Results from multiple experiments suggest that the ActA protein of Listeria, an integral membrane protein expressed on the cell surface, is the only bacterial protein that is required for the movement of the bacterium within its host cell. Which combination of the following observations shows that ActA is both, necessary and sufficient for bacterial movement?Please note: E. Coli is not an intracellular bacterium - it does not normally move inside the cytosol, and it does not express ActA.

I. ActA binds to the Arp2/3 complex

II. E. Coli in which ActA is expressed moves in host cell cytosol

III. Listeria lacking the ActA gene fail to move inside the host cell cytosol.

IV. ActA has 3 transmembrane domains

V. Arp2/3, when activated, nucleates a branched actin network

A. I and II

B. II and III

C. I and III

D. III and IV

E. I and V

Answers

Answer 1

Answer:

Answer:

The correct answer is option B. "II and III".

Explanation:

In order to prove that ActA protein of Listeria is necessary and sufficient for bacterial movement within its host cell a series of results in transformed Listeria and transformed E. Coli must be obtained. First, if Listeria lacking the ActA gene fail to move inside the host cell cytosol it will prove that ActA is necessary for bacterial movement. Second, a transformed E. Coli that expresses ActA should be able to move in host cell cytosol. Wild type E. Coli does not expresses ActA, if ActA alone makes E. Coli able to move in host cell cytosol it will prove that ActA is sufficient for bacterial movement.

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Rapid cell growth in preparation for cell divisionoccurs during
the G1 phase
the S phase.
O the G2 phase.
mitosis.

Answers

Final answer:

The G2 phase of the cell cycle is when rapid cell growth in preparation for cell division occurs. This is after DNA replication and before mitosis.

Explanation:

Rapid cell growth in preparation for cell division occurs during the G2 phase of the cell cycle. The cell cycle is a four-stage process consisting of the G1 phase (gap 1), the S phase (synthesis), the G2 phase (gap 2), and mitosis. During the G1 phase, cells grow and prepare for DNA replication which happens in the S phase. In the G2 phase, the cell experiences rapid growth in preparation for cell division, cell size increases dramatically, and proteins necessary for cell division are synthesised.

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Fructose‑2,6‑bisphosphate is a regulator of both glycolysis and gluconeogenesis for the phosphofructokinase reaction of glycolysis and the fructose‑1,6‑bisphosphatase reaction of gluconeogenesis. In turn, the concentration of fructose‑2,6‑bisphosphate is regulated by many hormones, second messengers, and enzymes. Classify each condition according to its effect on glycolysis and gluconeogenesis Activates glycolysis and inhibits gluconeogenesis Activates gluconeogenesis and inhibits glycolysis

Answers

Answer: Activates glycolysis and inhibits gluconeogenesis

Explanation:

Fructose‑2,6‑bisphosphate coordinates glucose breakdown in glycolysis generates by modulating the action of phosphofructokinase 1 and at the same time inhibits gluconeogenesis.

The Isotope calcium-41 decays Into potassium-41, with a half-life of 1.03 x 105 years. There is a sample of calcium-41 containing 5 x 10⁹ atoms.How many atoms of calcium-41 and potassium-41 will there be after 4.12 x 105 years?
OA 3.125 x 108 atoms of calcium-41 and 4.375 x 10⁹ atoms of potassium-41
OB. 6.25 x 108 atoms of calcium-41 and 4.6875 x 10° atoms of potassium-41
OC. 6.25 x 108 atoms of calcium-41 and 4.375 x 109 atoms of potassium-41
OD. 3.125 x 108 atoms of calcium-41 and 4.6875 x 10° atoms of potassium-41
hts reserved.
Reset
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Answers

The correct answer for this question will be option D.

Calcium-41 = 3.12× $10^8$ atoms

Potassium-41 = 4.69× $10^9$ atoms

As stated in the question,

Half life of isotope of Ca-14 which decays into K-14 = 1.03 × $10^5$ years

Therefore, after 1.03 × $10^5$ years (1 half life)

Ca-41 will be 50%

and, K-41 will be 50%

After 2.06 × $10^5$ years (2 half lives)

Ca-41 will be 25%

K-41 will be 75%

After 3.09 × $10^5$ years (3 half lives)

Ca-41 will be 12.5%

K-41 will be 87.5%

After 4.12 × $10^5$ years (4 half lives)

Ca-41 will be 6.25%

K-41 will be 93.75%

After 4.12 × $10^5$ years,

Ca-41 = 6.25/100 × 5× $10^9$

= 3.12 × $10^8$ atoms

K-41 = 93.75/100 × 5× $10^9$

= 4.69 × $10^9$ atoms

Therefore, option D is correct.

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Answers

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Blood type is coded by two alleles, which have three possible type: A, B and O. The O alleles is recessive, so you need two O allele(from both parent) to have type O blood child.
A person with type AB blood type allele should be one A and one B, so it is impossible to get O allele him.

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The southerly lowlands of the island average around 0 °C (32 °F)

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