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# helix angle formula

From Wikipedia

Alpha helix

A common motif in the secondary structure of proteins, the alpha helix (Î±-helix) is a right-handed coiled or spiral conformation, in which every backbone N-H group donates a hydrogen bond to the backbone C=O group of the amino acid four residues earlier (i+4 \rightarrow i hydrogen bonding). This secondary structure is also sometimes called a classic Pauling-Corey-Branson alpha helix (see below). Among types of local structure in proteins, the Î±-helix is the most regular and the most predictable from sequence, as well as the most prevalent.

## Historical development

In the early 1930s, William Astbury showed that there were drastic changes in the X-rayfiber diffraction of moist wool or hair fibers upon significant stretching. The data suggested that the unstretched fibers had a coiled molecular structure with a characteristic repeat of ~5.1|Ã…|nm.

Astbury initially proposed a kinked-chain structure for the fibers. He later joined other researchers (notably the American chemist Maurice Huggins) in proposing that:

• the unstretched protein molecules formed a helix (which he called the Î±-form); and
• the stretching caused the helix to uncoil, forming an extended state (which he called the Î²-form).

Although incorrect in their details, Astbury's models of these forms were correct in essence and correspond to modern elements of secondary structure, the Î±-helix and the Î²-strand (Astbury's nomenclature was kept), which were developed by Linus Pauling, Robert Corey and Herman Branson in 1951 (see below); that paper showed both right- and left-handed helices, although in 1960 the crystal structure of myoglobin showed that the right-handed form is the common one. Hans Neurath was the first to show that Astbury's models could not be correct in detail, because they involved clashes of atoms. It is interesting to note that Neurath's paper and Astbury's data inspired H. S. Taylor, Maurice Huggins and Bragg and collaborators to propose models of keratin that somewhat resemble the modern Î±-helix.

Two key developments in the modeling of the modern Î±-helix were (1) the correct bond geometry, thanks to the crystal structure determinations of amino acids and peptides and Pauling's prediction of planarpeptide bonds; and (2) his relinquishing of the assumption of an integral number of residues per turn of the helix. The pivotal moment came in the early spring of 1948, when Pauling caught a cold and went to bed. Being bored, he drew a polypeptide chain of roughly correct dimensions on a strip of paper and folded it into a helix, being careful to maintain the planar peptide bonds. After a few attempts, he produced a model with physically plausible hydrogen bonds. Pauling then worked with Corey and Branson to confirm his model before publication. In 1954 Pauling was awarded his first Nobel Prize "for his research into the nature of the chemical bond and its application to the elucidation of the structure of complex substances"[http://nobelprize.org/nobel_prizes/chemistry/laureates/1954/] (such as proteins), prominently including the structure of the Î±-helix.

## Structure

### Geometry and hydrogen bonding

The amino acids in an Î± helix are arranged in a right-handed helical structure where each amino acid residue corresponds to a 100Â° turn in the helix (i.e., the helix has 3.6 residues per turn), and a translation of 1.5|Ã…|nm|abbr=on along the helical axis. (Short pieces of left-handed helix sometimes occur with a large content of achiral glycine amino acids, but are unfavorable for the other normal, biological D-amino acids.) The pitch of the alpha-helix (the vertical distance between two consecutive turns of the helix) is 5.4|Ã…|nm|abbr=on which is the product of 1.5 and 3.6. What is most important is that the N-H group of an amino acid forms a hydrogen bond with the C=O group of the amino acid four residues earlier; this repeated i+4 \rightarrow i hydrogen bonding is the most prominent characteristic of an Î±-helix. Official international nomenclature [http://www.chem.qmul.ac.uk/iupac/misc/ppep1.html] specifies two ways of defining Î±-helices, rule 6.2 in terms of repeating Ï†,Ïˆ torsion angles (see below) and rule 6.3 in terms of the combined pattern of pitch and hydrogen bonding. The alpha-helices can be identified in protein structure using several computational methods, one of which is DSSP (Dictionary of Protein Secondary Structure).

Similar structures include the 310 helix (i+3 \rightarrow i hydrogen bonding) and the Ï€-helix (i+5 \rightarrow i hydrogen bonding). The Î± helix can be described as a 3.613 helix, since the i + 4 spacing adds 3 more atoms to the H-bonded loop compared to the tighter 310 helix. The subscripts refer to the number of atoms (including the hydrogen) in the closed loop formed by the hydrogen bond.

Residues in Î±-helices typically adopt backbone (Ï†, Ïˆ) dihedral angles around (-60Â°, -45Â°), as shown in the image at right. In more general terms, they adopt dihedral angles such that the Ïˆ dihedral angle of one residue and the Ï† dihedral angle of the next residue sum to roughly -105Â°. As a consequence, Î±-helical dihedral angles, in general, fall on a diagonal stripe on the Ramachandran diagram (of slope -1), ranging from (-90Â°, -15Â

Rodrigues' rotation formula

In the theory of three-dimensional rotation, Rodrigues' rotation formula (named after Olinde Rodrigues) is an efficient algorithm for rotating a vector in space, given an axis and angle of rotation. By extension, this can be used to transform all three basis vectors to compute a rotation matrix from an axis-angle representation. In other words, the Rodrigues formula provides an algorithm to compute the exponential map from so(3) to SO(3) without computing the full matrix exponent.

If v is a vector in \mathbb{R}^3 and k is a unit vector describing an axis of rotation about which we want to rotate v by an angle Î¸ (in a right-handed sense), the Rodrigues formula is:

\mathbf{v}_\mathrm{rot} = \mathbf{v} \cos\theta + (\mathbf{k} \times \mathbf{v})\sin\theta + \mathbf{k} (\mathbf{k} \cdot \mathbf{v}) (1 - \cos\theta).

## Derivation

Given a rotation axis z represented by a unit vectork = (kX, kY, kZ), and a vector v = (vX, vY, vZ) that we wish to rotate about z, the vector

\mathbf{v}_z = (\mathbf{k} \cdot \mathbf{v}) \mathbf{k}

is the component of v parallel to z, also called the vector projection of v on k, and the vector

\mathbf{v}_x = \mathbf{v} - \mathbf{v}_z = \mathbf{v} - (\mathbf{k} \cdot \mathbf{v}) \mathbf{k}

is the projection of v onto the xy plane orthogonal to z, also called the vector rejection of v from k.

Notice that we chose to define a reference framexyz in which the z axis is aligned with the rotation axis, and the x axis with the rejection of v from k. This simplifies the demonstration, as it implies that v lies on the xz plane, and its component vy is null (see figure). However, xyz does not coincide with the reference frame XYZ in which vectors v, k, vx, and vz are actually represented. For instance, v = (vX, vY, vZ) â‰  (vx, vy, vz). In other words, Rodrigues' formula is independent of the orientation in space of the reference frame XYZ in which v and k are represented.

Next let

\mathbf{w} = \mathbf{k}\times\mathbf{v}.

Notice that vx and w have the same length. By definition of the cross product, the length of w is:

|\mathbf{w}| = |\mathbf{k} \times \mathbf{v}| = |\mathbf{k}| \, |\mathbf{v}| \sin \phi \

where Î¦ denotes the angle between z and v. Since k has unit length,

|\mathbf{w}| = |\mathbf{k} \times \mathbf{v}| = |\mathbf{v}| \sin \phi.

This coincides with the length of vx, computed trigonometrically as follows:

|\mathbf{v}_x| = |\mathbf{v}| \cos(\pi/2-\phi) = |\mathbf{v}| \sin \phi.

So w can be viewed as a copy of vx rotated by 90Â° about z. Using trigonometry, we can now rotate vx by Î¸ around z to obtain vx rot. Its two components with respect of x and y are vxcosÎ¸ and wsinÎ¸, respectively. Thus,

\begin{align} \mathbf{v}_{x\ \mathrm{rot}} &= \mathbf{v}_x\cos\theta + \mathbf{w}\sin\theta\\ &= (\mathbf{v} - (\mathbf{k} \cdot \mathbf{v}) \mathbf{k})\cos\theta + (\mathbf{k} \times \mathbf{v})\sin\theta. \end{align}

vx rot can be also described as the projection on xy (or rejection from z) of the rotated vector vrot (see figure). Since vz is obviously not affected by a rotation about z, the other component of vrot (i.e. its projection on z) coincides with vz. Thus,

\begin{align} \mathbf{v}_{\mathrm{rot}} &= \mathbf{v}_{x\ \mathrm{rot}} + \mathbf{v}_{z\ \mathrm{rot}} \\ &= \mathbf{v}_{x\ \mathrm{rot}} + \mathbf{v}_z \\ &= (\mathbf{v} - (\mathbf{k} \cdot \mathbf{v}) \mathbf{k}) \cos\theta + (\mathbf{k} \times \mathbf{v})\sin\theta + (\mathbf{k} \cdot \mathbf{v}) \mathbf{k} \\ &= \mathbf{v} \cos\theta + (\mathbf{k} \times \mathbf{v})\sin\theta + \mathbf{k} (\mathbf{k} \cdot \mathbf{v}) (1 - \cos\theta), \end{align}

as required.

### Matrix notation

By representing v and k as column matrices, and defining a matrix [\mathbf{k}]_\times as the "cross-product matrix" for \mathbf{k}, i.e.,

[\mathbf{k}]_\times \mathbf{v} = \mathbf{k}\times\mathbf{v} =

\left[\begin{array}{ccc} 0 & -k_3 & k_2 \\ k_3 & 0 & -k_1 \\ -k_2 & k_1 & 0 \end{array}\right]\mathbf{v} ,

Rodrigues' formula can be written in matrix notation:

\begin{align} \mathbf{v}_{\mathrm{rot}} &= \mathbf{v} \cos\theta + ([\mathbf{k}]_\times \mathbf{v}) \sin\theta + \mathbf{k} (\mathbf{k}^\top \mathbf{v}) (1 - \cos\theta) \\ &= \mathbf{v} \cos\theta + [\mathbf{k}]_\times \mathbf{v} \sin\theta + \mathbf{k} \mathbf{k}^\top \mathbf{v} (1 - \cos\theta). \end{align}

## Conversion to rotation matrix

The equation can be also written as

\begin{align} \mathbf{v}_{\mathrm{rot}} &= (I\cos\theta) \mathbf{v} + ([\mathbf{k}]_\times \sin\theta) \mathbf{v} + (1 - \cos\theta) \mathbf{k} \mathbf{k}^\top \mathbf{v} \\ &= \left( I \cos\theta + [\mathbf{k}]_\times \sin\theta + (1 - \cos\theta) \mathbf{k} \mathbf{k}^T \right) \mathbf{v}\\ &= R\mathbf{v} \end{align}

where I is the 3Ã—3 identity matrix. Thus we have a formula for the rotation matrix R corresponding t

Question:How do you use the exact angle formula to find cos (105)? thank you!!!

Answers:cos(105) = cos(90 + 15) = cos90cos15 - sin90sin15 = -sin15 We now need to find the exact value of sin15 Now sin15 = sin(45 - 30) = sin45cos30 - cos45sin30 = (1/sqrt(2))*(sqrt(3)/2) - (1/sqrt(2))*(1/2) = (sqrt(3)/2sqrt(2)) - 1/2sqrt(2) = (sqrt(6) - sqrt(2)) / 4 Hence -sin15 = (sqrt(2) - sqrt(6))/4 But since cos105 = -sin15 ==> cos105 = (sqrt(2) - sqrt(6)) / 4

Question:1. how do i show an evaluation of sin 120 degrees using the trigonometric addition formulas? 2. how do i show an evaluation of sin 120 degrees using the double-angle formulas?

Answers:sin (120) = sin (60 + 60) = sin 60 cos 60 + cos 60 sin 60 You can evaluate that, right ? sin (120) = sin (2[60]) = 2 sin 60 cos 60 Go again ! sin (120) = ( 3) / 2 QED

Question:If A is the acute angle such that sin A= 3/5 and B is the obtuse angle such that sin B= 5/13, find without using a calculator the values of cos(A+B) and tan (A-B)

Answers:Clearly cosA = 4/5 and cosB = - 12/13. Cos(A+B) = (4/5)((-12/13) - (3/5)(5/13) = - 63/65 tan(A-B) = {(3/4)-(-5/12)}/{1+(3/4)(-5/12)} now simplify it.

Question:Given an isosceles triangle with a base side a and two equal sides b and b' and a height of h, what is the formula to calculate the angle between b and b', using a and h?

Answers:h is perpendicular to a, so they form 2 right triangles. The 2 sides are congruent, and h is congruent to itself. Thus, the 2 triangles are congruent by HL Theorem. Using CPCTC and def'n of midpoint, the intersection of h and a is the midpoint of a, and using CPCTC again, the "upper angles" are congruent and are half the vertex angle of the isosceles triangle. Let x = angle between b and b' Using trigonometric functions, tan (x/2) = (a/2)/h tan (x/2) = a/2h x/2 = arctan (a/2h) x = 2 arctan (a/2h) QED Hope this helps^_^