Concept explainers
Alkaptonuria is a human autosomal recessive disorder caused by mutation of the HAO gene that encodes the enzyme homogentisic acid oxidase. Restriction mapping
of the HAO gene region reveals four BamHI restriction sites
DNA samples taken from a mother (M), father (F), and two children
a. Using A to represent the
b. In a separate figure, draw the autoradiograph patterns for all the genotypes that could be found in children of this couple.
c. Explain how the DNA sequence change results in a
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Genetic Analysis: An Integrated Approach (3rd Edition)
- You are studying a protein that contains the peptide sequence RDGSWKLVI. The part of the DNA encoding this peptide is included in the sequence shown below. 5'-CGTGACGGCTCGTGGAAGCTAGTCATC-3' 3'-GCACTGCCGAGCACCTTCGATCAGTAG-5' This sequence does not contain any BamHI restriction enzyme sites. The target sequence for the BamHI restriction nuclease is GGATCC. Your goal is to create a BamHI site on this plasmid by manipulating the DNA sequence, without changing the coding sequence of the protein. How would you do this, ie what would the new sequence be?arrow_forwardMouse genomic DNA is treated with a restriction endonuclease and electrophoresed in an agarose gel. A radioactive probe made from the human gene rxr-1 is used to perform a Southern blot. The experiment was repeated three times. Explain the results of these repeated experiments:arrow_forwardA 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_forward
- Genomic DNA from a family where sickle-cell disease is known to be hereditary, is digested with the restriction enzyme MstII and run in a Southern Blot. The blot is hybridised with two different 0.6 kb probes, both probes (indicated in red in the diagram below) are specific for the β-globin gene (indicated as grey arrow on the diagram below). The normal wild-type βA allele contains an MstII restriction site indicated with the asterisk (*) in the diagram below; in the mutated sickle-cell βS allele this restriction site has been lost. What size bands would you expect to see on the Southern blots using probe 1 and probe 2 for an individual with sickle cell disease (have 2 βS alleles)? Probe 1 Probe 2 (a) 0.6kb 0.6kb and 1.2kb (b) 0.6kb and 1.8kb 0.6kb, 1.2kb and 1.8kb (c) 1.2kb 0.6kb (d) 1.8kb 1.8kb a. (a) b. (b) c. (c) d. (d)arrow_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_forwardTable 21.3 describes the cleavage sites of five different restrictionenzymes. After these restriction enzymes have cleaved the DNA, four of them produce sticky ends that can hydrogen bond with complementary sticky ends, as shown in Figure 21.1. The efficiency of sticky ends binding together depends on the number of hydrogen bonds; more hydrogen bonds makes the ends “stickier” and more likely to stay attached. Rank these four restriction enzymes from Table 21.3 (from best to worst)with regard to the efficiency of their sticky ends binding to each other.arrow_forward
- For a restriction enzyme that recognizes the restriction site GGCC, Which of the following statements is/are true?arrow_forwardThe following is a section of the gene coding for bovine rhodopsin along with several restriction endonucleases, their recognition sequences, and their hydrolysis sites. Which endonucleases will catalyze cleavage of this section of DNA? 5-GCCGTCTACAАСССGGTCATCTAАСТАТСАТGATCААСАAGCAGTTCCGGAACT-3' Recognition Sequence Recognition Sequence Enzyme Enzyme AG | CT TGG | CCA CG I CG GG | CC CI CGG | GATC GC | GGCCGC GAGCT | C Alul Hpall Ball Mbol FnuDII NotI HealII Sadarrow_forwardTable 1 shows a list of restriction endonucleases with their recognition sequence and the sites of cleavage indicated by arrows. Table 1 Enzyme name Recognition sequence and position of cut 5'GIAATTC3 5'G!GATCC3' 5'GIGTACC3 5'GCIGGCCGC3' 5'IGATC3' 5'GGTACIC3' 5'ALGATCT3 EcoRI ВатHI Аcс651 Notl Sau3A Kpnl BglII (i) Which restriction enzyme(s) produce blunt ends? (ii) Are there any pair of neoschizomers in the list? Explain. (iii) Are there any pair of isocaudomers in the list? Explain.arrow_forward
- Single and double digestion of plasmid pMCS326 were performed using the restriction enzymes AluIII and EcoRV. DNA fragments were shown in an electrophoretogram below. Construct a restriction map of plasmid pMCS326 for enzymes AluIII and EcoRV. (Create restriction mapping with explanation)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) 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_forwardRefer to the diagram of pUC18 (Fig.) to determine which restriction enzymes you could use to insert a gene that would interfere with production of β-galactosidase by the host cell.arrow_forward
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