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3 years ago

Why is a blood transport system to carry oxygen to all the cells important in multi-cellular organisms?

Biology

1 answer:

marishachu [46]3 years ago

5 0

Organisms need the oxygen to respire in order to produce energy needed for various processes to take place in the body

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The amino acids for beta hemoglobin found in five species were compared to the amino acids found in human (Homo sapiens) beta he

ollegr [7]

Answer:

The correct answer should be option A.

It is because there is only one difference in the sequence of protein synthesized in the Gorilla gorilla and Homo sapiens.

As they have the least difference in the nucleotide or amino acid sequences of the protein they would the most closely related as compared to others.

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3 years ago

As with all multicellular organisms, homeotic genes in plants control morphogenesis (how the tissues and organs of the plant are

N76 [4]

<span>Answer: In symmetrical cell division, the cytoplasm of the parent cell is distributed equally between the two daughter cells, resulting in daughter cells of equal size. In asymmetrical cell division, the cytoplasm of the parent cell is distributed unequally between the two daughter cells, resulting in daughter cells of different sizes.</span>

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3 years ago

In a population of birds, the average beak size is favored. What type of selection is this?

alexira [117]

Answer:

C. Stabilizing selection

Explanation:

Natural selection, is of three different types namely: disruptive selection, directional selection and stabilizing selection. Disruptive selection occurs when the two extremes of a population are being favored over the average or intermediate.

Directional occurs when one extreme of a population is favored. However, in STABILIZING SELECTION, the average or intermediate population is favored over both extremes of a population. Example is when the average-sized beaks of a bird population is being favored over the long and short ones (extremes).

4 0

3 years ago

Pseudomonas putida is used for fermentation of the lactose present in cheese whey. The bacteria are cultivated in a steady-state

fgiga [73]

Answer:

Explanation:

Given that:

The dilution rate D = 0.28 h⁻¹

The concentration of lactose in the feed $S_o = 2.0 \ g/L$

The effluent S = 0.10 g/L

Also;

$Y_{X/S} = 0.45 g\ X/g \ S , \\ \\ Y_{X/O2 }= 0.25 g \ X/g \ O2, \\ \\$

Saturation C* = 8 mg/l

To calculate the steady-state biomass, we use the formula:

$X = Y_{X/S}(S_o-S_e) \\ \\ X = 0.45(2.0 -0.10) \ g/L \\ \\ X= 0.45 (1.9) \ g/L \\ \\ X = 0.855\ g/L \\ \\ X = 855 \ mg/L$

The biomass is 0.855 g/L

For a steady-state condition, the oxygen uptake rate can be illustrated by using the formula:

$q_{o_2}X =\dfrac{\mu_X}{Y_{X/O_2}}$

where;

$\mu =$ dilution rate (D)

Thus, the steady-state can be expressed as:

$q_{o_2}X =\dfrac{D}{Y_{X/O_2}}$

$q_{o_2}X =\dfrac{0.28}{0.25}$

$q_{o_2}X =1.12 \ h^{-1}$

The specific rate of oxygen consumption $q_{o_2}X =1.12 \ h^{-1}$

In the fermentation medium, if the desired DO concentration $C_L$ = 2 mg/L

Here, the oxygen transfer is regarded as the rate-limiting step.

As such, the oxygen transfer rate(OTR) is equivalent to the oxygen uptake rate.

In this scenario, let's determine the oxygen transfer coefficient $(K_{La})$ by using the formula:

$OTR = K_{La}(C^* - C_L)$

where;

$K_{La}$ = coefficient of oxygen transfer

C* = saturation

Since $OTR = q_{O_2}X$

$q_{o2}X = K_{La}(C^*-C_L) \\\\ (1.12 )(855) = K_{La}(8-2) \\ \\ 957.6 = K_La (6) \\ \\ K_{La}= \dfrac{957.6}{6} \\ \\ K_{La} = 159.6 \ h^{-1}$

Thus, the oxygen transfer coefficient $K_{La}$ = 159.6 h⁻¹

4 0

3 years ago

....... bacteria as in absence of air

navik [9.2K]

Clostridium, Bacteroides, and the methane-producing archaea (methanogens), are called obligate anaerobes because their energy-generating metabolic processes are not coupled with the consumption of oxygen.

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3 years ago