Answer: Prefix: Preobservation
Sufix: Observations
Explanation:
Well to find the answer for force, the formula is mass X acceleration.
So, 20m/s² X 0.5kg = 10kg-m/s².
But, because we measure force in Newtons, we have to change the units (kg/m/s²) to Newtons.
1N (newton) equals to 1kg-m/s² and because we had 10, your answer is
10N
<u>Colony morphology on agar</u> is NOT a phenotypic test.
Examples of phenotypic tests are;
- Antibody reaction
- Antibiotic resistance profile
- Gram staining
Explanation:
Colony morphology is the visible shape taken by a colony of microorganisms. Phenotype, on the other hand, is the external traits of an individual organism based on its genotype and environmental influence. Colony morphology is not determined by the genotype of the organisms hence can't be considered a phenotype trait.
Learn More:
For more on phenotype check out;
brainly.com/question/12075985
brainly.com/question/11850730
#LearnWithBrainly
Protists are unicellular eukaryotes, whereas Eubacteria and Archaebacteria are unicellular prokaryotes.
Eubacteria and Archaebacteria belong to kingdom Monera; whereas Protists belong to kingdom Protista.
All Monerans have prokaryotic cell structure. Protists have eukaryotic cell structure and are unicellular.
Protists either lack cell wall or have cell wall made up of cellulose.
Eukaryotes have cell wall made up of peptidoglycan or murein.
In Archaebacteria cell wall lacks peptidoglycan but contains proteins and non-cellulosic polysaccharides.
Protists have typical sexual reproduction involving fusion of gametes. In Eubacteria and Archaebacteria typical sexual reproduction is absent.
Cell division is mitotic type in Protists and amitotic in Eubacteria and Archaebacteria.
To calculate the frequency of the heterozygote genotype (Pq) for this gene we must use the Hardy-Weinberg equation ( p2 + 2pq + q2 = 1 ). This equation relies on the Hardy-Weinberg principle, a model in population genetics that states that the frequency of the alleles in a population is never changing, only the combinations (the genotypes) are changing.
If there are only two alleles (variations) of this gene in a population, then their frequencies should add up to 1 (100%). From this, we can calculate the frequency of the q allele.
p +q=1
0,3 +q=1
q= 1-0,3
q= 0,7
Now hat we have the frequency of the q allele we can use the HW equation to calculate the frequency of the heterozygotes.
0,09 + 2pq +0.49= 1
2pq +0,58= 1
2pq= 1-0.58
2pq=0,42
The freqency of the heterozygotes in this population is 0.42
