGCTCCCATGGTTAGCCTGAAGC----TTC---GA 5’ –bacterial DNA

9 nucleotides in first fragment, 26 nucleotides in second fragment, 3 nucleotides in third fragment and 2 nucleotides in last fragment.

Number of nucleotides in every fragment of human DNA are

3’ CTTTGTGTCGCACCAT---GGCAGGTACCGAACCTCAACTTTCCTC 5’

16 nucleotides in first fragment and 27 nucleotides in second fragment

f.

The order of fragments that appear in Agarose gel electrophoresis from bacterial DNA is 40, 26, 9, 3, 2

The order of fragments that appear in Agarose gel electrophoresis from human DNA is 43, 27, 16

bacterial DNA uman DNA 43bp 40bp 27bp 26bp 16bp 9bp 3bp 2bp

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