BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
9th Edition
ISBN: 2818000069358
Author: BERG
Publisher: MAC HIGHER
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Chapter 4, Problem 14P
Interpretation Introduction
Interpretation:
The number of missing base pairs from a mutant where the DNA of a deletion mutant of gamma bacteriophage has a length of 15 micrometers instead of 17 should be determined.
Concept introduction:
When the deletion of DNAs is small, a few base pairs are removed whereas larger deletion can lead to the removal of larger base pairs. Small deletions are not too risky whereas larger ones can be quite fatal.
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Classifying mutations
A certain section of the coding (sense) strand of some DNA looks like this:
$-ATGTATATCTCCAGTTAG-3"
It's known that a very small gene is contained in this section.
Classify each of the possible mutations of this DNA shown in the table below.
mutant DNA
5- ATGTATCATCTCCAGTTAG-3'
S-ATGTATATCTCCAGTTAG-3
5- ATGTATATATCCAGTTAG-3'
type of mutation
(check all that apply)
insertion
deletion
point
silent
noisy
insertion
O deletion
point
silent
noisy
insertion
O deletion
point
silent
Onoisy
X
G
Genome for C. diphtheriae have about 2,500,000 nucleotides, 87% of them are coding. This ingle circular chromosome contains 2,389 genes from which 2,272 proteins are coded. It does not contain any plasmids. The genome contains Pathogenicity Islands (PAIs), which C. diphtheriae has 13. What is a PAI and what are their characteristics?
Chapter 4 Solutions
BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
Ch. 4 - Prob. 1PCh. 4 - Prob. 2PCh. 4 - Prob. 3PCh. 4 - Prob. 4PCh. 4 - Prob. 5PCh. 4 - Prob. 6PCh. 4 - Prob. 7PCh. 4 - Prob. 8PCh. 4 - Prob. 9PCh. 4 - Prob. 10P
Ch. 4 - Prob. 11PCh. 4 - Prob. 12PCh. 4 - Prob. 13PCh. 4 - Prob. 14PCh. 4 - Prob. 15PCh. 4 - Prob. 16PCh. 4 - Prob. 17PCh. 4 - Prob. 18PCh. 4 - Prob. 19PCh. 4 - Prob. 20PCh. 4 - Prob. 21PCh. 4 - Prob. 22PCh. 4 - Prob. 23PCh. 4 - Prob. 24PCh. 4 - Prob. 25PCh. 4 - Prob. 26PCh. 4 - Prob. 27PCh. 4 - Prob. 28PCh. 4 - Prob. 29PCh. 4 - Prob. 30PCh. 4 - Prob. 31PCh. 4 - Prob. 32PCh. 4 - Prob. 33PCh. 4 - Prob. 34PCh. 4 - Prob. 35PCh. 4 - Prob. 36PCh. 4 - Prob. 37PCh. 4 - Prob. 38PCh. 4 - Prob. 39PCh. 4 - Prob. 40PCh. 4 - Prob. 41PCh. 4 - Prob. 42PCh. 4 - Prob. 43PCh. 4 - Prob. 44PCh. 4 - Prob. 45PCh. 4 - Prob. 46PCh. 4 - Prob. 47PCh. 4 - Prob. 48PCh. 4 - Prob. 49PCh. 4 - Prob. 50PCh. 4 - Prob. 51PCh. 4 - Prob. 52P
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- THE MOLECULAR GENETICS OF SICKLE CELL ANEMIA The following is the base sequence of DNA that codes for first eight amino acids of the ß chain of hemoglobin. The ß chain of hemoglobin contains a total of 147 amino acids so this is a small part of the entire gene. mone formed DNA Template Strand: 3'CACGTGGACTGAGGACTCCTC5' 1. What is the minimum number of DNA nucleotides in this whole gene? 2. What is the sequence of bases on the strand of DNA that is complementary to the template strand? 4. What amino acids will this mRNA code for? 3. What mRNA will be formed from the template strand of DNA? 5. If the 17th 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 this? 67arrow_forwardAll are correct about DNA gyrase in E. coli EXCEPT: It works to remove positive supercoiling introduced by the DnaB protein (helicase). It is a topoisomerase that hydrolyzes ATP during its reaction mechanism. Its mechanism involves the breaking of a single phosphoester bond in one strand of dsDNA. It works to relieve supercoiling in DNA to overcome the torsion stress imposed upon unwinding.arrow_forwardGiven the following double-stranded fragment of DNA: 5'- ACTTGGCAGGCCTTCGATCC-3' 3'- TGAАССGTCСGGAAGCTAGG-5' A hypothetical restriction endonuclease recognizes a 6bp sequence with two-fold symmetry (typical for restriction enzymes) found in this fragment and catalyzes cleavage of this DNA on both strands between GG nucleotides within the recognition sequence. This nuclease exhibits b-type cleavage (atypical for restriction enzymes). Draw the double-stranded sequence of each fragment after cleavage showing any phosphates left on the ends.arrow_forward
- e. four-base, not overlapping4. An example of a portion of the T4 rIIB gene in whichCrick and Brenner had recombined one + and one −mutation is shown here. (The RNA-like strand of theDNA is shown.)wild type 5′ AAA AGT CCA TCA CTT AAT GCC 3′mutant 5′ AAA GTC CAT CAC TTA ATG GCC 3′a. Where are the + and − mutations in the mutant DNA?b. The double mutant produces wild-type plaques.What alterations in amino acids occurred in thisdouble mutant?c. How can you explain the fact that amino acids aredifferent in the double mutant than in the wild-typesequence, yet the phage has a wild-type phenotype?arrow_forward5' 3' ORF for gene X +1 You find a mutation elsewhere in the genome (not within the sequence above) that you decide to call Mutation #2. When Mutation #1 in gene X (from the above question) is combined with Mutation #2 in the same organism, you get the phenotype in the blots below. This second mutation is most likely in what specific factor? Wild Type Northern Gene X O ribosome O sigma factor O DNA polymerase RNA polymerase Double mutant Wild Type Western Gene X direction of transcription Double mutant 3' 5'arrow_forwardA number of yeast-derived elements were added to thecircular bacterial plasmid pBR322. Yeast that requireuracil for growth (Ura− cells) were transformed withthese modified plasmids and Ura+ colonies were selected by growth in media lacking uracil. For plasmidscontaining each of the elements listed in parts (a) to(c), indicate whether you expect the plasmid to integrate into a chromosome by recombination, or insteadwhether it is maintained separately as a plasmid. If theplasmid is maintained autonomously, is it stably inherited by all of the daughter cells of subsequent generations when you no longer select for Ura+ cells (that is,when you grow the yeast in media containing uracil)?a. URA+ geneb. URA+ gene, ARS c. URA+ gene, ARS, CEN (centromere)d. What would need to be added in order for these sequences to be maintained stably in yeast cells as alinear artificial chromosome?arrow_forward
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