The answer is the Ribosomes of mitochondria and plastids are very similar in their structure and function to bacterial ribosomes.Mitochondria, chloroplasts, and bacteria are alike in size. Bacteria also have DNA and ribosomes alike to those of mitochondria and chloroplasts. Based on this and other proof, experts ponder host cells and bacteria shaped endosymbiotic relationships precedent, when separate host cells took in oxygen-using and photosynthetic bacteria but did not put an end to them.
"A plants vascular system is made up of specialized cells that for straw-like tubes" is the one among the following that <span>allows
a plant's vascular system to work as a transport system. The correct
option among all the options that are given in the question is the third
option or option "C". I hope this helps.</span>
<em><u>A.SURFACE </u></em><em><u>RUNOFF</u></em><em><u> </u></em><em><u>FROM </u></em><em><u>RAIN.</u></em>
<em><u>C.MELTING</u></em><em><u> </u></em><em><u>ICE</u></em>
<em><u>B.OCEAN </u></em><em><u>CURRENTS </u></em><em><u>(</u></em><em><u> </u></em><em><u>I </u></em><em><u>guess?</u></em><em><u>)</u></em>
<em><u>[that's it]:)</u></em>
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.
4 mps increase in 10 seconds = 2 mps increase in 5 seconds = .4meters/2sec