Mutation analysis of GCK gene in patients with diabetes revealed a c.114 TA (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. ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGCATGGCAGCTCTAATGACAGGATGGTCACCCCTGC TGAGGCCACTCCTGGTCACCATGACAACCACAGGCCCTCTCAGTATCACAGTAAGCCCTGGCAGGAG AATCCCCCACTCCACACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAGCAG GCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Restriction enzyme Recognition seguence Nar I GG/CGCC Dde I c/TNAG www Hae III NGCGC/n Cc/GG Hpall Alul AG/CT Smal ССС/GGG Mbol /GATC Mae III wwww /GTNAC Bsp 1286 I GnGCn/C ww Hind III A/AGCTT EcoR I G/AATTC wwm n: any nucleotide /: cutting site

Mutation analysis of GCK gene in patients with diabetes revealed a c.114 TA (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. ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGCATGGCAGCTCTAATGACAGGATGGTCACCCCTGC TGAGGCCACTCCTGGTCACCATGACAACCACAGGCCCTCTCAGTATCACAGTAAGCCCTGGCAGGAG AATCCCCCACTCCACACCTGGCTGGAGCACGAAATGCCGAGCGGCGCCTGAGCCCCAGGGAAGCAG GCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Restriction enzyme Recognition seguence Nar I GG/CGCC Dde I c/TNAG www Hae III NGCGC/n Cc/GG Hpall Alul AG/CT Smal ССС/GGG Mbol /GATC Mae III wwww /GTNAC Bsp 1286 I GnGCn/C ww Hind III A/AGCTT EcoR I G/AATTC wwm n: any nucleotide /: 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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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:…
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) 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…
E. coli K12 genomic DNA (expected PCR product was 3075bp). For expression the gene was cloned into the expression plasmid pET28a (Novagen), where expression is under control of the inducible T7 promoter.Give information about the enzyme being purified, its expression in E. coli and the property being used for its purification. please provide the references for the answer.
The recognition sequence for the restriction enzyme TaqI is T↓CGA. Indicate the products of the reaction of TaqI with the DNA sequence shown.
The cellulase gene of Bacillus licheniformis was successfully cloned into the pET21a vector and expressed in Escherichia coli BL21. The pET21a vector consists of ampicillin resistant gene. To screen for the successful transformants, E. coli BL21 was cultivated on LB agar containing ampicillin (100 pg/mL) and 0.5% (w/v) carboxymethylcellulose, and incubated at 37 0C for 6 hours. After that the agar plate was stained with Congo red solution for 15 minutes and washed twice with sodium chloride solution and the observation is as shown in Figure 3. Answer the following: (i.) Briefly explain why ampicillin was added to the LB agar. (ii) Indicate the function of carboxymethylcellulose in the LB agar. (iii) Conclude how the researchers were able to identify the E. coli BL21 that carried the cellulase gene.
Compatible ends generated by restriction enzyme can be connected by ligase. Determine which of the following enzymes can generate compatible ends for enzyme 1. The recognition site for enzyme 1 is 5'-G^GTACC-3' (^ is the cleavage site). The recognition sites for available enzymes are given as the choices. Select one: a. GGTAC^C b. CGTAC^G c. C^CTAGG d. C^GATCO e. C^GTACG✔ f. GGT^ACC g. CGT^ACG Your answer is correct. The correct answer is: C^GTACG
In the Ames test shown in Figure 16-17, what is the reason for adding the liver extract to each sample?
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)
For 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).
Decide on two restriction sites that you can use to clone this into pL4440’s MCS. Identify their sequence and explain why. Tip: The plasmid map is showed below, details of restriction site sequences can be found at https://enzymefinder.neb.com/#!#nebheader
CTP synthetase catalyzes the glutamine-dependent conversion of UTP to CTP. The enzyme is allosterically inhibited by the product, CTP. Mamma- lian cells defective in this allosteric inhibition are found to have a complex phenotype: They require thymidine in the growth medium, they have unbal- anced nucleotide pools, and they have an elevated spontaneous mutation rate. Explain the likely basis for these observations.