Based on equal probability of any base occurrence in a genome, what should be the
minimum length of a probe to bind specifically on a single locus on a bacterial
genome of 1 Mbp?
Answers
Problem Set 5 Answers
1.a. The frequency of cutting in a random DNA sequence for a given restriction enzyme is once per every 4n, where n is the number of bases in the restriction enzymes recognition sequence. The “4” derives from the fact that there are four different possible nucleotides that may be inserted at any one position (G, A, T, or C).
Eco RI and Hin dIII have a six-base recognition site, so they will cut once per every 46, or 4096 bases.
Hin dII has a five-base recognition site, so it will cut once per every 45, or 1024 bases.
Hpa II has a four-base recognition site, so it will cut once per every 44, or 256 bases.
b. To determine the number of sites, divide the number of bases in the DNA by the average length of the restriction fragments resulting from the given enzyme. The size of l is 48,502 base pairs, so the number of expected sites are:
Site
Eco RI
Hin dIII
Hin dII
Hpa II
Expected (A+T=C+G)
12
12
48
>50
2a. Cutting with Hin dIII will yield fragments of 1, 2, and 2.5 kilobases. Cutting with Eco RI will yield fragments of 1.5, 2, and 3.5 kilobases. Cutting with both Hin dIII and Pvu II will yield fragments of 1, 1.5, and 2 kilobases.
b. The cDNA in pBR322 will only consist of the two exons, which are flanked at the ends by EcoRI. When cut by EcoRI, the cDNA will yield 0.5, 1.5, and 2 kilobase fragments. Digesting with both EcoRI and HindIII, will yield 0.5, 1, and 1.5 kilobase fragments.
c. The 1 kb radiolabelled probe is from the second exon in the cDNA. If hybridized to genomic DNA that has been digested to EcoRI, it will hybridize to the 3.5 kilobase fragment.
d. The radiolabelled probe derived from the wild-type cDNA will only be able to hybridize to the section of genomic DNA that contains the first exon. When the genomic DNA is cut with HindIII, this fragment is 2.5 kilobases long.
e. Because the length of the PvuII restriction fragment in the mutant is the same as that of the wild-type, there does not appear to be any deletion between the two PvuII sites. Therefore, the mutation is not the same as that in part d. The mutation here is likely to be a point mutation.
3. a.
E H M H E
|_________|_____|__|________________|
2.5 1.5 0.5 4.5
b. There is at least one intron, because the radiolabelled cDNA did not hybridize to the 1.5 and 0.5 fragments, but did hybridize to the flanking fragments. Other introns may reside in the 2.5 and 4.5 fragments.
c. Two methods that can be used to labed cDNAs are primer extension and nick translation. Primer extension uses an oligonucleotide primer and Klenow fragment to synthesize a labeled DNA strand. In nick translation, a small amount of DNase is used to create nicks in the dDNA. Labeled oligonucleotides are then added with DNA Polymerase I in order to fill in at the nicks.
4.a. The "sticky ends" resulting from digesting with the enzymes are the same, but the actual restriction sites are different.
b. After isolating the plasmid, digest it with both enzymes, purify the 4.5 kb fragment, self-ligate the fragment, transform the new plasmid into bacteria.
c. There will be no BamHI or BglII sites.
d. You can cut with both enzymes again after ligation – this would ensure that only a plasmid lacking both sites would transform the bacteria.
5. Walk to a clone that overlaps the end of a translocation or deletion (this will show difference in Southern blots from organisms with and without the genetic abnormality). Use a clone from wild-type to pick up a junction fragment in the mutant strain. Use this fragment to pick up a clone at the other end of the chromosomal abnormality from a wild library.
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