Who performed classic experiments that supported the semiconservative model of DNA replication? (A) Watson and Crick (B) Meselso
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1a. The parent is homozygote dominant and homozygote recessive where red(R) is the dominant trait and white(r) is the recessive trait. The genotype of the parents should be:
homozygote dominant= RR
homozygote recessive= rr
1b. To find the possible offspring, you need to know the possible gametes from each parent and make a punnet square with it.
RR
possible gamete= 100% R
rr
possible gamete= 100% r
punnet square of the offspring:
R
r Rr
The answer is 100% Rr
1c. Red flower is a phenotype. This because red is dominant, phenotype will be expressed when the gene is homozygote dominant or heterozygote. The offspring from the cross is 100% heterozygote, so the chance for red flower would be 100% too.
1d. White flower is a phenotype. This because white is recessive, phenotype will be expressed when the gene is homozygote recessive only. The offspring from the cross is 100% heterozygote which mean 0% chance for homozygote recessive. So the chance for white flower would be 0%.
2.
The purple stem(P) is dominant of green stem(p) and red fruit(R) is dominant over yellow fruit(r).
Both parents for the crossing is heterozygote for both traits, so the phenotype and genotype would be green stem and yellow fruit(Pp Rr)
PpRr possible gamete: PR, Pr, pR, pr
The punnet square should be 4x4 with the possible gamete above. It is hard to draw in here but the phenotype should be at 9:3:3:1 ratio
3a. Brown allele is dominant to all other traits. That means you need at least 1 brown allele to make the cat has brown eyes phenotype. The possible phenotype would be:
The total is 4 possible genotype
3b. The grey allele is dominant to blue, but recessive to brown and green. Grey eye phenotype would need at least one grey allele, without the brown or green allele. The possible genotype would be:
The total is 2 possible genotype
3c. Green eye cat will need at least green allele without brown allele. The punnet square should be:
W Y
N WN YN
Y WY YY
The phenotype of the offspring is WN(brown), WY(brown), YN(green), YY(grey). So, the chance for green eye would be 1/4 = 25%
