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Chapter 29: Problem 5

DNA-binding proteins may recognize specific DNA regions either by reading the base sequence or by "indirect readout." How do these two modes of protein:DNA recognition differ?

Short Answer

Expert verified
The two modes of protein:DNA recognition vary as such; Direct readout involves the protein recognizing and binding to specific base sequences of the DNA, whereas indirect readout involves the protein binding to the DNA based on the physical shape or conformation of the DNA, not the base sequence.

Step by step solution

01

Understanding Direct and Indirect Readout

Firstly, the terms 'direct readout' and 'indirect readout' need to be understood. 'Direct Readout' refers to the process where proteins recognize and bind to specific DNA sequences by forming chemical bonds with the bases themselves. This usually involves hydrogen bonds, salt bridges, or hydrophobic interactions, making direct contacts between the protein and the major grooves in the DNA. ‘Indirect Readout', on the other hand, is a recognition process in which the protein does not primarily recognize the base sequence but recognizes and binds to the specific shape or conformation of the DNA helix that results from that sequence.
02

Direct Readout Process

Understanding the process of direct readout involves recognizing that the protein directly binds to the base pairs of the DNA sequence, thus implying that the protein has to have a complementarity shape to the base pairs of the DNA sequence. This is possible due to the sequence-specific interactions between protein side chains and bases upon a major groove of the DNA helix.
03

Indirect Readout Process

For the indirect readout, the protein recognizes and binds to the specific physical conformation or shape of the DNA helix, which is a reflection of the sequence. This conformations may result from characteristic bends, kinks, or flexibility of the DNA at the binding site and the protein binds by adapting to these deformations in the DNA structure.

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Most popular questions from this chapter

RNA polymerase has two binding sites for ribonucleoside triphosphates: the initiation site and the elongation site. The initiation site has a greater \(K_{m}\) for \(\mathrm{NTPs}\) than the elongation site. Suggest what possible significance this fact might have for the control of transcription in cells.

Draw the structures that comprise the lariat branch point formed during mRNA splicing: the invariant A, its \(5^{\prime}\) - \(\mathrm{R}\) neighbor, its \(3^{\prime}-\mathrm{Y}\) neighbor, and its \(2^{\prime}\) - \(G\) neighbor.

Chromatin decompaction is a preliminary step in gene expression (Figure \(29.46) .\) How is chromatin decompacted?

Make a list of the ways that transcription in eukaryotes differs from transcription in prokaryotes.

\(\mathrm{C} / \mathrm{EBP} \beta\) is a \(b Z I P\) transcription factor in neuronal differentiation, learning and memory process, and other neuronal and glial functions. The structure of the \(b Z I P\) domain of \(C / E B P \beta\) bound to DNA is shown in pdb file \(1 \mathrm{GU} 4 .\) Explore this structure to discover the leucine zipper dimerization domain and the DNA-binding basic regions. On the left side of the www.pdb.org \(1 \mathrm{GU} 4\) page under "Display Files," click "pdb file" to see the atom-by-atom coordinates in the three-dimensional structure (scroll down past "Remarks" to find this information). Toward the end of this series, find the amino acid sequence of the \(\mathrm{C} / \mathrm{EBP} \beta\) domain used in this study. Within this amino acid sequence, find the leucine residues of the leucine zipper and the basic residues in the DNA-binding basic region.
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