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.
The answer would be B, because they have subcellular organelles.
<span>The genetic material of HIV consists of two molecules of single-stranded RNA.
</span>
These viral particles, also known as virions, consist of two or three parts: the genetic material made from either DNA or RNA, long molecules that carry genetic information, a protein coat, called the capsid, which surrounds and protects the genetic material.
A. The following statements are hypotheses:
1. Glucose may cause an increase in bacterial growth.
2. Increased glucose may lead to capsule formation in bacteria.
B. The following statements are observations:
1. The bacterial colony grown without glucose do not have capsules around their cells.
2. Bacteria growth in the glucose environment increase rapidly and then stopped after sometime.