Chapter 29, Problem 18P

Interpretation Introduction

Interpretation:

The structure of a DNA-Bound bZIP transcription factor should be explained.

Concept introduction:

The structure is determined by five transcription factors. Each transcription factor is a member of special structural category. Representation of novel and unlooked-for DNA-binding motifs is done by TFIID TATA-binding super molecule. Determination of the structure of -bound super molecule is done by the exception of TFIID.Acknowledgement of specific polymer sequences is done by the every employed fold.

Answer to Problem 18P

C/EBP $\beta$ has 2 identical macromolecule chains. The sequence of those chains is as follows:

  $\begin{array}{l}\text{VAL LYS SER LYS ALA LYS LYS THR VAL ASP LYS HIS SER ASP GLU}\\ \text{TYR LYS ILE ARG ARG GLU ARG ASN ASN ILE ALA VAL ARG LYS SER }\\ \text{ARG ASP LYS ALA LYS MET ARG ASN LEU GLU THR GLN HIS LYS VAL }\\ \text{LEU GLU LEU THR ALA GLU ASN GLU ARG LEU GLN LYS LYS VAL GLU }\\ \text{GLN LEU SER ARG GLU LEU SER THR LEU ARG ASN LEU PHE LYS GLN}\\ \text{LEU PRO GLU}\end{array}$

Separating the sequence of the macromolecule before every essential amino acid residue gives:

  $\begin{array}{l}VALLYSSERLYSALALYSLYSTHRVALASPLYSHISSER\\ ASPGLUTYRLYSILEARGARGGLUARGASNASNILE\\ ALAVALARGLYSSERARGASPLYSALALYSMETARG\\ ASNLEUGLUTHRGLNHISLYSVAL\end{array}$

  $\text{LEU GLU}$

  $\begin{array}{l}\text{LEU THR ALA GLU ASN GLU}\\ \text{ARG LEU GLN LYS LYS VAL }\\ \text{GLU GLNLEU SER ARG GLU}\end{array}$

  $\begin{array}{c}\text{LEU SER THR}\\ \text{LEU ARG ASN}\\ \text{LEU PHE LYS GLN}\\ \text{LEU PRO GLU}\end{array}$

Rearranging the sequence to allow 2 additional seven residue sequence gives:

  $\begin{array}{l}\text{VAL LYS SER LYS ALA }\\ \text{LYS LYS THR VAL ASP LYS HIS SER ASP GLU TYR }\\ \text{LYS ILE ARG ARG GLU ARG ASN ASN ILE ALA VAL ARG }\\ \text{LYS SER ARG ASP LYS ALA LYS MET ARG ASN}\end{array}$

  $\text{LEU GLU THR GLN HIS LYS VAL}$

  $\begin{array}{l}\text{LEU GLU}\\ \text{LEU THR ALA GLU ASN GLU ARG}\\ \text{LEU GLN LYS LYS VAL GLU GLN}\\ \text{LEU SER ARG GLU LEU SER THR}\\ \text{LEU ARG ASN LEU PHE LYS GLN}\\ \text{LEU PRO GLU}\end{array}$

Explanation of Solution

An essential amino acid zipper can also be represented in coiled-coil domain. It contains $\alpha$ -helices command along with hydrophobic interactions of essential amino acid residues. Each seven residues are repeated by coiled-coil domains. Thus, we tend to expect one domain of the 1GU4 $\alpha$ -helices to possess essential amino acid residues repetition for each seven residues. Essential amino acid zipper domain contains twenty eight amino acids, thus we are able to expect four such repeats.

C/EBP $\beta$ has 2 identical macromolecule chains. The sequence of those chains is as follows:

  $\begin{array}{l}\text{VAL LYS SER LYS ALA LYS LYS THR VAL ASP LYS HIS SER ASP GLU}\\ \text{TYR LYS ILE ARG ARG GLU ARG ASN ASN ILE ALA VAL ARG LYS SER }\\ \text{ARG ASP LYS ALA LYS MET ARG ASN LEU GLU THR GLN HIS LYS VAL }\\ \text{LEU GLU LEU THR ALA GLU ASN GLU ARG LEU GLN LYS LYS VAL GLU }\\ \text{GLN LEU SER ARG GLU LEU SER THR LEU ARG ASN LEU PHE LYS GLN}\\ \text{LEU PRO GLU}\end{array}$

Separating the sequence of the macromolecule before every essential amino acid residue gives:

  $\begin{array}{l}VALLYSSERLYSALALYSLYSTHRVALASPLYSHISSER\\ ASPGLUTYRLYSILEARGARGGLUARGASNASNILE\\ ALAVALARGLYSSERARGASPLYSALALYSMETARG\\ ASNLEUGLUTHRGLNHISLYSVAL\end{array}$

  $\text{LEU GLU}$

  $\begin{array}{l}\text{LEU THR ALA GLU ASN GLU}\\ \text{ARG LEU GLN LYS LYS VAL }\\ \text{GLU GLNLEU SER ARG GLU}\end{array}$

  $\begin{array}{c}\text{LEU SER THR}\\ \text{LEU ARG ASN}\\ \text{LEU PHE LYS GLN}\\ \text{LEU PRO GLU}\end{array}$

This is often followed by a two residue repeats and 2 additional seven residue repeats. These 2 seven residue repetition begin the twenty eight residue essential amino acid zipper. Rearranging the sequence to allow 2 additional seven residue sequence gives:

  $\begin{array}{l}\text{VAL LYS SER LYS ALA }\\ \text{LYS LYS THR VAL ASP LYS HIS SER ASP GLU TYR }\\ \text{LYS ILE ARG ARG GLU ARG ASN ASN ILE ALA VAL ARG }\\ \text{LYS SER ARG ASP LYS ALA LYS MET ARG ASN}\end{array}$

  $\text{LEU GLU THR GLN HIS LYS VAL}$

  $\begin{array}{l}\text{LEU GLU}\\ \text{LEU THR ALA GLU ASN GLU ARG}\\ \text{LEU GLN LYS LYS VAL GLU GLN}\\ \text{LEU SER ARG GLU LEU SER THR}\\ \text{LEU ARG ASN LEU PHE LYS GLN}\\ \text{LEU PRO GLU}\end{array}$

The essential amino acid residues of the essential amino acid zipper are residues 48, 55, 62, and 69. Essential amino acid residue 76 might also participate within the essential amino acid zipper. All of those residues are on the C-terminal aspect of the chain along with the essential amino acid zipper.

Conclusion

An essential amino acid zipper could be a coiled-coil domain. Coiled-coil domains repeat each seven residues, thus we tend to expect one domain of the 1GU4 -helices to possess essential amino acid residues repetition each seven residues. Essential amino acid zipper domain contains 28 amino acids, thus we can expect four such repeats. The first essential amino acid repeat could be a seven residue repeat.

This is often followed by a two residue repeat and 2 additional seven residue repeats.