Out of a population of 800 individuals in generation of a cross between yellow round and green wrinkled Pea plants, what would be number of yellow and wrinkled seeds

800

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400

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200

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150

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Solution

The correct option is D
150

Phenotypic ratio of dihybrid cross is 9 : 3 : 3 : 1

So 9 + 3 + 3 + 1 = 16, Total population is 800

Yellow and wrinkled seeds number $= 800 \times \frac{3}{16} = 150$

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Similar questions

Q. Out of a population of 800 individuals in F

_{2}

generation of a cross between yellow round and green wrinkled pea plant, what would be the number of yellow and wrinkled seeded plants?

Q. Total population of 800 individuals which formed F

_{2}

population (9 : 3 : 3 : 1) of a cross between yellow round and green wrinkled. Find the number of plants with yellow and wrinkled seeds.

Mendel first crossed pure-bred pea plants having round -yellow seeds with pure-bred pea plants having wrinkled -green seeds and found that only round- yellow seeds were produced in the $F_{1}$ generation. When $F_{1}$ generation pea plants having round-yellow seeds were cross-bred by self-pollination, then peas having round-yellow seeds, round green seeds, wrinkled-yellow seeds and wrinkled -green seeds were produced. Mendel collected a total of 2160 seeds.

(a) What will be the number of (i) round green seeds (ii) wrinkled green seeds (iii) round yellow seeds, and (iv) wrinkled-yellow seeds?

(b) Which 'ratio' as established by Mendel have you made use of in answering the part (a) above?

Question 5
If a round, green-seeded pea plant (RRyy) is crossed with wrinkled, yellow-seeded pea plant (rrYY), the seeds produced in F1 generation are

(a) round and yellow
(b) round and green
(c) wrinkled and green
(d) wrinkled and yellow
( Round-R, Green-y, Wrinkled-r, Yellow-Y)

Q. A cross between parent pea plants having green round seeds with those having yellow wrinkled seeds would produce

F_{1}

plants with ____.

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Out of a population of 800 individuals in generation of a cross between yellow round and green wrinkled Pea plants, what would be number of yellow and wrinkled seeds

The correct option is D 150 Phenotypic ratio of dihybrid cross is 9 : 3 : 3 : 1 So 9 + 3 + 3 + 1 = 16, Total population is 800 Yellow and wrinkled seeds number =800×316=150

The correct option is D
150

Phenotypic ratio of dihybrid cross is 9 : 3 : 3 : 1

So 9 + 3 + 3 + 1 = 16, Total population is 800

Yellow and wrinkled seeds number $= 800 \times \frac{3}{16} = 150$