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### Problem 6
#### Part a
# Explanation:
## Step1: Recall incomplete dominance
In incomplete dominance, heterozygous shows intermediate phenotype. Given $ RR $ (homozygous red, since $ R $ is red allele), so phenotype is red.
# Answer:
Red (assuming $ RR $ is red, as heterozygous $ Rr $ is pink, so $ RR $ is red)

#### Part b
# Explanation:
## Step1: Identify genotype $ Rr $
Given heterozygous $ Rr $ is pink (from problem statement: "the heterozygous genotype is expressed as pink").
# Answer:
Pink

#### Part c
# Explanation:
## Step1: Analyze genotype $ rr $
$ rr $ is homozygous for white allele ($ r $ is white), so phenotype is white.
# Answer:
White

### Problem 7
# Explanation:
## Step1: Determine genotypes
Pink - flowered plant: $ Rr $ (heterozygous, incomplete dominance). White - flowered plant: $ rr $.
## Step2: Set up Punnett square
|   | $ R $ | $ r $ |
|---|---|---|
| $ r $ | $ Rr $ | $ rr $ |
| $ r $ | $ Rr $ | $ rr $ |
## Step3: Calculate probability
Out of 4 offspring, 2 are $ Rr $ (pink). Probability $ = \frac{2}{4} \times 100 = 50\% $.
# Answer:
50

### Problem 8
# Explanation:
## Step1: Identify possible crosses
Possible crosses: $ RR \times Rr $, $ Rr \times Rr $, $ RR \times rr $, $ Rr \times rr $.
## Step2: Analyze $ Rr \times Rr $ cross (Punnett square)
|   | $ R $ | $ r $ |
|---|---|---|
| $ R $ | $ RR $ | $ Rr $ |
| $ r $ | $ Rr $ | $ rr $ |
## Step3: Count pink ($ Rr $) offspring
Out of 4, 2 are $ Rr $ (pink). For $ RR \times Rr $:
|   | $ R $ | $ R $ |
|---|---|---|
| $ R $ | $ RR $ | $ RR $ |
| $ r $ | $ Rr $ | $ Rr $ |
Here, 2 out of 4 are pink (same as $ Rr \times Rr $)? Wait, no, wait: Wait, original problem: in snapdragons, $ RR $ is red, $ Rr $ is pink, $ rr $ is white. Wait, no, wait the problem says "the two alleles are red (R) and white (r). The heterozygous genotype is expressed as pink." So $ RR $: red, $ Rr $: pink, $ rr $: white.

Wait, so cross to get most pink: $ RR \times rr $ gives all $ Rr $ (pink)? Wait no: $ RR $ (red) x $ rr $ (white) gives $ Rr $ (pink) all? Wait, no: $ RR $ x $ rr $:
|   | $ R $ | $ R $ |
|---|---|---|
| $ r $ | $ Rr $ | $ Rr $ |
| $ r $ | $ Rr $ | $ Rr $ |
Oh! Wait, I made a mistake earlier. $ RR $ x $ rr $ gives 4 $ Rr $ (all pink). Wait, that's different. Wait, let's re - evaluate.

Wait, the problem says "the two alleles are red (R) and white (r). The heterozygous genotype is expressed as pink." So:

- $ RR $: red (homozygous red)
- $ Rr $: pink (heterozygous)
- $ rr $: white (homozygous white)

So cross $ RR $ (red) x $ rr $ (white):

Punnett square:

|   | $ R $ | $ R $ |
|---|---|---|
| $ r $ | $ Rr $ | $ Rr $ |
| $ r $ | $ Rr $ | $ Rr $ |

All 4 offspring are $ Rr $ (pink).

Cross $ Rr $ x $ Rr $:

|   | $ R $ | $ r $ |
|---|---|---|
| $ R $ | $ RR $ | $ Rr $ |
| $ r $ | $ Rr $ | $ rr $ |

Here, 2 out of 4 are pink ($ Rr $), 1 red ($ RR $), 1 white ($ rr $).

Cross $ RR $ x $ Rr $:

|   | $ R $ | $ R $ |
|---|---|---|
| $ R $ | $ RR $ | $ RR $ |
| $ r $ | $ Rr $ | $ Rr $ |

2 out of 4 are pink ($ Rr $), 2 red ($ RR $).

Cross $ Rr $ x $ rr $:

|   | $ R $ | $ r $ |
|---|---|---|
| $ r $ | $ Rr $ | $ rr $ |
| $ r $ | $ Rr $ | $ rr $ |

2 out of 4 are pink ($ Rr $), 2 white ($ rr $).

So the cross $ RR \times rr $ produces all pink - flowered plants (4 out of 4), which is more than the other crosses. Wait, but the problem says "produce the most pink - flowered plants". So $ RR \times rr $ cross:

Punnett square:

|   | $ R $ | $ R $ |
|---|---|---|
| $ r $ | $ Rr $ | $ Rr $ |
| $ r $ | $ Rr $ | $ Rr $ |

All offspring are $ Rr $ (pink), so this cross produces the most pink - flowered plants.

# Answer:
The cross between a homozygous red - flowered plant ($ RR $) and a homozygous white - flowered plant ($ rr $) will produce the most pink - flowered plants. The Punnett square for $ RR\times rr $ is as shown above, and all offspring have the genotype $ Rr $, which is expressed as pink flower color.

### Problem 9
# Explanation:
## Step1: Determine genotypes and phenotypes of parents
Homozygous black bird: genotype $ BB $, phenotype: black. Homozygous white bird: genotype $ bb $, phenotype: white.
## Step2: Set up Punnett square for $ BB\times bb $ cross
|   | $ B $ | $ B $ |
|---|---|---|
| $ b $ | $ Bb $ | $ Bb $ |
| $ b $ | $ Bb $ | $ Bb $ |
## Step3: Determine genotypes and phenotypes of offspring
Offspring genotype: $ Bb $, phenotype: bluish - gray (incomplete dominance, as heterozygous shows intermediate phenotype between black and white).
# Answer:
- Cross: $ BB $ (homozygous black) $ \times $ $ bb $ (homozygous white)
- Parents: Genotype - $ BB $ (black) and $ bb $ (white); Phenotype - black and white
- Offspring: Genotype - $ Bb $; Phenotype - bluish - gray. The Punnett square is as shown above, and all offspring have genotype $ Bb $ with bluish - gray phenotype.

### Problem 10
# Explanation:
## Step1: Determine genotypes
Black individual: $ BB $, Bluish - gray individual: $ Bb $ (from problem 9, heterozygous is bluish - gray).
## Step2: Set up Punnett square
|   | $ B $ | $ B $ |
|---|---|---|
| $ B $ | $ BB $ | $ BB $ |
| $ b $ | $ Bb $ | $ Bb $ |
## Step3: Determine genotypes and phenotypes of offspring
- Genotypes: $ BB $ and $ Bb $
- Phenotypes: $ BB $ - black, $ Bb $ - bluish - gray
- Ratio: 2 (black, $ BB $) : 2 (bluish - gray, $ Bb $) or 1:1 ratio.
# Answer:
When a black individual ($ BB $) is crossed with a bluish - gray individual ($ Bb $), the offspring genotypes are $ BB $ and $ Bb $ in a 1:1 ratio. The phenotypes are black (for $ BB $) and bluish - gray (for $ Bb $) in a 1:1 ratio. The Punnett square is as shown above, with two $ BB $ (black) and two $ Bb $ (bluish - gray) offspring.

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QUESTION IMAGE

Question

Explanation:

Problem 6

Part a

Step1: Recall incomplete dominance

Step1: Identify genotype \( Rr \)

Step1: Analyze genotype \( rr \)

Answer:

Part b