Answer:
a. Marigold with no fertilizer
Explanation:
An experiment contains variables including independent and dependent variables. The independent variable is the variable that is changed by the experimenter, which is the TYPE OF FERTILIZER in this case, while the dependent variable is the measured variable, which is the GROWTH OF MARIGOLDS in this experiment.
However, a control group is a group that does not receive any experimental treatment. In other words, the independent variable is unchanged in the control group. The control group is regarded as the "normal" used to compare with the experimental group (group that receives treatment). In this experiment, the group of marigolds that DOES NOT RECEIVE FERTILIZER is the control group.
Hi,
Let us get an introduction of the problem first, Incomplete dominance is a pattern of inheritance in which, offspring from the cross of two parents shows a phenotype that is not like either parent but in between both.
For example: Let us take the example of Oompah's hair that can be either red or purple or blue. As described in the question, Red is dominant so let us denote allele for Red hair with R. The other hair color is blue that is recessive, let us denote its allele with r. Now, there is another intermediate phenotype that is purple, let us denote it with Rr because it is derived through a combination of R and r.
Now, here is the key for all possible genotypes and phenotypes of Oompahs. There are only three possible genotypes and phenotypes for the hair color.
Genotype: Phenotype
RR : Red hair
Rr : purple hair
rr : Blue hair
This is because no allele can fully express itself over the other. So is we cross a red hair Oompah with Blue hair Oompah, let us see what will happen:
RR x rr
Gametes: r and R
rr: Rr : Rr : RR
It means that 25 % of the offspring will have red hair, 50% will have contain purple hair and 25 % will have blue hair.
RR: Red hair 25%
Rr: Purple hair 50 %
rr: Blue hair 25%
Hope it helps!
Answer:
Toxic sediment and nutrient that's the answer
Explanation:
Answer:
0.153
Explanation:
We know the up-thrust on the fish, U = weight of water displaced = weight of fish + weight of air in air sacs.
So ρVg = ρ'V'g + ρ'V"g where ρ = density of water = 1 g/cm³, V = volume of water displaced, g = acceleration due to gravity, ρ'= density of fish = 1.18 g/cm³, V' = initial volume of fish, ρ"= density of air = 0.0012 g/cm³ and V" = volume of expanded air sac.
ρVg = ρ'V'g + ρ"V"g
ρV = ρ'V'g + ρ"V"
Its new body volume = volume of water displaced, V = V' + V"
ρ(V' + V") = ρ'V' + ρ"V"
ρV' + ρV" = ρ'V' + ρ"V"
ρV' - ρ"V' = ρ'V" - ρV"
(ρ - ρ")V' = (ρ' - ρ)V"
V'/V" = (ρ - ρ")/(ρ' - ρ)
= (1 g/cm³ - 0.0012 g/cm³)/(1.18 g/cm³ - 1 g/cm³)
= (0.9988 g/cm³ ÷ 0.18 g/cm³)
V'/V" = 5.55
Since V = V' + V"
V' = V - V"
(V - V")/V" = 5.55
V/V" - V"/V" = 5.55
V/V" - 1 = 5.55
V/V" = 5.55 + 1
V/V" = 6.55
V"/V = 1/6.55
V"/V = 0.153
So, the fish must inflate its air sacs to 0.153 of its expanded body volume
