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ioda
3 years ago
6

In the Kirby-Bauer disk diffusion test, the _______ of the zone of inhibition is measured and used for interpretation.

Biology
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

The correct option is : a. diameter    

Explanation:

The Kirby–Bauer test or the disk diffusion test, is a method to determine the antibiotic sensitivity of the given bacteria. This test involves the use of antibiotic discs to determine the effect of antibiotics on the bacteria.

In this test, the wafers having antibiotics and the bacteria are placed on the agar plate and incubated. If the antibiotics present stops the growth of the bacteria, there will be an area around wafer with no bacterial growth, such an area is known as the zone of inhibition.

<u>The </u><u>diameter of this zone of inhibition</u><u> is measured to determine the </u><u>antibiotic sensitivity of the given bacteria</u><u>.</u>

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A plant with purple flowers is allowed to self-pollinate. Generation after generation, it produces purple flowers. This is an ex
Virty [35]

Answer:

True breeding

Explanation:

True breeding is a breeding in which parents produce the offspring which carry same phenotype.  

The parents in true breeding are homozygous for every trait.  

<u>True breeding occurs in the plants when the plants produce offspring of same variety only when self pollination takes place.  </u>

<u>For example, if a plant has purple flowers will produce only seeds which will grow into plants which have purple flowers.</u>

7 0
3 years ago
Julio is studying the bald eagle, which has a binomial name of Haliaeetus leucocephalus. Using the modern system of
Lubov Fominskaja [6]

The kingdom is Animals. The genus is Haliaeetus. The species is leucocephalus are the statements which Julio can make about the bald eagle.

The option C is the correct answer.

Explanation:

Given that Bald eagle has binomial name as Haliaeetus leucocephalus.

Modern classification of binomial nomenclature is given by Carl Linnaeus.

From the modern system of classification the sequence is as kingdom, phylum, class, order, family, genus and species. The classification of bald eagle is:

Kingdom - Animalia

Phylum - chordata

class - Aves

Order- Accipitriformes

family - Accipitridae

Genus - Haliaeetus

species - leucocephalus.

The two word name of the species having genus first followed by species is binomial nomenclature which Julio has used for bald eagle as Haliaeetus leucocephalus.

4 0
3 years ago
What correction needs to be made to the information on the table
Vedmedyk [2.9K]

Answer:

Igneous rocks are the cooling and hardening of magma or lava and sedimentary rock is made of sediment cemented together

Explanation:

The three types of rock are igneous, metamorphic and sedimentary. Igneous rock is formed by the cooling and hardening of magma and lava. Examples include granite and basalt. Metamorphic rock is formed when other rocks are exposed to high heat and pressure, causing a chemical change in the rock. Examples include marble and gneiss. Sedimentary rock is formed when sediment is cemented together by minerals. Examples include sandstone and limestone.

Hope this Helps! ::)

3 0
3 years ago
Could you please answer 10, 13, 14? Thanks!
amm1812
14. They get pollinated by animals, and birds that come to

4 0
3 years ago
Free palmitate is activated to it's coA derivative(palmitoyl-CoA) in the cytosol before it can be oxidized in the mitochondria.
n200080 [17]

Answer: hydrolysis of intermediate palmitoyl Co A ,with loss of labeled CoA.before reaching the matrix gives the answer

Explanation:

This is because when the labeled Coenzyme A of the Plamitate combines as Palmitoy-CoA with oxaloacetate to form intermediate (palmitoyl-CoA )in Citric Acid cycle:

CoA is hydrolysed with loss of the labelled CoA which returns to the cystosol. Therefore, the labelled CoA does not reach the matrix of the mitochondrial,but returns to the Cystosol.

Consequently, the CoA in the Cystosol will be labelled in palmitoylCoA and the one in the matrix of the liver mitochondrial will be non radioactive(,will not labelled).

7 0
3 years ago
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