Answer:
<u><em>Reproduction</em></u> of cells is one of the processes that help multicellular organisms maintain homeostasis.
Explanation:
In biology, homeostasis can be described as the capability of an organism to maintain its constant internal environment despite the changes in the external environment.
External factors like pathogens in the environment tend to kill cells in our body like the skin cells and the immune cells. Through the process of cell division, our cells can reproduce and hence, maintain an equilibrium of the internal conditions.
Reproduction of cells also causes growth. Hence, the process of reproduction of cells maintains homeostasis.
Answer:
Somatic cells
Explanation:
A somatic cell mutation in an organism is passed on to the daughter cells in an organism. But this type of mutation doesn't affect the future generations because only genes carried by sperm or ova can become part of offspring's gene material.
Answer:
mutualism and parasitism
Examples:
mutualism: crocodiles and birds
parasitism: mosquitoes and humans
Because they can be differentiated to form insulin-producing cells
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.