Mutation analysis of GCK gene in patients with diabetes revealed a c.114 T->A (shown inbold and underlined) substitution in heterozygote state. In order to check the mutation inhealthy individuals, restriction enzyme analysis will be used.a) which enzyme can we use to differentiate wild type and mutant sequence? Pleaseindicate which allele (wild type or mutant allele) will be cut with the restriction enzyme.Use table 1 shown below.b)ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGCATGGCAGCTCTAATGACAGGATGGTCACCCCTGСTGAGGCCACTCCTGGTCACCATGACAАССАCAGGCCCTCTТСAGTATCACAGTAAGCCCTGGCAGGAGAATCCCCCACTCCACACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAGCAGGCTAGGATGTGAFigure 1. GCK gene sequence. Length of the fragment is 213bp.Table1. The restriction enzymes and their recognition sequences.Bestriction enzymeRecognition sequenceNariGG/CGCCDdelC/TOAGHae IIDGCGC/nHpalcc/GGAlulAG/CTSmalccc/GGGMbol/GATCMae IIIGTDACBsp 1286 IGNGCn/cHind IIA/AGCTTECOR IG/AATTCD: any Ducleotide1: cutting site

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 CTGAGGCCACTCCTGGTCACCATGACAACCACAGGCCCTCTCAGTAICACAGTAAGCCCTGGCAGG AGAATCCCCCACTCCACACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAG CAGGCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Restriction enzyme Recognition sequence Nari GG/CGCC Ddei C/TOAG Hae II DGCGC/n Hpal cc/GG Alul AG/CT Smal ссс/GGG Mbol /GATC Mae II IGTDAC Bsp 1286I GNGCn/c Hind II A/AGCTT ECOR I G/AATTC D: any Ducleotide 1: cutting site

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 CTGAGGCCACTCCTGGTCACCATGACAACCACAGGCCCTCTCAGTAICACAGTAAGCCCTGGCAGG AGAATCCCCCACTCCACACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAG CAGGCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Restriction enzyme Recognition sequence Nari GG/CGCC Ddei C/TOAG Hae II DGCGC/n Hpal cc/GG Alul AG/CT Smal ссс/GGG Mbol /GATC Mae II IGTDAC Bsp 1286I GNGCn/c Hind II A/AGCTT ECOR I G/AATTC D: any Ducleotide 1: cutting site

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter13: An Introduction To Genetic Technology

Section: Chapter Questions

Problem 20QP: Analyzing Cloned Sequences A base change (A to T) is the mutational event that created the mutant...

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For Pet41 (choose Pet41 a, b, or c as provided in the image) how would you design the primers (forward and reverse) for the following gene of interest and what restriction enzymes would be used (as shown in the image)? Be sure to explain and elaborate on why selected and how. Gene of Interest: atgggc gacaaaggga 241 cccgagtgtt caagaaggcc agtccaaatg gaaagctcac cgtctacctg ggaaagcggg 301 actttgtgga ccacatcgac ctcgtggacc ctgtggatgg tgtggtcctg gtggatcctg 361 agtatctcaa agagcggaga gtctatgtga cgctgacctg cgccttccgc tatggccggg 421 aggacctgga tgtcctgggc ctgacctttc gcaaggacct gtttgtggcc aacgtacagt 481 cgttcccacc ggcccccgag gacaagaagc ccctgacgcg gctgcaggaa cgcctcatca 541 agaagctggg cgagcacgct taccctttca cctttgagat ccctccaaac cttccatgtt 601 ctgtgacact gcagccgggg cccgaagaca cggggaaggc ttgcggtgtg gactatgaag 661 tcaaagcctt ctgcgcggag aatttggagg agaagatcca caagcggaat tctgtgcgtc 721 tggtcatccg gaaggttcag tatgccccag agaggcctgg cccccagccc acagccgaga 781 ccaccaggca gttcctcatg tcggacaagc ccttgcacct…
Refer to the succeeding diagram of two DNA samples (A, B) with the specific restriction enzyme(RE) recognition sites marked with arrows. A spontaneous mutation (insertion) occurred in DNAsample A and introduced an additional restriction site. 1. Design two (2) DNA probes to be used in RFLP analysis involving these two DNA samples.Draw/illustrate the region(s) in the DNA molecule which you will use as reference(homologous) sequences in the design of the probes, THAT -a. Will reveal RFLP polymorphism between the DNA samples after probedetection/visualization; andb. Will not reveal RFLP polymorphism between the DNA samples
Table 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.
STEP BY STEP Exaplanation would be appreciated so I can understand for upcomming exam. In humans, the COL1A1 locus codes for a collagen protein found in bone. A recessive COLA1 allele believed to reduce bone density and increase risk of fractures differs from the wild type allele by the presence a GT in the first intron of the gene. This mutation can be easily screened by PCR, using COL1A1-specific primers followed restriction enzyme digest of the product with MscI. The normal allele is denoted as G and the recessive allele as T. A recent study of 894 women found that 570 were GG, 291 were GT and 33 were TT. (Assume for this exercise that these values reflect both male and female values for the sampled population. What are the frequencies of each allele? What are the frequencies of each genotype? c. Is this locus in Hardy-Weinberg equilibrium? Use the table below to evaluate the significance of any deviation from H-W expectations.
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)
E. coli strain BW25113 can grow on (D) Arg as the sole carbon source. A scientist working with you on this project has discovered that mice encode a protease that cancleave between (D)Arg‐(D)Arg. a. In a set of discrete steps outline how to construct this genomic library so that recombinant expression can be performed in E. coli b. Once the genomic library has been constructed, describe in DISCRETE steps how you would isolate the DNA sequence that encodes for the protease that cleaves between (D)Ag-(D)Arg
he gel presented here shows the pattern of bands of fragments produced with several restriction enzymes. (a) From your analysis of the pattern of bands on the gel, draw the correct restriction map and explain your reasoning. (b) The highlighted bands (magenta) in the gel hybridized with a probe for the gene pep during a Southern blot. Where in the gel is the pep gene located?
The restriction enzymes Kpn I and Acc 65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from Kpn I cleavage cannot be ligated directly to the sticky end formed from Acc 65I cleavage. Explain why.
Although a large number of mutagenic chemicals are known,none is known that induces mutations in only a single gene(gene-specific mutagenesis). From what you know aboutmutagens, explain why it is unlikely that a gene-specificchemical mutagen will be found. How then is site-specificmutagenesis accomplished?
The restriction enzymes KpnI and Acc65I recognize and cleave the same 6-bp sequence. However, the sticky end formed from KpnI cleavage cannot be ligated directly to the sticky end formed from Acc65I cleavage. Explain why.
A site-directed mutagenesis experiment was done on the catalytic triad of the serine protease subtilisin where all of the amino acids of the catalytic triad were mutated to alanine. The triple mutant enzyme was characterized kinetically and the mutant displayed a thousand-fold (103) rate enhancement over the uncatalyzed reaction. Explain the source of this rate enhancement. (The native wild-type subtilisin has a rate acceleration of 1010, when compared to the uncatalyzed reaction.)
A mutant has no activity for the enzyme isocitrate lyase.Does this result prove that the mutation is in the geneencoding isocitrate lyase?