Explore Related Concepts


Best Results From Wikipedia Yahoo Answers Youtube
From Wikipedia
A common motif in the secondary structure of proteins, the alpha helix (Î±helix) is a righthanded coiled or spiral conformation, in which every backbone NH 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 PaulingCoreyBranson 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 Xrayfiber 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 ~5.1Ã…nm.
Astbury initially proposed a kinkedchain 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 Î²form).
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 Î²strand (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 lefthanded helices, although in 1960 the crystal structure of myoglobin showed that the righthanded 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 righthanded 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 1.5Ã…nmabbr=on along the helical axis. (Short pieces of lefthanded helix sometimes occur with a large content of achiral glycine amino acids, but are unfavorable for the other normal, biological Damino acids.) The pitch of the alphahelix (the vertical distance between two consecutive turns of the helix) is 5.4Ã…nmabbr=on which is the product of 1.5 and 3.6. What is most important is that the NH 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 alphahelices can be identified in protein structure using several computational methods, one of which is DSSP (Dictionary of Protein Secondary Structure).
Similar structures include the 3_{10} 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 3.6_{13} helix, since the i + 4 spacing adds 3 more atoms to the Hbonded loop compared to the tighter 3_{10} 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
 This article is about the Rodrigues' rotation formula, which is distinct fromEulerâ€“Rodrigues parameters and The EulerRodrigues formula for 3D rotation.
In the theory of threedimensional 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 axisangle 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 righthanded 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 = (k_{X}, k_{Y}, k_{Z}), and a vector v = (v_{X}, v_{Y}, v_{Z}) 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 v_{y} is null (see figure). However, xyz does not coincide with the reference frame XYZ in which vectors v, k, v_{x}, and v_{z} are actually represented. For instance, v = (v_{X}, v_{Y}, v_{Z}) â‰ (v_{x}, v_{y}, v_{z}). 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 v_{x} 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 v_{x}, 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 v_{x} rotated by 90Â° about z. Using trigonometry, we can now rotate v_{x} by Î¸ around z to obtain v_{x rot}. Its two components with respect of x and y are v_{x}cosÎ¸ 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}
v_{x rot} can be also described as the projection on xy (or rejection from z) of the rotated vector v_{rot} (see figure). Since v_{z} is obviously not affected by a rotation about z, the other component of v_{rot} (i.e. its projection on z) coincides with v_{z}. 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 "crossproduct 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
From Yahoo Answers
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
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
Answers:Clearly cosA = 4/5 and cosB =  12/13. Cos(A+B) = (4/5)((12/13)  (3/5)(5/13) =  63/65 tan(AB) = {(3/4)(5/12)}/{1+(3/4)(5/12)} now simplify it.
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^_^
From Youtube