6. Mutation analysis of GCK gene in patients with diabetes revealed a c.114 TàA (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) (5p) 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) (10p) Draw the expected agarose gel result of a homozygous wild type, homozygous mutant and heterozygote individual after restriction enzyme analysis. A TGA

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Chapter15: From Dna To Protein
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6. Mutation analysis of GCK gene in patients with diabetes
revealed a c.114 TàA (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) (5p) 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) (10p) Draw the expected agarose gel result of a
homozygous wild type, homozygous mutant and
heterozygote individual after restriction enzyme
analysis.
ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGC
ATGGCAGCTCTAATGACAGGATGGTCACCCCTG
CTGAGGCCACTCCTGGTCACCATGACAACCACA
GGCCCTCTCAGTATCACAGTAAGCCCTGGCAGG
AGAATCCCCCACTCCACACCTGGCTGGAGCACG
AAATGCCGAGCGGCGCCTGAGCCCCAGGGAAGC
AGGCTAGGATGTGA
Figure 1. GCK gene sequence. Length of the fragment is
213bp.
Table1. The restriction enzymes and their recognition
sequences.
Restriction enzyme|Recognition sequence
GG/CGCC
Nar I
Dde I
|C/TNAG
NGCGC/n
|CC/GG
Hae III
Hpall
|Alul
AG/CT
CCC/GGG
Smal
Mbol
/GATC
Мae III
/GTNAC
|Bsp 1286 I
Hind III
EcoR I
GNGCN/C
A/AGCTT
G/AATTC
n: any nucleotide
: cutting site
Transcribed Image Text:6. Mutation analysis of GCK gene in patients with diabetes revealed a c.114 TàA (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) (5p) 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) (10p) Draw the expected agarose gel result of a homozygous wild type, homozygous mutant and heterozygote individual after restriction enzyme analysis. ATGAGGCTCTTTGCCACCAGTCCCAGTTTTATGC ATGGCAGCTCTAATGACAGGATGGTCACCCCTG CTGAGGCCACTCCTGGTCACCATGACAACCACA GGCCCTCTCAGTATCACAGTAAGCCCTGGCAGG AGAATCCCCCACTCCACACCTGGCTGGAGCACG AAATGCCGAGCGGCGCCTGAGCCCCAGGGAAGC AGGCTAGGATGTGA Figure 1. GCK gene sequence. Length of the fragment is 213bp. Table1. The restriction enzymes and their recognition sequences. Restriction enzyme|Recognition sequence GG/CGCC Nar I Dde I |C/TNAG NGCGC/n |CC/GG Hae III Hpall |Alul AG/CT CCC/GGG Smal Mbol /GATC Мae III /GTNAC |Bsp 1286 I Hind III EcoR I GNGCN/C A/AGCTT G/AATTC n: any nucleotide : cutting site
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