The light side of the eggs contains yolk in the vegetal cytoplasm. The upper half divides more fastly and is known as an animal hemisphere of the cell. The presence of yolk on the lower side allows space for the upper portion, for the movements of the embryo.
So the answers are. The P generation has yellow and green seeds. The F2 generation has yellow and green seeds. The F1 generation has all yellow seeds.
Suppose that the proportion of the white crest alleles (r) is given by w and that of the Red crest allele (R) is given by p. We have that p+w=1. The probability that an individual has 2 r alleles is given by w*w since for each allele position the probability is w. Only these individuals have a White phenotype. Hence, we get that w^2=

; the right hand side is the proportion of white birds in the total population. Doing the calculations, this yields that w=0.37. From this, we calculate that p=0.63. The possible ways we have heterozygous individuals are the combinations Rr and rR. The probability for each of those is p*w. Thus, the total probability is 2pw. This is equal to 0.466=0.47. This is the fraction of the future population that is going to be heterozygous assuming the conditions of the Handy-Weinberg equilibrium like random reproductive matching etc.
Answer is from dd parents in this simulation it very lucky to survive as a long as they did.