6. 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.

Principles Of Pharmacology Med Assist
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Author:RICE
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Chapter26: Medications Used In Treatment Of Endocrine Disorders
Section: Chapter Questions
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Otomatik Kaydet
General...
О Ara
Burak Fakir
BF
Dosya
Giriş
Ekle
Tasarım
Düzen
Başvurular
Posta Gönderileri
Gözden Geçir
Görünüm
Zotero
Yardım
3 Paylaş
P Açıklamalar
X Kes
- A A Aa v
AaççĞğth AaççĞğHh AAÇÇĞİ AaÇÇĞğt AaÇ
Ara v
Calibri (Gövde) 11
B Kopyala
V Biçim Boyacısı
Değiştir
A Seç v
Yapıştır
K T A - ab x, x A - erAv
1 Normal
1 Aralık Yok
Başlık 1
Başlık 2
Konu Başlı. -
Dikte
Duyarlılık
Düzenleyici
Dosyaları
Yeniden Kullan
Pano
Yazı Tipi
Paragraf
Stiller
Düzenleme
Ses
Duyarlılık
Düzenleyici
Dosyaları Yeniden Ku.
6. 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.
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 sequence
Nar I
GG/CGCC
Dde I
C/TNAG
Hae II
NGCGC/n
Hpall
CC/GG
Sayfa 1/2
322 sözcük
İngilizce (ABD)
Dodak
%130
Transcribed Image Text:Otomatik Kaydet General... О Ara Burak Fakir BF Dosya Giriş Ekle Tasarım Düzen Başvurular Posta Gönderileri Gözden Geçir Görünüm Zotero Yardım 3 Paylaş P Açıklamalar X Kes - A A Aa v AaççĞğth AaççĞğHh AAÇÇĞİ AaÇÇĞğt AaÇ Ara v Calibri (Gövde) 11 B Kopyala V Biçim Boyacısı Değiştir A Seç v Yapıştır K T A - ab x, x A - erAv 1 Normal 1 Aralık Yok Başlık 1 Başlık 2 Konu Başlı. - Dikte Duyarlılık Düzenleyici Dosyaları Yeniden Kullan Pano Yazı Tipi Paragraf Stiller Düzenleme Ses Duyarlılık Düzenleyici Dosyaları Yeniden Ku. 6. 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. 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 sequence Nar I GG/CGCC Dde I C/TNAG Hae II NGCGC/n Hpall CC/GG Sayfa 1/2 322 sözcük İngilizce (ABD) Dodak %130
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