All categories

2 years ago

Methanogens are able to live in anaerobic conditions. a. True b. False

Biology

2 answers:

Brrunno [24]2 years ago

8 0

ANSWER: The answer is true

Darina [25.2K]2 years ago

6 0

Answer:

True

Explanation:

Methanogens are prokaryotic microbes that produce methane as a by-product of metabolism in anoxic or anaerobic conditions. They are commonly found in the gastro-intestinal parts of ruminants, Marine sediments, and wetlands etc. They are responsible for the methane content released when cows burp/belch and the marsh gas of the wetlands.

Methanogens are strictly anaerobic (they thrive best in a no-oxygen condition) and play a vital ecological role by using up excess hydrogen as an energy source and other products of fermentation released during anaerobic respiration. Due to this, methanogens thrive in an environment which has all electron acceptors e.g. Oxygen, Sulphate, Nitrate etc. removed. This excludes CO2 because methanogens use CO2 as their carbon source.

You might be interested in

Madge” was found wandering the streets of New Jersey. She was brought to the attention of a licensed clinical social worker beca

Effectus [21]

Answer:

Madge is likely to have schizophrenia.

Explanation:

Schizophrenia is a mental disorder that causes an individual to suffer delusions and hallucinations that cause totally abnormal and parranoic behaviors.

As has already been said, schizophrenia causes altered and abnormal behaviors, confused thoughts and difficulties in adapting to reality. It is a very disabling disease and can cause an individual to completely lose control of his mental faculties.

Madge's behavior shows that he has confused thoughts, paranoid behaviors and loss of reality, so we can say that he is likely to have schizophrenia.

7 0

1 year ago

(c) find the longest interval during which no tree-cutting samples occurred. what is the first year that no tree-cutting samples

MaRussiya [10]

Answer:

Longest interval during which no tree-cutting sample occurred: 9 years (1203 AD - 1212 AD)

First year in that interval: 1203 AD

Explanation:

Lets complete the question first.

Q) A study uses tree rings to determine the year in which a tree was cut. The data is shown in the leaf and stem diagram in the ATTACHMENT.

The three digits number in the left column represents centuries and decades AD. The numbers on the right represents the years.

For example, consider the first input in the diagram

119 | 5 6

which means it represents 2 samples.

1195 AD, 1196 AD.

The longest interval with no tree-cutting samples is found from 1203 AD to 1212 AD.

3 0

2 years ago

Pseudomonas sp. has a mass doubling time of 2.4 h when grown on acetate. The saturation constant using this substrate is 1.3 g/l

erastova [34]

Answer:

a) Cell concentration when the dilution rate is one-half of the maximum is 0.598g cell/L

b) the substrate concentration when the dilution rate is 0.8 $D_{max}$ is 5.2g/l

c) the maximum dilution rate is : 0.41 h⁻¹

d) the cell productivity at 0.8 $D_{max}$ is 2.40g cell/L

Explanation:

Given data :

Mass doubling time of Pseudomonas sp. = 2.4 hr

Saturation constant = 1.3 g/L

Cell yield on acetate = 0.46g cell/g acetate

We are to find;

a. Cell concentration when the dilution rate is one-half of the maximum.

Here, cell yield =amount of cell produced / amount of substrate consumed.

[S] at 0.5D max is determined using the Monod's equation.

Using the formula :

$D = \frac {D_{max}[S] }{ks+[S]}$

, where D is the dilution rate,

[S] is substrate concentration; &

Ks is the saturation constant.

By replacing the values, we get :

$0.5 = \frac{S}{1.3+[S]}$

$\\\0.65=0.5[S]$

[S]=1.3g/L

The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.

=0.46×1.3

= 0.598g cell/L

Substrate concentration when the dilution rate is 0.8 $D_{max}$ is calculated as:

$D = \frac {D_{max}[S] }{ks+[S]}$

0.8=[S]/1.3+[S]

1.04+0.8[S]=[S]

[S]= 5.2g/L

Therefore , the substrate concentration when the dilution rate is 0.8 $D_{max}$ is 5.2g/l

Maximum dilution rate is calculated using the expression $D_{max} = \frac{1}{time}$

=1/2.4

=0.41 h⁻¹

So, the maximum dilution rate is : 0.41 h⁻¹

The cell productivity at 0.8 $D_{max}$ can be calculated by multiplying the amount of the cell yield with the amount of the substrate consumed at 0.8 $D_{max}$

Cell yield = $\frac {cell \ productivity \ at \ 0.8Dmax}{amount \ of \ substrate\ consumed \at\ 0.8 \D_{max}}$

Cell productivity at 0.8 $D_{max}$ = 0.46 × 5.2

=2.40g cell/L

Therefore, the cell productivity at 0.8 $D_{max}$ is 2.40g cell/L

5 0

2 years ago

Which of these statements describes some aspect of facilitated diffusion?a.Facilitated diffusion is another name for osmosis.b.F

frosja888 [35]

Answer:

Explanation:

Facilitated diffusion is a form of passive transport hence no energy is required by the cell. This means that while the molecules are moving down a concentration gradient – line normal diffusion – the movement of the molecules needs to be facilitated (in this case by a transmembrane protein) either because the molecule is polar and can't pass through the hydrophobic region of the cell membrane, or the molecule is too big to passively pass through the small natural pores of the cell membrane.

5 0

2 years ago

What are the coding segments of a stretch of eukaryotic DNA called? a. introns b. codons c. replicons d. exons e. transposons

AleksAgata [21]

Answer:

Explanation:

3 0

1 year ago