Lecture 3- Coding Capacity, Capsid Architecture, and Classification

Question 1

Viral classification based on structure

Answer 1

DNA:
helical-> naked or enveloped
cubical-> naked or enveloped
binal-> naked (two types of sum elements, T4 phage is an example)
DNA can be ss or ds, and then linear or circular

RNA:
helical-> naked or enveloped
cubical-> naked or enveloped
RNA can be ds or ss, not circular, ds all occur in segments
ss can be can be linear ss or segmented ss

Question 2

Key points concerning virus structure

Answer 2

1) Viruses have either helical or spherical (cubical) symmetry
2) Virus particles are uniform in shape and size (at least without an envelope; enveloped are irregular but genome and protein shell has symmetry)
3) Enveloped viruses often irregular, but nucleocapsid regular
4) Surface topology of capsids reflect morphological subunits

Question 3

General principle governing capsid construction

Answer 3

Viral capsids are composed of many copies of one or at most a few different kinds of protein molecules
Identical protein subunits called “protomers” or “structural subunits” (interchangeable)
protomers/structural units make up capsomeres or morphological units
capsomeres composed of 5-6 promoters (5= pentamer, 6= hexameter)

Question 4

Why does adenovirus have 12 spikes?

Answer 4

Each ball is a morphological subunit (can’t see promoters). Adenovirus has 12 spikes because there is one on each vertex/pentamer

Question 5

Basis for subunit theory of viral capsids

Answer 5

Coding capacity limitation and the crystallographer’s argument

Question 6

Coding capacity limitation

Answer 6

Viruses don’t have enough genetic potential to code for a protein that is large enough to enclose a genome

Question 7

The crystallographer’s argument

Answer 7

The most logical way to build a regular symmetrical structure (a virion) out of asymmetric subunits (protein molecules) is by the regular aggregation of many identical subunits. This is b/c viruses can crystallize easily; must be high level of reg to be able to crystallize easily

Question 8

Coding Capacity Argument

Answer 8

Coding capacity: the predicted sum of the molecular weights of all the proteins encoded in a viral genome. (if the entire genome coded for one protein, what would be its molecular weight?)
Coding capacity = how much protein a virus could code for

Question 9

Sample calculation coding capacity:

Answer 9

If molecular weight of a DNA is 10 x 10^6:
ds DNA: (10x 10^6) / 660 = 1.5 x 10^4 bp/DNA (nucleotides)
mRNA: divide result by 3 (b/c three nucleotides per codon)
get 5000 codons (5000 AA in protein that would be translated)
Average MW of an amino acid is approx 100; 100 x 5000 codons = 500,000
coding capacity = 500,000
visions greater than or equal to 50% protein by weight so sum of molecular weight of proteins must be greater than 10x10^6
Since coding capacity is only .5x10^6, need multiple copies of protein subunits

Question 10

What is the magic number?

Answer 10

60. Minimum number of viruses, at least w/ cubic symmetry

Question 11

3 points of the crystallographer’s argument

Answer 11

1) Viruses are highly regular structures. Naked capsid viruses tend to crystallize easily.
2) The protein structural subunits are by their very nature irregular or asymmetric
3) Crick and Watson reasoned from these observations that the easiest way to imagine how a virus is constructed is by the regular aggregation of many identical subunits.
Follows that each protein subunit will find itself in the same local environment as all of the other subunits (like a crystal)

Question 12

Construction of helical capsids

Answer 12

In TMV:
Regular in shape and distance from one part to next
RNA is inside coil; length determined by RNA
subunits
subunits make contact with each other; protein subunits on outside attached to RNA (hypothetical helical arrangement of protomers)

Question 13

Architecture of viruses with cubic symmetry

Answer 13

Viruses have adapted to another solid structure related to the icosahedron (12 vertices, 20 faces, and 30 sides):
The Icosadeltahedron
Derived from icosahedron by triangulating the 20 faces according to specific rules
The number of triangles into which each face is subdivided is called the triangulation number or T
allowable T values are: 1,3,4,7,9,12,13,16
Only viruses with small genomes use icosahedron

Question 14

Placement of structural subunits on surface of icosahedron

Answer 14

1) Structural subunits (protomers) are assembled into morphological subunits (capsomeres) containing 5 units (called pentamers)
2) Place a pentamer at each one of the 12 vertices of the icosahedron
T:1
Pentamers: 12
Hexamers: 0
Total capsomeres: 12
Total protomers: 60 (b/c 12 x 5)
T1 structure: Based on icosahedron, place pentamer at each of 12 pentamers

Question 15

Rules for deriving the family of icosadeltahedron (T=3)

Answer 15

Consider one triangular face of icosahedron
For T3, can divide the triangular face into three triangles
1) Place a pentamer at each one of the original 12 vertices of the icosahedron.
2) At each new vertex that is a generated place, place a hexamer
T:1,3
Pentamers: 12, 12
Hexamers: 0, 20
Total capsomers: 12, 32
Total protomers: 60, 180
(180 comes from 12 x 5, 20 x 60)

Question 16

Icosadeltahedron with T = 4

Answer 16

Again, with one triangular face of the icosahedron
Divide triangular face into 4 triangles
1) place a pentamer at each one of the original 12 vertices of the icosahedron
2) at each new vertex that is generated place, place a hexamer
T1 & 3: the same
T4: 
Pentamers: 12
Hexamers: 30
Total capsomers: 42
Total protomers: 
Total capsomers: (10 x T) + 2
Total protomers: (60 x T)

Question 17

General rule for constructing icosadeltahedra:

Answer 17

Pentamers: 12
Hexamers: x
Total capsomers: 10T + 2
Total protomers: 60T