BENSON'S MICROBIO....,CONC.(LL)W/ACCESS
BENSON'S MICROBIO....,CONC.(LL)W/ACCESS
14th Edition
ISBN: 9781264437276
Author: Brown
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 50, Problem 1SAQ
Summary Introduction

To explain: The property of DNA that is influenced by the temperature in PCR.

Introduction: PCR stands for polymerase chain reaction, which is an invitro laboratory technique used for the synthesis of multiple copies of specific DNA segments of interest. This technique is carried out in a thermocycler to generate more copies of DNA.

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Explanation of Solution

The PCR technique is performed under high temperatures to amplify the DNA fragments. The intertwined nature of DNA is influenced by the temperature in PCR. The double strands of the DNA are tightly bonded together by the hydrogen bonds between the bases of each strand. These strands provide stabilization to the DNA molecule. A higher temperature is required to break the bonding and to unwind the two helices of the DNA separately.

Summary Introduction

To explain: The reason for the denaturation and annealing steps to proceed at different temperatures.

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Explanation of Solution

Denaturation is the initial step of PCR in which the double-stranded DNA template is separated into two single strands by the application of extreme temperature. It occurs at a temperature of 95°C.

Annealing is the second step in which the primers are attached or joined or annealed to the separated DNA template. This step is carried out at the temperature between 48–72°C.

The denaturation and annealing steps are carried out under different temperatures. This is because the higher temperature in the first step facilitates the denaturation of DNA strands. Such a high temperature would not facilitate the annealing of primers to the DNA strands in the second step.

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For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…
3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Properties
4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2
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