Home
Class 12
BIOLOGY
In garden pea, round shape of seeds is d...

In garden pea, 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 seedling would have the parental phenotype?

A

400

B

1600

C

1200

D

800

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine how many seedlings will exhibit the parental phenotype when a heterozygous pea plant (Rr) is selfed. Here’s a step-by-step breakdown of the solution: ### Step 1: Understand the Genotypes In garden pea plants: - Round shape of seeds (dominant) is represented by the allele 'R'. - Wrinkled shape of seeds (recessive) is represented by the allele 'r'. - A heterozygous plant for round seeds has the genotype Rr. ### Step 2: Set Up the Punnett Square When a heterozygous plant (Rr) is selfed (Rr x Rr), we can set up a Punnett square to determine the possible genotypes of the offspring. | | R | r | |------|------|------| | **R** | RR | Rr | | **r** | Rr | rr | ### Step 3: Determine the Genotypic Ratio From the Punnett square, we can see the possible genotypes of the offspring: - 1 RR (homozygous round) - 2 Rr (heterozygous round) - 1 rr (homozygous wrinkled) This gives us a genotypic ratio of: - 1 RR : 2 Rr : 1 rr ### Step 4: Determine the Phenotypic Ratio Since both RR and Rr produce round seeds, the phenotypic ratio is: - 3 Round : 1 Wrinkled ### Step 5: Calculate the Number of Seedlings Given that a total of 1600 seeds are produced, we can calculate the number of seedlings with the parental phenotype (round seeds). - Total round seedlings = (3/4) * Total seeds - Total round seedlings = (3/4) * 1600 = 1200 ### Step 6: Conclusion Therefore, the number of seedlings that would have the parental phenotype (round seeds) is **1200**. ---
Promotional Banner

Similar Questions

Explore conceptually related problems

In Garden Pea, round shape is dominant over wrinkled shape. A pea plant heterozygous for round shape of seed is selfed and 1600 sedds produced during the cross are subsequently germinated. How many offspring will have parental phenotype

A tobacco plant heterozygous for recessive character is self-pollinated and 1200 seeds are subsequently germinated. How many seedings would have the parental genotype ?

A tobacco plant heterozygous for albinism is selfpollinated and 1200 seeds are subsequently germinated. How many seedings would have the parental genotype :-

A tobacco plant heterozygous for recessive trait of albinism is selfed and 1200 seeds are obtained. How many seedlings obtained from such seeds will have parent genotype?

In pea plants, round seed is dominant over wrinkled seed. A plant heteroxygous for roundseeds was crossed with a plant with wrinkled seeds, which one of the following progenies agrees with expected result?

In Garden Pea, yellow colour of cotyledons is dominant over green and round shape of seed is dominant over wrinkled. When a plant with yellow and round seeds is crossed with a plant having yellow and wrinkled seeds, the progeny showed segregation for all the four characters. The probability of obatining green round seeds in the progeny of this case is

In the garden pea, round seeds are dominant over wrinkled seeds. An investigator crosses a plant having round seeds with a plant having wrinkled seeds. He counts 400 offspring. How many of the offspring have wrinkled seeds if the plant having round seeds is a heterozygote?

A hybrid round and yellow seeded plant is selfed and total 32 seeds are obtained. How many seeds are wrinkled green ?

Round seed trait ( R) is dominant over wrinkled ( r) seed trait in Pea. Heterozygous round seede plant (Rr) is crossed with wrinkled seeded plant (rr). What is the possible progeny ?