Answer:
During active transport, substances move against the concentration gradient, from an area of low concentration to an area of high concentration. This process is “active” because it requires the use of energy (usually in the form of ATP). It is the opposite of passive transport.
Explanation:
This may help you out! :)
These groups are arranged in order from most inclusive (most general) to least inclusive (most specific) is gnathostomes, osteichthyans, lobe-fins, tetrapods, amphibians.
<h3>What is
gnathostomes?</h3>
The jawed vertebrates are called gnathostomata. The phrase comes from the Greek words "jaw" and "mouth." Approximately 60,000 species make up the diversity of the gnathostome, which represents 99% of all vertebrates still alive today.
<h3>What is
osteichthyans?</h3>
A broad taxonomic group of fish called osteichthyes, also known as the "bony fish," has skeletons that are predominantly made of bone tissue.
<h3>What is
lobe-fins?</h3>
The taxon Sarcopterygii, also known as Crossopterygii, is made up of bony fishes noted for having lobe-finned fishes as its members.
<h3>What is
tetrapods?</h3>
Four-legged vertebrates that make up the superclass Tetrapoda are known as tetrapods, which derives from the Ancient Greek (tetra-) "four" and "foot." It consists of synapsids, dinosaurs, and extinct as well as living amphibians, reptiles, and dinosaur-related birds (including mammals).
To learn more about Tetrapods visit:
brainly.com/question/15289594
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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.