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Goryan [66]
3 years ago
8

Who would win in a fight aang or korra

Biology
2 answers:
Kisachek [45]3 years ago
5 0
Korra would win in a fight
e-lub [12.9K]3 years ago
4 0

Answer:

korra

Explanation:

hes powerful

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What is the primary function of the highlighted vessel? a) to bring oxygen-rich blood to the liver b) to keep the cells function
Dennis_Churaev [7]

Answer:

"D" is the option that best answers the question

Explanation:

Nutrient-rich blood flows into the liver from the intestines through the hepatic portal vein.

Its primary function is to help with absorption to bring nutrient- and toxin-rich blood to the liver for processing.

6 0
3 years ago
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fomenos

Answer:

3:1

Explanation:

i hope it will be helpful

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Explain the difficulties in developing antiviral drugs against dna viruses, when compared to rna viruses.
Juli2301 [7.4K]
The  DNA   virus  uses  nucleus  of  the  host  to  replicate it  genetic  material since  it  lack reverse  transcriptase enzyme.The similarities between  the  DNA virus  with human  protein it  make  difficult  to  treat  this  virus without  damaging  the  host cell.
                    while
RNA  virus  has  reverse  transcriptase  enzyme which  is  used to  replicate  its  genetic material,  but  it  lack  ability  to check  errors,  which  led  to  higher mutation  rate.The  transcriptase  enzyme  is  used  to make vaccine  which  is  used  to treat  RNA  virus
5 0
3 years ago
The Go' for the reaction Citrate  Isocitrate is +6.64 kJmol-1 . The Go' for the reaction Isocitrate  α-Ketoglutarate is -267
xz_007 [3.2K]

Answer:

ΔG°' for the conversion of citrate to α-ketoglutarate = - 273.64  kJmol-1

Explanation:

ΔG°' for Citrate is +6.64 kJmol-1

ΔG°' for Isocitrate is -267 kJmol-1

ΔG°' for the conversion of citrate to α-ketoglutarate = ΔG°' for product - ΔG°' for reactant

ΔG°' for the conversion of citrate to α-ketoglutarate = -267 kJmol-1 - (+6.64 kJmol-1)

ΔG°' for the conversion of citrate to α-ketoglutarate = - 273.64  kJmol-1

7 0
3 years ago
The retinoblastoma protein, pRB, is a tumor-suppressor protein that controls the G1/S cell-cycle checkpoint. While pRB is presen
gladu [14]

Answer:

1. pRB phosphorylated: The cell will not remain in G1 phase i.e. it will enter S phase because pRB will be degraded.

2. pRB not phosphorylated: The cell will remain in G1 phase i.e. will not enter S phase because pRB will remain bound to E2F.

3. pRB bound to E2F: The cell will remain in G1 phase i.e. will not enter S phase.

4. pRB not bound to E2F: The cell will not remain in G1 phase i.e. it will enter S phase.

5. E2F bound to DNA: The cell will not remain in G1 phase i.e. it will enter S phase.

6. CDK4 bound to Cyclin D1:  The cell will not remain in G1 phase i.e. it will enter S phase.

Explanation:

Retinoblastoma protein, pRB supresses tumor by regulating cell cycle progression in G1 phase. When a transcription factor known as E2F is bound to pRB the cell remains in G1 phase and did not proceed further to enter S phase. But as soon as pRB becomes phosphorylated, E2F becomes free and causes gene expression of cyclin E and and cyclin A which are required in progression of cell cycle from G1 phase to S phase.

<u>Signaling involved in cell cycle progression from G1 phase to S phase. </u>

When a cell is unstimulated, pRB is bound to E2F but G1 phase specific growth factors induce the expression of cyclin D through Ras-MAP kinase pathway.

Synthesis of cyclin D leads to the activation of CDK-4 which is a serine-threonine kinase. This kinase causes phosphorylation of pRB which results in the degradation of pRB.

Because of degradation of pRB, <u>E2F becomes free</u> and leads to the expression of cyclin E and and cyclin A which are S phase specific cyclins. This is how cell progresses from G1 phase to S phase.

E2F is a transcription factor which causes expression of S phase specific cyclins i.e. cyclin E and and cyclin A. So, E2F bound to DNA will result in cell cycle progression from G1 phase to S phase.

7 0
3 years ago
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