Answer:
Explanation:
The rate limiting reaction of aerobic respiration involves the phosphorylation of fructose phosphate by the enzyme phosphofructokinase. The rate at which this enzyme makes product is [increased, decreased] when ATP levels rise because the molecule ATP binds to an allosteric site on the enzyme and acts as a direct inhibitor. When ATP levels fall AMP binds to the allosteric site of phosphofructokinase. This interaction leads to an increase of cellular ATP, so that this is an example of a negative feedback loop.
Which of the following species reproduces by internal fertilization?
a) cats
(b) Toads
(c) The fish
(d) alternative b and c
Answer:
(a) Los gatos
Explanation:
Only cats are capable of internal fertilization from the given choices above.
Internal fertilization is the fusion of gametes within an organism. This process occurs during sexual reproduction within the female.
- Toads and fishes exhibit external fertilization.
- External fertilization occurs where the male sperm externally fertilizes the eggs of the female.
- This usually takes place in a moist or wet place especially water bodies. The liquid medium will transport the sperm to the egg.
In cats, the female specie takes in reproductive cells from the male during coitus and it fertilizes the her eggs to produce a zygote.
Answer:
Limpets
Explanation:
Since the arrows represent flow of energy, and there is no path of energy from fish to limpets, limpets won't be affected at all.
Similarities:
They contain nematocyst (cells that sting)
some are sessil/free swimming
Some corals have a symbiotic relationship with algae, they need to grow in shallow waters so that light can penetrate for the algae to photosynthesize.
Answer:
In the F1 generation
RR = 0%
Rr = 50% (or 0.5)
rr = 50% (or 0.5)
Explanation:
A pink flowering plant has the genotype Rr. It is heterozygous for the allele. The alleles for this gene appears to show incomplete dominance, as the heterozygous phenotype is a blend of the two homozygous genotypes.
A white flowering plant has the genotype rr. It is homozyogous for the white allele
A punnet square of the cross is shown.
The resulting punnet square shows that only Rr and rr genotypes are possible, at a ratio of 50:50 (or 1:1). Therefore, the genotype frequency of Rr is 50%, and rr is 50% in the F1 generation. This can also be written as 0.5. It is not possible to get a red plant, as the genotype RR can not come from this cross
