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Snowcat [4.5K]
4 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]4 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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The answer to this is A
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The number of bacteria in a certain population is predicted to increase according to a continuous exponential growth model, at a
Brilliant_brown [7]

The given question is incomplete. The complete question is as follows;

The number of bacteria in a certain population is predicted to increase according to a continuous exponential growth model, at a relative rate of 16% per hour. Suppose that a sample culture has an initial population of 71 bacteria. Find the predicted population after three hours Do not round any intermediate computations, and round your answer to the nearest tenth bacteria .

Answer:

114.7

Explanation:

A (t) represent the population of the bacteria at the time t.

Since, the population grows exponentially, the population can be calculated as follows:

A (t) = Ao ×  e^{kt}

A (t) is teh final population, Ao is the initial population, e is the exponential, k is rate and t is time.

A (t) = 71  × e^{0.16\times t}

For t = 3 hours

A (t) = 71 × e^{0.16\times 3}

A (t) = 114.7.

The population of bacteria after 3 hours is 114.7.

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4 years ago
Summarize the four steps to natural selection
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3 years ago
According to the World Health Organization, persistent organic pollutants (POPs) are found in nearly all tested organism tissues
olga55 [171]

Answer:

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Explanation:

8 0
2 years ago
If A/A ⋅ B/B is crossed with a/a ⋅ b/b and the F1 is testcrossed, what percentage of the testcross progeny will be a/a ⋅ b/b if
Alex787 [66]

Answer:

a) 25%; b) 50%; c) 45%; d) 38%

Explanation:

<h3 /><h3>a) Unlinked genes</h3>

A/A B/B X a/a b/b

F1: A/a B/b

Testcross A/a B/b X a/a b/b

The homozygous recessive individual only produces <em>ab</em><em> </em>gametes.

The F1 produces four types of gametes (each of them with a frequency of 1/4):  <em>AB</em>, <em>Ab</em>, <em>aB </em>and <em>ab</em>.

25% of the progeny will be a/a b/b.

<h3>b) Completely linked genes</h3>

AB/AB X ab/ab

F1: AB/ab

Testcross AB/ab X ab/ab

The F1 produces only two types of gametes, the parentals (each of them with a frequency of 1/2):  <em>AB</em> and <em>ab</em>.

50% of the progeny will be ab/ab.

<h3>c) 10 m.u. apart</h3>

AB/AB X ab/ab

F1: AB/ab

Testcross AB/ab X ab/ab

The F1 produces four types of gametes, the parentals <em>AB</em> and <em>ab </em>and the recombinants <em>Ab</em> and <em>aB</em>.

Since the genes are 10mu apart, 10% of the produced gametes will be recombinant and 90% will be parentals. Since there are two types of parental gametes, each of them has a frequency of 45%.

45% of the progeny will be ab/ab.

<h3>d) 24 m.u. apart</h3>

This is very similar to c).

Since the genes are 24mu apart, 24% of the produced gametes will be recombinant and 76% will be parentals. Since there are two types of parental gametes, each of them has a frequency of 38%.

38% of the progeny will be ab/ab.

5 0
3 years ago
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