19. 【Capstone lab】 Who can see this fragment? · Prof. Kwon's TCR-HLA and CD4/CD8, First Steps

Last lab. It gathers the previous eighteen lessons into one question.

If you build a therapy against this fragment, how many people can use it?

Three pieces are needed.

What we need	Where from	Which lesson
Which plate this fragment rides on	VDJdb	7
What share of people carry that plate	cited population frequencies	this one
Whether a different plate really means not seeing it	the polymorphism of lessons 8 and 9	8

Let us stack them in order.

The same skew as lesson 7. This time we attach people to it.

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# Step 1 — recount the plate rankings (same calculation as lesson 7)
import thla

r = thla.restriction("I", top=8).rename(columns={"줄": "rows", "에피토프": "fragments"})
tot = len(thla.vdj("I"))
print(f"{'plate':<16}{'rows':>9}{'%':>7}{'fragments':>11}")
print("-" * 44)
for a, row in r.iterrows():
    print(f"{a:<16}{int(row['rows']):>9,}{row['%']:>7.1f}{int(row['fragments']):>11}")
print()
print(f"The single top plate is {r['%'].iloc[0]:.0f}% of all {tot:,} rows.")

A person carries two HLA-A copies, so 'people carrying' exceeds the allele frequency.

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# Step 2 — what share of people carry that plate (cited figures)
import thla

f = thla.facts()
print(f["cited_note_en"])
print()
print(f"{'allele':<16}{'people carrying %':>18}")
print("-" * 36)
for a, p in sorted(f["freq"].items(), key=lambda kv: -kv[1]):
    print(f"{a:<16}{p:>18.1f}  " + "█" * max(1, int(p)))

This is why TCR therapy starts at A*02:01, and why stopping there is not acceptable.

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# Step 3 — stack several plates: what share do they cover
import thla

order = ["HLA-A*02:01", "HLA-A*01:01", "HLA-A*03:01", "HLA-A*24:02", "HLA-A*11:01"]
print(f"{'plate added':<16}{'cumulative coverage %':>24}")
print("-" * 42)
for i in range(1, len(order) + 1):
    print(f"{order[i-1]:<16}{thla.covered(order[:i]):>24.1f}")
print()
print("One is not even half. It takes five to scrape past it.")
print("This assumes the alleles are independent of each other - see the box below.")

Pick plates and stack them. Count for yourself how many it takes to pass half.

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It shows how coarse a single line of 'A*02 positive' is as a trial criterion.

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# Step 4 — the same 'A*02' is not the same plate
import thla

h = thla.hla("A")
groove = h[h["domain"].isin(["α1", "α2"])]
hot = groove.nlargest(8, "entropy")[["pos", "domain", "n_aa", "entropy"]]
print("the most divided positions in the HLA-A groove")
print(hot.to_string(index=False))
print()
f = thla.facts()
print(f"people carrying A*02:01 {f['freq']['HLA-A*02:01']}%")
print(f"people carrying A*02:07 {f['freq']['HLA-A*02:07']}%  (common in East and Southeast Asia)")
print()
print("The two names are the same A*02 family, but positions inside the groove differ.")
print("A TCR built on A*02:01 is not guaranteed to work on A*02:07.")

So what this one picture says

This is the shape of the table

More than half the records pile onto one plate. Most of the 1,946 fragments are things loaded on A*02:01.

This is the shape of people

About a quarter carry A*02:01 — and that share varies widely by ancestry.

So here is the work

Somebody has to run experiments finding fragments on the less-studied plates. More algorithm will not do it — what is not in the table cannot be computed out of it.

This is where lesson 6's sentence arrives. The skew of this table is the skew of researchers, not of the immune system. And that skew is not merely a statistical matter; it becomes a question of who gets treated.

🤔 Check yourself

A*02:01 alone covers well under half of people. Why is it more than half the table?

🤔 Check yourself

What goes wrong when you multiply and add coverage across several alleles?

🤔 Check yourself

What is the risk of using 'A*02 positive' as a trial criterion?