Ecosystems can be large and are made of biomes. Some examples of ecosystems are deserts and forests.
Answer:
Soap is a basic
Explanation:
Soap is a combination of a weak acid (fatty acids) and a strong base (lye), which results in what is known as “alkalai salt,” or a salt that is basic on the pH scale.
Substrate level phosphorylation is the formation of ATP to ADP. Due to substrate level phosphorylation, glycolysis forms 4 ATP.
Answer: The angle of refraction is 32°
Explanation:
By snell law, we have that:
n₁*sen(θ₁) = n₂*sen(θ₂)
Where n₁ = 1.5
n₂ = 2.0
θ₁ = 45°.
then we want to find the angle of refraction.
1.5*sin(45°) = 2*sin(x)
x = Asin(1.5*sin(45°)/2) = 32°
Answer:
c. 1:2:1
The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.
Explanation:
If flower color were determined by a gene showing incomplete dominance, the possible genotypes and phenotypes are as follows:
- RR- red
- ww - white
- Rw - pink
If pink sweet peas are self-pollinated, then a cross between two heterozygous individuals is done (Rw x Rw).
<u>From this cross the expected ratios are:</u>
- 1/4 RR (red)
- 2/4 Rw (pink)
- 1/4 ww (white)
So the null hypothesis is that the observed results exhibit a 1:2:1 ratio.
<h3><u>Chi square test</u></h3>
<u>The observed frequencies were:</u>
Total 150
<u>The expected frequencies for our null hypothesis are:</u>
- 1/4 x 150 = 37.5 Red
- 2/4 x 150 = 75 Pink
- 1/4 x 150 = 37.5 white
The degrees of freedom (DF) are calculated as number of phenotypes - 1; in this case DF = 3-1 = 2.
If we look at the Chi square table, for 2 DF and a probability of p0.05, the critical value is 5.991
Our X^2 value of 0.5067 is less than the critical value, so we do not reject the null hypothesis. The results are consistent with incomplete dominance for this trait, with pink flowers being heterozygous.