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Eva8 [605]
3 years ago
7

Explain how the daily energy needs are different between the zebra and its predator, the lion. Be certain to include any pertine

nt differences in the digestibility of these organisms’ food sources, along with differences in digestive tracts and energy spent on digesting. Cite any behaviors you feel may play a role as well.
Biology
1 answer:
Snezhnost [94]3 years ago
5 0

Answer:

Both animals eats different types of food so they have different digestive system.

Explanation:

The daily energy needs are different between the zebra and its predator, the lion because they have difference in daily activities and they eat different types of food and. Zebra is herbivorous animal while the lion is carnivorous so there is highly difference occurs in their digestive system. Zebra activities is to eat plants and run from the prey which needs more calories but the lion needs very high amount of calories for catching their pray for food.

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An anthropologist discovers the fossilized heart of an extinct animal. The evidence indicates that the organism's heart was larg
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A reasonable conclusion supported by these observations is that the animal was endothermic and had a high metabolic rate.

<u>Explanation:</u>

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If a variation occurs, i.e. if the heat produced is less than the heat loss, the metabolic rate increases to compensate for that loss. Shivering is another method used by people and many other warm-blooded organisms to boost their body temperature. But in case the heat loss is smaller than the heat produced, the process of panting or suddening occurs to increase the heat loss.

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When is a hypothesis considered scientific?
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8 0
3 years ago
How many possible open reading frames (frames without stop codons) are there that extend through the following sequence? 5'... C
fenix001 [56]

Answer:

There are 4 reading frames without stop codons.

Explanation:

1) To find the possible reading frames you need to separate your sequence in codons (trios)  for the 5'-> 3' sense, first, excluding the first nucleotide, then excluding 2 nucleotides and then not excluding nucleotides.

5'... C TTA CAG TTT ATT GAT ACG GAG AAG G... 3'  NO STOP CODON

5'... CT TAC AGT TTA TTG ATA CGG AGA AGG... 3'  NO STOP CODON

5'... CTT ACA GTT TAT <u>TGA</u> TAC GGA GAA GG... 3'

2)Then you find the complementary sequence (which you already have)

3'... GAA TGT CAA ATA ACT ATG CCT CTT CC... 5'

3'... GA ATG TCA AAT AAC TAT GCC TCT TCC... 5'

3'... G AAT GTC AAA TAA CTA TGC CTC TTC C... 5'

3) Next, you have to obtain the reverse complementary

CCT TCT CCG TAT CAA TAA ACT GTA AG NO STOP CODON

CC TTC TCC GTA TCA ATA AAC TGT AAG NO STOP CODON

C CTT CTC CGT ATC AAT AAA CTG <u>TAA</u> G

4)Now you need to find start (AGT in terms of DNA) and stop codons(<u>TAA</u>, <u>TAG,</u> <u>TGA</u>, also in terms of DNA), in 5->3 and de reverse complementary.

<em />

<em>As you can see, you have for open reading frames that lack of stop codon, unfortunately, they also lack of AGT, the start codon.</em>

I added a table where you can find thestart and stop codons as well as the proteins that they translate to in terms of RNA.

I hope you find this information interesting and useful, good luck!

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