A tobacco plant heterozygous for a recessive character is self-pollina...
Given information:
- A tobacco plant is heterozygous for a recessive character.
- The plant is self-pollinated.
- 1200 seeds are germinated.
To solve the problem, we need to use the principles of Mendelian genetics and Punnett squares.
Step-by-step solution:
1. Write down the genotype of the parent plant:
- Let's use A to represent the dominant allele and a to represent the recessive allele. The parent plant is heterozygous, so its genotype is Aa.
2. Create a Punnett square to show the possible gametes and offspring:
| | A | a |
|----|---|---|
| A | AA | Aa |
| a | aA | aa |
- Each box represents a possible offspring genotype that can result from the combination of the parent's gametes.
- The probability of each genotype occurring can be calculated by multiplying the probabilities of the corresponding gametes.
3. Determine the probability of each genotype:
- In this case, we want to know the probability of the parental genotype (Aa) occurring.
- We can see from the Punnett square that there are two Aa offspring genotypes out of four possible genotypes, so the probability is 2/4 or 1/2.
4. Calculate the number of offspring with the parental genotype:
- We know that there are 1200 seeds germinated, so we can use the probability we calculated in step 3 to find the expected number of Aa offspring.
- Expected number of Aa offspring = probability of Aa genotype x total number of offspring
- Expected number of Aa offspring = 1/2 x 1200 = 600
5. Choose the correct answer:
- The question asks how many seedlings would have the parental genotype, which is Aa.
- From step 4, we know that the expected number of Aa offspring is 600. Therefore, the correct answer is B, 600.
A tobacco plant heterozygous for a recessive character is self-pollina...
1250