Gene mapping worksheet featuring tables with crossover frequencies for different gene combinations, used to construct chromosome maps.

Let's solve each problem step-by-step in the Gene Mapping Worksheet. The goal is to use recombination frequencies (which reflect genetic distance between genes) to construct chromosome maps.

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Problem 1: Drosophila Gene Mapping

We are given recombination frequencies between genes A, B, C, and S:

| Gene Combination | Recombination Frequency |
|------------------|-------------------------|
| A/B | 2.5% |
| A/C | 3.0% |
| B/C | 5.5% |
| B/S | 5.5% |
| A/S | 8.0% |
| C/S | 11.0% |

#### Step 1: Understand the relationships
Recombination frequency ≈ map distance in centimorgans (cM).
We want to place genes A, B, C, and S on a linear map based on these distances.

#### Step 2: Use pairwise distances

Let’s look at the smallest distances:
- A/B = 2.5%
- A/C = 3.0%
- B/C = 5.5%

Note that A/B + A/C = 2.5 + 3.0 = 5.5 → exactly matches B/C.
This suggests that A lies between B and C, with:
- B — 2.5 cM — A — 3.0 cM — C
So: B — A — C, total B–C = 5.5 cM ✔

Now check other distances:
- B/S = 5.5%
- A/S = 8.0%
- C/S = 11.0%

We already have:
- B — A — C (with B–A = 2.5, A–C = 3.0)

Now consider where S fits.

Try placing S relative to B and C.

We know:
- B/S = 5.5% → same as B/C → so S could be near C or on opposite side?
But C/S = 11.0%, which is larger than B/C.

Also A/S = 8.0%. Since A is 2.5 from B and 3.0 from C, let’s try placing S beyond C.

Suppose order: B — A — C — S
Then:
- A/S = A–C + C–S = 3.0 + ? = 8.0 → C–S = 5.0? But we have C/S = 11.0% → too small.

Wait: C/S = 11.0%, so if C–S = 11.0, then A/S = A–C + C–S = 3.0 + 11.0 = 14.0 ≠ 8.0 → no.

Try: B — A — S — C?

Then:
- A/S = ?
- B/S = ?
- C/S = ?

But we don’t have direct info.

Alternatively, try placing S relative to B.

We have:
- B/S = 5.5%
- A/S = 8.0%
- A/B = 2.5%

So if A and B are 2.5 apart, and A/S = 8.0, B/S = 5.5 → this suggests S is not on the same side of A and B.

Let’s suppose order: B — A — S, then:
- B/A = 2.5
- A/S = 8.0 → B/S = 2.5 + 8.0 = 10.5 ≠ 5.5 → too big.

What if S is between B and A?

Then B/S + S/A = B/A → 5.5 + x = 2.5 → impossible.

So S cannot be between B and A.

Try: S — B — A — C

Then:
- B/S = 5.5
- B/A = 2.5 → so S–B = 5.5 → S–A = 5.5 + 2.5 = 8.0 → matches A/S = 8.0 ✔
- A/C = 3.0 → so A–C = 3.0 → C–S = S–B + B–A + A–C = 5.5 + 2.5 + 3.0 = 11.0 → matches C/S = 11.0 ✔

Perfect!

So the order is: S — B — A — C

Distances:
- S–B = 5.5 cM
- B–A = 2.5 cM
- A–C = 3.0 cM

Total S–C = 5.5 + 2.5 + 3.0 = 11.0 cM → matches C/S = 11.0%

All values match.

✔ Final map for Problem 1:

```
S — 5.5 — B — 2.5 — A — 3.0 — C
```

---

Problem 2: Construct a chromosome map for genes P, Q, R, S

Given crossover frequencies:

| Genes | Crossover Frequency |
|-----------|---------------------|
| P & Q | 5% |
| P & R | 8% |
| P & S | 12% |
| Q & R | 13% |
| Q & S | 17% |

We need to find the correct order.

Start with the closest pair: P & Q = 5%

Next: P & R = 8% → R is farther from P than Q is.

Now, Q & R = 13%

Check: If P–Q = 5%, P–R = 8%, then Q–R should be |8 – 5| = 3% if they are on same side, or 8+5=13% if on opposite sides.

We observe Q–R = 13% → so Q and R are on opposite sides of P

Thus: Q — 5 — P — 8 — R → Q–R = 5 + 8 = 13% ✔

Now add S.

P & S = 12% → S is 12 cM from P.

Q & S = 17% → Q–S = 17%

Current positions:
- Q at 0
- P at 5
- R at 13

Now place S.

If S is on the side of R (beyond R), then P–S = 12 → S at 5 + 12 = 17 → then Q–S = 17 → matches Q&S = 17% ✔

So S is at 17.

Check:
- P–S = 12 → yes (from 5 to 17)
- Q–S = 17 → yes (0 to 17)
- P–R = 8 → R at 13 → 13 - 5 = 8 → yes
- Q–R = 13 → 13 - 0 = 13 → yes

So order: Q — 5 — P — 8 — R — 4 — S
(S is 4 cM beyond R)

Wait: P–S = 12 → from P at 5 to S at 17 → difference = 12 → yes.

But R is at 13 → so R–S = 17 - 13 = 4 cM

Final map:

```
Q — 5 — P — 8 — R — 4 — S
```

Or equivalently:

```
Q — 5 — P — 8 — R — 4 — S
```

All distances match.

✔ Problem 2 Map:

```
Q — 5 — P — 8 — R — 4 — S
```

---

Problem 3: Linkage group W, X, Y, Z

Given recombination frequencies per 100 eggs (i.e., %):

| Genes | W | X | Y | Z |
|-------|---|---|---|---|
| W | – | 5 | 7 | 8 |
| X | 5 | – | 2 | 3 |
| Y | 7 | 2 | – | 1 |
| Z | 8 | 3 | 1 | – |

We interpret this table: the number is the recombination frequency between two genes.

So:
- W–X = 5%
- W–Y = 7%
- W–Z = 8%
- X–Y = 2%
- X–Z = 3%
- Y–Z = 1%

Step 1: Find the closest pair → Y–Z = 1% → very close.

Then X–Y = 2%, X–Z = 3% → so X is near Y and Z.

W–X = 5%, W–Y = 7%, W–Z = 8%

Let’s try to build around Y–Z.

Since Y–Z = 1%, assume order: Y — 1 — Z

Now X–Y = 2%, X–Z = 3%

If X is on one side of Y-Z, say left of Y:

Then X–Y = 2 → X at 0, Y at 2, Z at 3 → then X–Z = 3 → matches ✔

Now W–X = 5 → W is 5 cM from X

W–Y = 7 → W–Y = 5 + 2 = 7 → so W is on same side as X → W–X = 5 → W at 0 - 5 = -5?

Set positions:

Let’s fix X at position 0.

Then:
- X–Y = 2 → Y at 2
- X–Z = 3 → Z at 3
- Y–Z = 1 → 3 - 2 = 1 → yes ✔

Now W–X = 5 → W at 5 or -5

W–Y = 7 → distance from W to Y must be 7

If W at 5 → W–Y = |5 - 2| = 3 ≠ 7 → no

If W at -5 → W–Y = |-5 - 2| = 7 → yes ✔

W–Z = |-5 - 3| = 8 → matches ✔

So positions:
- W: -5
- X: 0
- Y: 2
- Z: 3

Order: W — 5 — X — 2 — Y — 1 — Z

✔ Final map:

```
W — 5 — X — 2 — Y — 1 — Z
```

---

Problem 4: Genes A, B, C, D

Recombination frequencies:

| Genes | A | B | C | D |
|-------|---|---|---|---|
| A | – |12 |15 | 4 |
| B |12 | – | 3 | 8 |
| C |15 | 3 | – |11 |
| D | 4 | 8 |11 | – |

So:
- A–B = 12%
- A–C = 15%
- A–D = 4%
- B–C = 3%
- B–D = 8%
- C–D = 11%

Step 1: Closest pair → B–C = 3% → very close.

Then A–D = 4% → also close.

Now see how others relate.

Try to place B and C first.

Assume: B — 3 — C

Now A–B = 12%, A–C = 15%

If A is on one side of B–C:

Say A–B = 12 → then A–C = 12 + 3 = 15 → matches ✔

So A is on the side of B away from C.

So order: A — 12 — B — 3 — C

Now D: A–D = 4%, B–D = 8%, C–D = 11%

Try placing D.

If D is near A: A–D = 4 → D at 4 from A.

Current positions:
- A: 0
- B: 12
- C: 15

Then D at 4 → D–B = |12 - 4| = 8 → matches B–D = 8% ✔

D–C = |15 - 4| = 11 → matches C–D = 11% ✔

Perfect!

So positions:
- A: 0
- D: 4
- B: 12
- C: 15

Order: A — 4 — D — 8 — B — 3 — C

Wait: D to B is 12 - 4 = 8 → yes

So final map:

```
A — 4 — D — 8 — B — 3 — C
```

✔ All distances match:
- A–D = 4 ✔
- D–B = 8 ✔
- B–C = 3 ✔
- A–B = 12 ✔
- A–C = 15 ✔
- C–D = 11 ✔

---

✔ Final Answers:

#### 1. Drosophila gene map:
```
S — 5.5 — B — 2.5 — A — 3.0 — C
```

#### 2. Organism Z gene map:
```
Q — 5 — P — 8 — R — 4 — S
```

#### 3. Linkage group W,X,Y,Z:
```
W — 5 — X — 2 — Y — 1 — Z
```

#### 4. Linkage group A,B,C,D:
```
A — 4 — D — 8 — B — 3 — C
```

Each map is consistent with all given recombination frequencies.

Let me know if you'd like visual diagrams!