Question

A homozygous pea plant with red flowers and round seeds is crossed with a plant showing white flowers and wrinkled seeds. With the help of checker board diagram, show the phenotypes and genotypes of F2 plants.

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Answer 1

Answer:

(ii) Genotype ratio is 1:2:1, RR : Rr : rr

Phenotype ratio is 3:1, round seeds : wrinkled seeds

(iii) There are two sex chromosomes in human, X and Y. Females have XX chromosomes and males have XY chromosomes.

(iv) Mutation refers to the changes in the nucleotide sequence of DNA which can be inherited.

(v) The number of chromosomes in human beings are 23 pairs, that is 22 pairs of autosomes and one pair of sex chromosome. Thus, the gamete will have 23 chromosomes.

Answer 2

Answer:

Tall pea plant bearing red flowers - TTRR

Short pea plant bearing white flowers - ttrr

F1 generation - TtRr

Selfing of F1 - TtRr X TtRr

Gametes - TR,Tr, tR, tr TR, Tr, tR, tr

F2 generation:

9:3:3:1 is the phenotypic ratio.

9 Tall red : 3 Tall white: 3 Dwarf red: 1 Dwarf white

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Explanation:

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