Biochemistry
9th Edition
ISBN: 9781319114671
Author: Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher: W. H. Freeman
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Question
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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The DNA of a deletion of alpha bacteriophage has a length of 15 micrometers instead of 17 micrometers? How many base pairs are missing from this mutant?
Restriction sites of Lambda (A) DNA - In base pairs (bp)
The sites at which each of the 3 different enzymes will cut the same strand of lambda DNA
are shown in the maps (see figure 3 B-D), each vertical line on the map is where the respective
enzymes will cut.
A DNA
A
(bp)
48502
10 000
20 000
30 000
40 000
9162
17 198
B
Sal I
7059
14 885
28 338
35 603
42 900
(bp)
Hae III
11 826
21 935
29 341
38 016
(bp)
11648
29,624
Eco R1
(bp)
10 592 16 246
28 915
41 864
Figure 3: Restrictrion site map showing the following A) inear DNA that is not cut as reference B) DNA CLt with Sal L C) DNA cut with Hae , D)
DNA cut with Eco RI
1. Calculate the size of the resulting fragments as they will occur after digestion and write
the sizes on the maps below. Note that linear DNA has a total size of 48 502 bp (see
figure 3A).
Page 3 of 7
9162
17 198
Sal i
(bp)
7059
14 885
28 338
35 603
42 900
Hae I
(bp)
11 826
21 935
29 341
38 016
11648
29,624
Eco R1
(bp)
10 592
16 246
28 915
41 864
Cynt
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
Chapter 4 Solutions
Biochemistry
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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- 5'-[seq]-3' The diagram shows the results of gel electrophoresis for Sanger sequencing. The wells are represented by open boxes and the DNA bands are represented by black boxes. The wells are labeled to show which dideoxy reaction was loaded into each. Write the sequence of the original template strand used for this sequencing reaction, with the 5’ end on the left and the 3’ end on the right.arrow_forwardIn the following gel showing stained bands of the Alu insertion sequence, what is the genotype of individual 2? 941 bp 641 bp->>> 1 2 3 4 5 6 Homozygous for the 641 bp sequence that does not contain in the Alu insertion Heterozygous, containing one 941 bp sequence and one 641 bp sequence O Homozygous for the 941 bp sequence containing the Alu insertionarrow_forwardConsider the structure of Cro repressor protein from bacteriophage lambda E. It is a DNA binding protein, and like many sequence- specific DNA binding proteins, it must function as a homodimer Ex. Notice the mutual docking of a phenylalanine residue from each subunit into a hydrophobic pocket of the partner subunit. These hydrophobic interactions are required for dimerization. The noncovalent interactions highlighted in yellow are also required for dimerization. These interactions represent examples of: Osecondary structure O tertiary structure O quaternary structure O secondary AND quaternary structure Ⓒ tertiary AND quaternary structurearrow_forward
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