That is not true you will get pulled over immediately and given a ticket
Answer:
- If the plant is homozygous, the percentage of purple-flowered progeny is expected to be 100%.
- If the plant is heterozygous, the percentage of purple-flowered progeny is expected to be 50%.
Explanation:
A test-cross is performed between an individual for whom you want to know the genotype (to find out if it is heterozygote or homozygote) and a recessive homozygote individual.
If the first individual is a heterozygote, the progeny phenotypic and genotypic proportions are 50% heterozygote and 50% recessive homozygote. But if the first individual is a homozygote, the whole progeny will be expressing one of the traits.
This is, from the proposed example, we know that:
- The allele for purple flower color is dominant to the allele for white flower color.
- A test-cross was performed to determine the genotype of a purple-flowered plant
We will represent the allele expressing purple color as P, and the allele expressing white color as p.
The purple-flowered plant genotype might be PP or Pp, so the options are:
Gametes) P P p p
Punnet square) P P
p Pp Pp
p Pp Pp
F1 Genotype) 100% Purple-flowered plants. The genotype of the tested individual was homozygote for the purple trait.
Gametes) P p p p
Punnet square) P p
p Pp pp
p Pp pp
F1 Genotype) 50% Purple-flowered plants and 50% white-flowered plants. The genotype of the tested individual was heterozygote for the purple trait.
1 is c 2 is d 3 is a 4 is b
CH2O5 +603-6CO2 + 6H2O + energy isn't the evidence of conservation of mass in cellular respiration.
Option B
<h3><u>Explanation:</u></h3>
Law of conservation of mass states that mass can neither be created nor be destroyed. And this law holds good for all sorts of chemical reactions except the nuclear reactions.
In case of cellular respiration, one molecule of glucose reacts with 6 molecules of oxygen to produce 6 molecules of carbon dioxide and 6 molecules of water and energy. Now this energy that is produced isn't produced in expense of mass, but in expense of chemical bonds that are present in glucose molecules.
Also if we calculate the number of atoms on each side of the reaction, the number of atoms remain same as well as number of atoms of individual elements also remain same.