Concept explainers
A geneticist searching for mutations uses the restriction endonucleases
a. What common feature do
b. What process is the researcher intending to detect with the use of these restriction enzymes?
c. Explain why CpG dinucleotides are hotspots of mutation.
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Genetic Analysis: An Integrated Approach (3rd Edition)
- A molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. HindIII (recognition sequence is AAGCTT)arrow_forwardA molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. HpaII (recognition sequence is CCGG)arrow_forwardWhen the restriction endonuclease EcoRI is used to digest a 10 kb DNA fragment, it produces 4 kb and 6 kb-sized fragments. Digesting the 10 kb fragment with BamHI yields three fragments, each ranging in size from one to three and a half kilobytes. Four pieces of 0.5, 1, 3 and 5.5 kb are formed after using both enzymes. Create a restriction map for this 10 kb piece of DNA using the information you have collected. Make a note of where the two enzymes cut, as well as the distances between the enzymes.arrow_forward
- A 2.0kb bacterial plasmid ‘BS1030’ is digested with the restriction endonuclease Sau3A; the plasmid map is depicted in the diagram below and the Sau3A (S) restriction sites are indicated. Which of the following DNA fragments do you expect to see on an agarose gel when you run Sau3A-digested plasmid ‘BS1030’ DNA? a. 250 bp, 450 bp, 550 bp, 1.1 kb, 1.5 kb and 2.0 kb b. 2.0kb c. 250 bp, 400 bp, 450 bp, 500 bp and 550 bp d. 100 bp, 200 bp, 250 bp, 400 bp, 500 bp and 550 bparrow_forwardA molecule of double-stranded DNA that is 5 million base pairs long has a base composition that is 62% G + C. How many times, on average, are restriction sites for the following restriction enzymes likely to be present in this DNA molecule? a. BamHI (recognition sequence is GGATCC)arrow_forwardAfter restriction enzymes cut, they contain unpaired bases. Type II restriction enzymes leave ends that may be 5' overhanging, 3' overhanging, or blunt. In all cases each end is left with a 3' OH and a 5' phosphate. All blunt ends, and any complementary overhanging ends may be re-ligated with T4 DNA ligase, as long as at least one 5'- phosphate is present. In the tables below G^AATTC means that the end after cutting with enzyme will be: -----G 3' -----CTTAA 5' GTGCA^C means that the end will be: -----GTGCA 3' -----C 5' Which RE’s from table below have a 5’ overhang? Which ones have a 3’ Overhang? AccI GT^CGAC BamHI G^GATCC ClaI AT^CGAT NsiI ATGCA^T PstI CTGCA^G BglII A^GATCT TaqI T^CGAarrow_forward
- A microbiologist discovers a new type II restriction endonuclease. When DNA is digested by this enzyme, fragments that average 1,048,500 bp in length are produced. What is the most likely number of base pairs in the recognition sequence of this enzyme?arrow_forwardRestriction sites are palindromic; that is, they read the same in the5' to 3' direction on each strand of DNA. What is the advantage ofhaving restriction sites organized this way?arrow_forwardGive the recognition sequences for each of the restriction enzymes (a) PagI, (b) AluI, (c) PstI, and (d) RcaI. Show the sequence in double-stranded DNA (dsDNA) format, indicate the cleavage position with a ^, and mention which types of ends are generated in each case (blunt or sticky; 5’ or 3’ overhang)? Recognition sequence for a)PagI cleaves DNA at the recognition sequence 5' ---T ^CATGA--- 3' b) AluI cleaves DNA at the recognition sequence 5' AG^CT 3' c) PstI cleaves DNA at the recog.arrow_forward
- Consider the ends of the DNA fragments shown below. They have been produced by digestion of a single sequence of DNA using a number of restriction endonucleases. 1. 5'A 3' 3'TTCGA5' 2. 5'G 3' 3'CAGCT5' 3. 5'AATTC3' 3' G5 4. 5'TCGAC3' 3' G5' 5. 5'GGG 3' 3'CCC 5' Which of these ends are capable of annealing and being joined by DNA ligase?arrow_forwardWhich of the following sequences, when combined with itscomplement, would be clipped by a restriction endonuclease?a. ATCGATCGTAGCTA b. AAGCTTCGAA c. GAATTC d. ACCATTGGAarrow_forwardFor the DNA sequence shown, indicate the products of its cleavage with the following restriction endonucleases (AKA restriction enzymes):5′-ACAGCTGATTCGAATTCACGTT-3′3′-TGTCGACTAAGCTTAAGTGCAA-5′a) EcoRI (the recognition sequence and cleavage site is G↓AATTC);b) AluI (the recognition sequence and cleavage site is AG↓CT).arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning