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Snowcat [4.5K]
3 years ago
8

Which of the following statements concerning Francisella is false?A. An attenuated vaccine is used for high-risk individuals B.

It resists phagocytosisC. Humans are its only known hostD. It is extremely infectiousE. It cannot be treated with penicillins or cephalosporins
Biology
1 answer:
Veronika [31]3 years ago
6 0

Answer:

Humans are its only known host

Explanation:

Humans are not the only known host for bacterium<em> Francisella</em>. For example, <em>Francisella</em> species such as <em>Francisella tularensis</em> uses many other organisms as its host.

Some of the non-human host organisms for <em>Francisella tularensis </em>are wild rabbits, most of the domestic and wild animals. The <em>Francisella tularensis </em>serve as causative agent of tularemia in these organisms which is a plague-like disease.

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Features that are homologous are similar due to __________.
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<span>The answer is shared an evolutionary history. Homologous structures evolved from the same ancestor structure. An example of homologous features is the four limbs of all tetrapods. In birds, the forelimbs evolved into wings while in crocodiles they still used for walking, and in humans, they evolved into hands. </span>






3 0
3 years ago
Which of the following is not a contributing factor to ecosystem stability species, genetic,ecosystem, or abiotic diversity.
insens350 [35]

Answer:

Ecosystem.

Explanation:

Ecosystem is not a contributing factor to ecosystem stability whereas, species, genetic, and abiotic diversity are the factors that contributes in the stability of an ecosystem. The diversity of species greatly affected the ecosystem stability if more species are present in he ecosystem. The abiotic factors also plays an important role in the stability of an ecosystem because these abiotic factors causes changes in the ecosystem environmental conditions.

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2 years ago
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7 0
2 years ago
A prokaryote's capsule is: A. A layer of amino acids made from polysaccharides. B. A layer of mucous made of phospholipids. C. A
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4 0
3 years ago
The nucleus of "Lead-208", 208 82 Pb, has 82 protons within a sphere of radius 6.34×10-15 m. Each electric charge has a value of
jarptica [38.1K]

Answer:

2.94 × 10²⁰ N/C

Explanation:

Given that:  

The nucleus of "Lead-208 has 82 protons,

with a radius (r) 6.34×10-15 m, &

each electric charge has a value of 1.60218 × 10^-19 C

∴ The formula for calculating an electrical field at the surface of the nucleus is:

 E=\frac{k*q}{r^2}  

Substituting our values into the equation above, we have;

 E = \frac{8.98755*10^8*82(1.60218*10^{-19C)}}{(6.34*10^{-15}_m)^2}

E = 2.93870499×10²⁰ N/C

E ≅ 2.94 × 10²⁰ N/C

6 0
3 years ago
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