A tabacoo plant hetrozygous for recessive trait of albunism is selfed and 1200 seeds are ontained. How many seedlings obtained from such seeds will have prevent genotype.
1) 900
2) 300
3) 600
4) 1200
What is "prevent genotype"?
To determine the number of seedlings with the homozygous recessive genotype, we need to understand the principles of Mendelian genetics and use the Punnett square.
In this case, we have a tobacco plant that is heterozygous (Aa) for the recessive trait of albinism. When it is selfed, meaning it is cross-pollinated with itself, the offspring will have the genotypes AA, Aa, or aa based on the laws of inheritance.
Given that the recessive trait of albinism is represented by the genotype aa, we need to determine the probability of obtaining aa from this cross.
Let's use the Punnett square to illustrate this:
| A a
-----|-------
A | AA Aa
a | Aa aa
As per the Punnett square, there is a 25% chance (1 out of 4) of obtaining aa (albino genotype) from this cross.
Now, since we have 1200 seeds obtained from this selfing process, we can calculate the number of seeds that will have the homozygous recessive genotype.
1200 seeds * 0.25 (25% probability) = 300 seeds with the homozygous recessive genotype (aa).
Therefore, the correct answer to the question is option 2) 300 seedlings.