When a red flowered pea plant was crossed with a white flowered pea plant, it was found that all plants in the F1generation carried only red flowers. If these plants are self crossed, i) Find the genotypic ratio that we can expect in F2 generation ii) What is the phenotype and genotype of F1 generation. Draw Punnett square for F2

The correct answer is -

i) genotype of F2 - 1:2:1 = dominant:heterozygous:recessive

ii) genotype of F1 - all heterozygous dominant

phenotype - all red

Explanation:

In the cross of red- flowered plant and white flowered plant the offspring produces red flowers that s only possible if the parent red flower plant has both alleles in dominant condition.

For the F1 the gametes will be R, R and r, r

from the Punnett square, we can see all the offspring are heterozygous (Rr) in genotype and red in phenotype

For F2 the gametes will be - R, r, and R, r

From the punnet square, the genotype we get is 1:2:1 = dominant:heterozygous:recessive

the transfer for conduction is the heat transfering from the liquid to the spoon

the transfer for convection is the heat rising from the liquid to the air

the transfer for radiation is the heat going through the sides of the cup and to the air

at least that is what i would do

looks like the answer is     c     to me!