
In garden pea, the round shape of seeds is dominant over wrinkled shape. A pea plant heterozygous for round shape of seed is selfed and 1600 seeds produced during the cross are subsequently germinated. How many seedlings would have the parental phenotype?
(A) 1600
(B) 800
(C) 400
(D) 1200
Answer
576.9k+ views
Hint: Mendel conducted artificial pollination or cross-pollination experiments using several true-breeding pea lines. A true breeding line is one that having undergone continuous self-pollination, shows the stable trait inheritance and expression for several generations.
Complete answer:
An individual having two different allele as R for round seeds and r for wrinkles seeds is called as heterozygous plants
He made self-cross between
$\mathop R\nolimits_r \times \mathop R\nolimits_r $
He get
RR ( homozygous round seeds)
Rr ( heterozygous round seeds) Rr ( heterozygous round seeds) rr ( homozygous wrinkled seeds)
Now , 1600 will be divided into four parts
1600/4 = 400 seed will be produced on each type listed above
Now parents have round seeds that is the phenotype of individual
Here three pairs show round seeds so \[400 \times {\text{ }}3{\text{ }} = {\text{ }}1200\]
1200 seeds will have the same phenotype as that of parents.
Now let us match this value with the given options :-
> 1600 :- 1600 plants will be the total number of seeds but all of them will not have the same phenotype as that of parents. Hence this option is not correct.
> 800 :- 800 seeds will be having the same genotype as that of the parent that is (Rr) but we have asked for phenotype and not genotype. Hence this option is not correct.
> 400 :- this number of seeds will be wrinkled which is different from parental type. Hence this option is not correct.
> 1200 :- this number of seeds will have the same phenotype as that of parents, that is they will be round. Thus this option is correct.
Our required answer is d that is 1200.
Note: This type of cross is known as monohybrid cross in which only a single character is taken as the shape of the seeds that are round or wrinkled. Phenotype of the organism is defined as observable morphological appearance.
Complete answer:
An individual having two different allele as R for round seeds and r for wrinkles seeds is called as heterozygous plants
He made self-cross between
$\mathop R\nolimits_r \times \mathop R\nolimits_r $
He get
RR ( homozygous round seeds)
Rr ( heterozygous round seeds) Rr ( heterozygous round seeds) rr ( homozygous wrinkled seeds)
Now , 1600 will be divided into four parts
1600/4 = 400 seed will be produced on each type listed above
Now parents have round seeds that is the phenotype of individual
Here three pairs show round seeds so \[400 \times {\text{ }}3{\text{ }} = {\text{ }}1200\]
1200 seeds will have the same phenotype as that of parents.
Now let us match this value with the given options :-
> 1600 :- 1600 plants will be the total number of seeds but all of them will not have the same phenotype as that of parents. Hence this option is not correct.
> 800 :- 800 seeds will be having the same genotype as that of the parent that is (Rr) but we have asked for phenotype and not genotype. Hence this option is not correct.
> 400 :- this number of seeds will be wrinkled which is different from parental type. Hence this option is not correct.
> 1200 :- this number of seeds will have the same phenotype as that of parents, that is they will be round. Thus this option is correct.
Our required answer is d that is 1200.
Note: This type of cross is known as monohybrid cross in which only a single character is taken as the shape of the seeds that are round or wrinkled. Phenotype of the organism is defined as observable morphological appearance.
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