Biology
4th Edition
ISBN: 9781259188121
Author: Peter Stiling, Robert Brooker, Linda Graham, Eric Widmaier
Publisher: McGraw-Hill Education
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Textbook Question
Chapter 12.2, Problem 1BC
Core Skill: Connections Look back at the role of DNA polymerase shown in Figure 11.15. What are similarities and differences between the function of DNA polymerase and that of RNA polymerase?
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Objective: Get a sense of how genomics, the study of the genome in its entirety,needs to think about how to go about its research.
Geonomic DNA is broken up into fragments. The 5’ and 3’ ends of each fragment(a “read”) are sequenced. The sequenced reads are assembled together intocontiguous sequences (“contigs”) based on sequence similarity.
The idea is to sequence enough random fragments so that every nucleotide in thegenome is represented on some read. The number of such fragments needed iscalled the coverage, c.
The coverage c can be calculated by the formula RL/G, where R is the number ofreads sequenced, L is the average length of a read and G is the total length of thegenome. The units of length are bases (b) or base pairs (bp).
Consider a genome whose length is 1000 bp. “Shotgun” sequencing techniquesare applied to the genome, resulting in 20 reads, with an average length of 50 bp.A very important point is that, even though 20 x 50 = 1000, there is no guaranteethat ALL…
Protein Synthesis and Mutation Practice
• Complete the lines below by determining the mRNA transcript and amino acid sequence.
• Compare the mutant DNA strands to the wild type strand.
⚫ Circle the mutation in the mutant DNA strands and describe the type of mutation (frameshift -
insertion, frameshift - deletion, point - missense, point - silent, or point-nonsense). Not all of these will
be used in this assignment!
Wild type DNA template: 3' TACGCGTGCACGATGCAGTAGTACATC5'
mRNA transcript sequence:
Amino acid sequence:
Mutation #1 DNA template: 3' TACGCGTGCACGATCCAGTAGTACATC5'
mRNA transcript sequence:
Amino acid sequence:
Type of mutation:
Mutation #2 DNA template: 3' TACGCGTGCTCGATGCAGTAGTACATC5'
mRNA transcript sequence:
Amino acid sequence:
Type of mutation:
TOPIC: PCR and Gene Cloning Basics
Question: What are 2 possible roles of CaCl2 in the transformation process?
Chapter 12 Solutions
Biology
Ch. 12.1 - What disease would result if a person inherited...Ch. 12.1 - Prob. 2CCCh. 12.1 - What is the direction of flow of genetic...Ch. 12.2 - Prob. 1CCCh. 12.2 - Core Skill: Connections Look back at the role of...Ch. 12.3 - Prob. 1CCCh. 12.4 - Prob. 1CCCh. 12.4 - Prob. 2CCCh. 12.4 - Prob. 1EQCh. 12.4 - Prob. 2EQ
Ch. 12.4 - Prob. 3EQCh. 12.5 - Prob. 1CCCh. 12.5 - Core Skill: Connections Look back at Figure 6.3,...Ch. 12.5 -
Figure 12.17 Comparison of small subunit rRNA...Ch. 12.6 - Prob. 1CCCh. 12 - Which of the following best represents the central...Ch. 12 - A mutation prevents a gene from being transcribed...Ch. 12 - Prob. 3TYCh. 12 - Prob. 4TYCh. 12 - If a eukaryotic mRNA failed to have a cap attached...Ch. 12 - Prob. 6TYCh. 12 - Prob. 7TYCh. 12 - During the initiation of translation, the first...Ch. 12 - Prob. 9TYCh. 12 - Prob. 10TYCh. 12 - Prob. 1CQCh. 12 - Prob. 2CQCh. 12 - Prob. 3CQCh. 12 - Prob. 1COQCh. 12 - Prob. 2COQ
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biology and related others by exploring similar questions and additional content below.Similar questions
- Exercise In this exercise we will practice transcribing and translating sample DNA. Using your sample DNA, unzipped and second strand removed, you will first create an RNA strand (transcription). DNA is read in the 5' → 3' direction, so when you create RNA, a 3'end pairs with the 5' end of DNA. From your RNA strand, you will need to create codons; remember that a codon is a group of 3 bases that codes for a specific amino acid. Your codons are read in the 5' → 3' direction (hint: it might be right to left!). You will then need to convert your codons into amino acids in the 5' → 3' direction (translation). DNA strand 3' TAC-TTA-CGA-TGG-TAC-ACG-CAA-TCT-ATA-CTC-AAA-TAT-AGG-ACC-TTG-ACG-TCG-AAT-CTC-CAC-TGT-ACC-TTG-AAC-CTG-ACT 5' RNA strand 5'-AUG-AAU Amino acid sequence G ne (9)arrow_forward4. Working backwards challenge You have been hired as a genetic engineer. Your patient, Mr. Williams, is a 42-year-old male who has diabetes. To regulate his blood sugar levels, Mr. Williams must test his blood daily and give himself insulin injections when necessary. Mr. Williams has contacted you to genetically engineer bacteria to produce insulin. To genetically engineer bacteria, the section of DNA that codes for insulin is inserted into the bacterial plasmid. You must work backwards to determine the DNA sequence. Insulin consists of 51 amino acids. A small segment of the protein is provided below. Use the mRNA codon charts underneath to determine the DNA sequence that would code for this segment of insulin. Amino Acid Start Glycine Leucine Valine Glutamic Acid Proline Stop MRNA AUG AVAV UGA DNA ТАС Gv TVTV ACT Second Base P Type here to search aarrow_forwardSolve 21arrow_forward
- Home Work: • Suppose you perform a PCR that begins with one double-strand of the following DNA template: +5' -СТАССТСCGGGTTGACTGСТАССТТССССGGATGCCCAAAAТТСТСGAG-3— :::::::::::: :::::::::::: :::: +3'-GATGGACССССААСТGACGATGGAAGGGCCCТАССGGTTTTAAGAGCTC-5'+ A. Draw one cycle of PCR reaction below the following diagram. B. Label the template DNA, the primers, and what is happening at each step. (1) température cycle #1arrow_forwardPractice: DNA Structure and Replication 1. Label each part of the model to the right. Include specific nitrogen base pairs in your labeling. 2. What molecule is it? 3. What is its purpose? 4. Where can it be found in a prokaryotic cell? 5. Where can it be found in a eukaryotic cell? 6. It gets copied during a process called replication. When does this happen? 7. What is the result of DNA replication? 8. Why is DNA replication necessary? 10. What would the chromosome to the right look like after DNA replication? 11. What would the chromosome to the right be called after DNA replication? 9. Why is DNA replication said to be semi-conservative? Draw a picture to support your answer. TAACCGAGTTCAGA b. TTAACCGAGTTCAGA Genetics Unit Sol Sol Dal 12. Replicate the following four DNA strands using what you know about complementary base pairs. TACOTCCAGATITT a. AATACGTCCAGATTTT c. CCCGCGGAATATACA O book It's Not Rocket Science 2016 d. AGGGCTACTTCAGAC J 7arrow_forwardGivearrow_forward
- Solve number 1 In this exercise, you analyzed DNA sequences that were 72 base pairs (bp) long. The entire human genome is a little more than three billion bp. Number 1. Describe the challenges scientists would face if they were to carry out DNA analysis by hand, and What advantages do computers provide to bioinformatics?arrow_forward12arrow_forwardSelect all that applyarrow_forward
- Please help witht this homework question A- What are the topological parameters (linking number, twist, and writhe)for a relaxed, 4,200 base pair circular double-stranded DNA plasmid? LK=? Tw= ? Wr=? The CRISPR-Cas9 protein first forms a protein-RNA complex with a guide RNA, and then binds to DNA sequences that match a 20-base target sequence within the guide. When the CRISPR-Cas9 complex binds to DNA, it unwinds about 20 base pairs of the DNA double helix. B- Why is DNA unwinding required for CRISPR-Cas9 protein-RNA complexto recognize target sites? C-How would the topological parameters (linking number, twist, andwrithe) of the 4,200 bp plasmid in (A) change when the CRISPR-Cas9 complex binds the DNA and unwinds about 20 bp? Lk=? Tw=? Wr=? D- What is the linking number after the CRISPR-Cas9 complex cleaves theDNA?arrow_forwardGive typed full explanationarrow_forward Proofreads each nucleotide its template as soon as it is added to the growing strand. A) DNA Ligase B) Helicase C) DNA Polyerase D) Primase The genetic code A) has no redundancy but does have ambiguity B) has both redundancy and ambiguity C) has redundancy and not ambiguity D) has ambiguity E) has redundancyarrow_forward
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