2. Two kinds of groove — one closed, one open · Prof. Kwon's TCR-HLA and CD4/CD8, First Steps

There is not one kind of plate. There are two. And the difference between them is half the reason CD4 and CD8 exist as separate things.

Class I — HLA-A · B · C

One chain makes the whole groove (α1+α2).
A small β2-microglobulin props it up from below.
Present on almost every nucleated cell.
Coreceptor CD8

Class II — HLA-DR · DQ · DP

Two chains contribute half each (α1+β1).
Both chains cross the membrane.
Mostly on dendritic cells, macrophages, B cells.
Coreceptor CD4

You do not have to memorise the architecture. One sentence is enough.

The class I groove is capped at both ends; the class II groove is open at both ends

A capped groove dictates length

Conserved residues sit like plugs at both ends of the class I groove, and those plugs grip the first and last letter of the fragment. A fragment that is too long will not fit; one that is too short cannot reach both ends.

Class II has no plugs. The fragment hangs out at both ends, like a sausage longer than its bun. So the length constraint is far looser.

That is not a description of a cartoon. It is a measurable prediction. Let us check it against real data.

Class I is nailed to 9 residues; class II spreads out.

Python Runs in your browser

# How fragment length differs by class — counting distinct epitopes only
import thla

for cls in ["I", "II"]:
    L = thla.epitope_len(cls)
    print(f"class {cls}  {len(L):,} epitopes  median {int(L.median())} residues")
    h = L.value_counts().sort_index()
    for n, c in h.items():
        bar = "█" * max(1, int(60 * c / h.max()))
        print(f"   {n:>2} aa {c:>5}  {bar}")
    print()

For class I a single length, 9, takes 84%. That is what a plug looks like in data.

Python Runs in your browser

# What share is 9 residues
import thla

F = thla.facts()
for cls in ["I", "II"]:
    h = F["ep_len"][cls]
    tot = sum(h.values())
    top = max(h, key=lambda k: h[k])
    print(f"class {cls}: most common length {top} aa = {h[top]:,}/{tot:,} = {100*h[top]/tot:.1f}%")
    lo = min(int(k) for k in h)
    hi = max(int(k) for k in h)
    print(f"          length range {lo}-{hi} aa")

Make the fragment longer and shorter and drop it into both grooves. The capped one refuses.

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🤔 Check yourself

What is the most direct reason class I fragments cluster near 9 residues?

🤔 Check yourself

Why are class II fragments longer?

🤔 Check yourself

Which statement about where class I and class II are expressed is correct?