Concept explainers
Using this map of pLAB30 (18 kb), give the number and lengths of the restriction fragments that would result from digesting pLAB30 with EcoR1, with BamHI, and with both enzymes together.
Which enzyme will give the smallest piece containing the tetracycline-resistance gene?
To determine:
The size and number of DNA Fragments given by restriction enzymes Eco RI, Bam HI, and both these enzymes together.
Introduction:
Restriction enzyme is an enzyme that cuts the DNA fragment at the specific site by recognizing the DNA sequence at that site. The site at which the DNA is cut is termed as the restriction site.
Explanation of Solution
pLAB30 is an 18 kb map, which is cut by two restriction enzymes, Eco RI, and Bam HI. When Eco RI acts, then it cuts the DNA fragment at two sites, giving an 11 kb fragment and 7 kb fragment of the DNA.
When Bam HIacts, then it cuts DNA fragment at three sites, giving three fragments. The size of these fragments is 9 kb, 3 kb, and 6 kb.
When both these restriction enzymes act together, then 5 DNA fragments are obtained. The size of these fragments is 1 kb, 3 kb, 3 kb, 3kb, and 8 kb.
The table showing the size and number of DNA fragments after the DNA is acted upon by restriction enzymes is shown below:
Restriction enzyme | Number of DNA fragments obtained | Size of DNA fragments |
Eco RI | 2 | 11 kb and 7 kb |
Bam HI | 3 | 9 kb, 3 kb, and 6 kb |
Eco RI + Bam HI | 5 | 1 kb, 3 kb, 3 kb, 3kb, and 8 kb |
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Chapter 30 Solutions
Laboratory Experiments in Microbiology (12th Edition) (What's New in Microbiology)
- Mutation analysis of GCK gene in patients with diabetes revealed a c.114 T->A (shown in bold and underlined) substitution in heterozygote state. In order to check the mutation in healthy individuals, restriction enzyme analysis will be used. a) which enzyme can we use to differentiate wild type and mutant sequence? Please indicate which allele (wild type or mutant allele) will be cut with the restriction enzyme. Use table 1 shown below. b) ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGCATGGCAGCTCTAATGACAGGATGGTCACCCCTG СTGAGGCCACTCCTGGTCACCATGACAАССАCAGGCCCTCTТСAGTATCACAGTAAGCCCTGGCAGG AGAATCCCCCACTCCACACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAG CAGGCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Bestriction enzyme Recognition sequence Nari GG/CGCC Ddel C/TOAG Hae II DGCGC/n Hpal cc/GG Alul AG/CT Smal ccc/GGG Mbol /GATC Mae II IGTDAC Bsp 1286 I GNGCn/c Hind II A/AGCTT ECOR I G/AATTC D: any Ducleotide 1:…arrow_forwardMutation analysis of GCK gene in patients with diabetes revealed a c.114 T→A (shown in bold and underlined) substitution in heterozygote state. In order to check the mutation in healthy individuals, restriction enzyme analysis will be used. a) Which enzyme can we use to differentiate wild type and mutant sequence? Please indicate which allele (wild type or mutant allele) will be cut with the restriction enzyme. Use table 1 shown below. b) Draw the expected agarose gel result of a homozygous wild type, homozygous mutant and heterozygote individual after restriction enzyme analysis. ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGCATGGCAGCTCTAATGACAGGATGGTCACCCCTGCTGAGGCC ACTCCTGGTCACCATGACAACCACAGGCCCTCTCAGTATCACAGTAAGCCCTGGCAGGAGAATCCCCCACTCCAC ACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAGCAGGCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Restriction enzyme Recognition seguence www wwwtw ww Nar I GG/CGCC…arrow_forwardAssume that a circular plasmid is 3200 base pairs in length and has restriction sites for HindIII restriction enzyme at the following locations: 400, 700, 1400, 2600. Give the expected sizes of the restriction fragments following complete digestion.arrow_forward
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