Genetics: Analysis and Principles
6th Edition
ISBN: 9781259616020
Author: Robert J. Brooker Professor Dr.
Publisher: McGraw-Hill Education
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Chapter 24, Problem 9EQ
Summary Introduction
To review:
The working of the BACKTRANSLATE program, its genetic principle, and the sequence generated by this program.
Introduction:
Amino acids are the organic compounds that form proteins, which are the building blocks of the body. An amino acid contains carboxyl and
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A researcher sequences the genome of a variety of bacterial and eukaryotic cells. She finds that the bacterial genome is smaller, but that there are more genes for a given number of base pairs in the eukaryotic cells. In other words, there are fewer genes per unit of length of DNA in the eukaryotic cells. What do you predict she will find if she examines the DNA more closely?
A. All of the bacterial DNA consists of coding sequences, but this is not true of the eukaryotic DNA.
B. There are more repetitive sequences in the eukaryotic DNA than in the bacterial DNA.
C. There are densely packed genes in the eukaryotic DNA that were not immediately distinguishable during the first analysis.
D. The bacteria have larger quantities of noncoding DNA than the eukaryotic cells.
Which one of the following options would be a good way to identify the location of the poly A tail in the published DNA sequence for a gene?
Look for AATAAA near the 3’ UTR of the DNA sequence.
Look for TTTTTTTTTTTTTT… (many Ts) near the 3’ UTR of the DNA sequence.
Look for AAAAAAAAAAA… (many As) near the 3’ UTR of the DNA sequence.
Look for a stop codon.
DNA from actively dividing bacteria was isolated and examined to find two groups of DNA. One group DNA included very large molecules (thousands or even millions of nucleotides long), and the other included short stretches of DNA (several hundred to a few thousand nucleotides in length). What was the researcher actually seeing in these two groups?
Group of answer choices
mRNA molecules and siRNA molecules
Okazaki fragments and RNA primers
leading strands and Okazaki fragments
leading strands and RNA primers
Chapter 24 Solutions
Genetics: Analysis and Principles
Ch. 24.1 - 1. A DNA microarray is a slide that is dotted...Ch. 24.1 - 2. The purpose of a ChIP-chip assay is to...Ch. 24.1 - 3. For the method of RNA sequencing (RNA-Seq),...Ch. 24.1 - A gene knockout is a gene a. whose function has...Ch. 24.2 - Prob. 1COMQCh. 24.2 - Prob. 2COMQCh. 24.2 - Prob. 3COMQCh. 24.2 - Prob. 4COMQCh. 24.3 - Prob. 1COMQCh. 24.3 - 2. Homologous genes
a. are derived from the same...
Ch. 24.3 - Prob. 3COMQCh. 24 - 1. Give the meanings of the following terms:...Ch. 24 - Prob. 2CONQCh. 24 - What is a database? What types of information are...Ch. 24 - Prob. 4CONQCh. 24 - Prob. 5CONQCh. 24 - Prob. 6CONQCh. 24 - Prob. 7CONQCh. 24 - Prob. 8CONQCh. 24 - Prob. 1EQCh. 24 - In the procedure called RNA sequencing (RNA-Seq),...Ch. 24 - 3. Can two-dimensional gel electrophoresis be used...Ch. 24 - Prob. 4EQCh. 24 - 5. Describe the two general types of protein...Ch. 24 - 6. Discuss the bioinformatics approaches that can...Ch. 24 - 7. What is a motif? Why is it useful for computer...Ch. 24 - Discuss why it is useful to search a database to...Ch. 24 - Prob. 9EQCh. 24 - In this chapter, we considered a computer program...Ch. 24 - Prob. 11EQCh. 24 - Prob. 12EQCh. 24 - Prob. 13EQCh. 24 - Refer to question 3 in More Genetic TIPS before...Ch. 24 - Prob. 15EQCh. 24 - Prob. 16EQCh. 24 - 1. Let’s suppose you are in charge of organizing...
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- Complete the table below 6. Below are several DNA sequences that are mutated compared with the wild-type sequence: 3’-T A C T G A C T GA C G A T C-5’. Envision that each is a section of a DNA molecule that has separated in preparation for transcription, so you are only seeing the template strand. Construct the complementary DNA sequences (indicating 5’ and 3’ ends) for each mutated DNA sequence, then transcribe (indicating 5’ and 3’ ends) the template strands, and translate the mRNA molecules using the genetic code, recording the resulting amino acid sequence (indicating the N and C termini). What type of mutation is each?6.a. Mutated DNA Template Strand #1: 3’-T A C T G T C T G A C G A T C-5’Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of mutation: 6.b. Mutated DNA Template Strand #2: 3’-T A C G G A C T G A C G A T C-5’Complementary DNA sequence:mRNA sequence transcribed from template:Amino acid sequence of peptide:Type of…arrow_forwardThe following is the base sequence of DNA that codes for first eight amino acids of the β chain of hemoglobin. The β chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. DNA Template Strand: TACCACGTGGACTGAGGACTCCTC 1. What is the minimum number of DNA nucleotides in this whole gene? 3 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 3' TACCACGTGGACTGAGGACTCCTC 5' . 5' ATGGTGCACCTGACTCCTGAGGAG 3' 3. What mRNA will be formed from the template strand of DNA? Sequence of mRNA formed from DNA template strand is shown below: 3' TACCACGTGGACTGAGGACTCCTC 5' . 5'AUGGUGCACCUGACUCCUGAGGAG 3 4. What amino acids will this mRNA code for? 5. If the 20th base in the template strand of the DNA is changed from T to A, rewrite the new template strand below. 6. When the template strand of the DNA is changed, this is referred to as a mutation. What kind of mutation is…arrow_forwardA piece of linear DNA is 700000 base pairs in length and is digested with a restriction enzyme that recognizes a six base pairs site. How many fragments do you predict would be produced? Show all work.arrow_forward
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