Gene mapping worksheet for ap biology | Study Guides, Projects ... - Free Printable

Absolutely! Let’s solve each of the four gene mapping problems step by step. In gene mapping, recombination frequency (RF) is directly proportional to the physical distance between two genes on a chromosome — 1% RF = 1 map unit (centimorgan, cM). The goal is to arrange the genes in linear order based on their pairwise distances.

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## Problem 1: Drosophila Genes A, B, C, S

Given recombination frequencies:
- 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: Find the largest distance → likely the ends

C/S = 11.0% → probably the two outermost genes.

Step 2: Use additive property

If genes are in order X–Y–Z, then distance X–Z = X–Y + Y–Z.

Check if any three genes satisfy this.

Try A–B–C:
- A/B = 2.5, B/C = 5.5 → A/C should be 2.5 + 5.5 = 8.0, but actual A/C = 3.0 → ✘ doesn’t fit.

Try B–A–C:
- B/A = 2.5, A/C = 3.0 → B/C should be 2.5 + 3.0 = 5.5 → ✔ matches!

So order so far: B – A – C with distances:
- B to A = 2.5 cM
- A to C = 3.0 cM
- B to C = 5.5 cM

Now place S.

We know:
- B/S = 5.5%
- A/S = 8.0%
- C/S = 11.0%

If we put S after C:
- C to S = 11.0 - 5.5 = 5.5? Wait, let’s check:

From B to C is 5.5 cM. If S is beyond C, then B to S = B to C + C to S → 5.5 + x = 5.5 → x=0? No.

Wait — B/S = 5.5%, same as B/C. So S could be on the other side of B?

Try placing S before B.

Order: S – B – A – C

Check distances:
- S to B = 5.5 → given ✔
- B to A = 2.5 → given ✔
- A to C = 3.0 → given ✔
- S to A = S-B + B-A = 5.5 + 2.5 = 8.0 → matches given A/S = 8.0% ✔
- S to C = S-B + B-A + A-C = 5.5 + 2.5 + 3.0 = 11.0 → matches C/S = 11.0% ✔

Perfect!

✔ Final Gene Map for Problem 1:

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

Total length: 11.0 cM

---

## Problem 2: Organism Z Genes P, Q, R, S

Given crossover frequencies:
- P & Q = 5%
- P & R = 8%
- P & S = 12%
- Q & R = 13%
- Q & S = 17%

Step 1: Largest distance = Q & S = 17% → likely ends.

Step 2: Try to build order using additive rule.

Try P – Q – R – S

Check:
- P-Q = 5
- Q-R = 13 → P-R should be 5+13=18, but given P-R=8 → ✘

Try Q – P – R – S

Distances:
- Q-P = 5
- P-R = 8 → Q-R = 5+8=13 ✔
- Now P-S = 12 → so R-S = ?
If P-R = 8, and P-S = 12, then R-S = 12 - 8 = 4? But we don’t have R-S directly.

Wait — Q-S = 17.

If Q–P–R–S:
- Q to P = 5
- P to R = 8 → Q to R = 13 ✔
- R to S = ? Let’s say x.
Then Q to S = Q-P + P-R + R-S = 5 + 8 + x = 13 + x = 17 → x = 4

But we don’t have R-S given. Is that acceptable? Yes, if it fits all known data.

Check P-S: P to R = 8, R to S = 4 → P to S = 12 ✔

Check Q-S: Q to P = 5, P to R = 8, R to S = 4 → total 17 ✔

All match!

✔ Final Gene Map for Problem 2:

```
Q —— 5 —— P —— 8 —— R —— 4 —— S
```

Total length: 17 cM

---

## Problem 3: Genes W, X, Y, Z

Recombination frequencies (per 100 eggs = %):

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

Step 1: Smallest distances indicate closest genes.

Look at Y–Z = 1 → very close.
X–Y = 2 → next closest.
X–Z = 3 → consistent: X-Y-Z should be 2 + 1 = 3 → ✔

So likely order: X – Y – Z with distances:
- X-Y = 2
- Y-Z = 1
- X-Z = 3

Now place W.

W-X = 5
W-Y = 7 → which is 5 + 2 → so W–X–Y → W to Y = W-X + X-Y = 5 + 2 = 7 ✔
W-Z = 8 → W-X + X-Y + Y-Z = 5 + 2 + 1 = 8 ✔

Perfect!

✔ Final Gene Map for Problem 3:

```
W —— 5 —— X —— 2 —— Y —— 1 —— Z
```

Total length: 8 cM

---

## Problem 4: Genes A, B, C, D

Recombination frequencies:

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

Step 1: Smallest distance = B–C = 3% → closest pair.

Next smallest: A–D = 4%

Then D–B = 8%, D–C = 11%, A–B = 12%, A–C = 15%

Try placing B–C together.

Suppose order: A – D – B – C

Check:
- A-D = 4 ✔
- D-B = 8 → A-B = A-D + D-B = 4 + 8 = 12 ✔
- B-C = 3 → D-C = D-B + B-C = 8 + 3 = 11 ✔
- A-C = A-D + D-B + B-C = 4 + 8 + 3 = 15 ✔

Perfect! All values match.

✔ Final Gene Map for Problem 4:

```
A —— 4 —— D —— 8 —— B —— 3 —— C
```

Total length: 15 cM

---

## 🧬 Summary of All Maps:

Problem 1:

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

Problem 2:

```
Q —— 5 —— P —— 8 —— R —— 4 —— S
```

Problem 3:

```
W —— 5 —— X —— 2 —— Y —— 1 —— Z
```

Problem 4:

```
A —— 4 —— D —— 8 —— B —— 3 —— C
```

Each map is constructed using the principle that recombination frequency ≈ distance in map units, and the additive property holds for linear gene order.

Let me know if you’d like these drawn visually or explained further!