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3 years ago

Which enzyme is used to break down bacteria, yeast, and other microorganisms?

Biology

1 answer:

Llana [10]3 years ago

7 0

Answer:

Explanation:

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I need help with #4<br> I have no Idea what it is or what I need to do

hoa [83]

I need the passage to read it

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2 years ago

An experiment was designed to test the effect of ultraviolet light on the growth of mold spores. A sample of mold spores was div

jenyasd209 [6]

Answer:

Explanation:

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2 years ago

Phosphatases are a family of enzymes that remove phosphate groups from specific proteins; these phosphate groups had been added

guapka [62]

Answer:

Kinases would stop working too.

Explanation:

Cells work constantly to maintain homeostasis or equilibrium within themselves. One of the mechanisms used for this purpose is feedback. As mentioned in the question, kinases put phosphates on proteins, and then, phosphatases take them off. Therefore, for phosphatases to work, kinases must be working at the same pace. By inhibiting the phosphatases the number of phosphate-proteins will be too high. This can be perceived by the kinases and they will automatically stop working since there's an excess of product. That process is called feedback.

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3 years ago

You can recognize the process of pinocytosis when _____.

Lena [83]

You can recognize the process of pinocytosis when THE CELL IS ENGULFING EXTRACELLULAR FLUID.
Pinocytosis is one of the forms of endocytosis; it involves the movement of large particle into the cell by an in-folding of the plasma membrane. It is an active form of transport used to move materials into the cells.

6 0

3 years ago

In hamsters, we can examine the genes for coat color and eye color which are linked:B = brown coat, b = white coat, G = green ey

Inessa [10]

Answer:

10.2 cM

Explanation:

Given,

pure breeding parent 1 : BBgg = brown coat and pink eyes

pure breeding parent 2 : bbGG = white coat and green eyes

BBGG X bbgg = F1 = BbGg ( All are brown with green eyes )

When F1 progeny is test crossed with white coat and pink eyes hamster:

BbGg X BbGg = F2 =

brown, pink eyes = 99 = parental

white, green eyes = 103 = parental

brown, green eyes = 12 = recombinant

white, pink eyes = 11 = recombinant

Recombination frequency =

(total number of recombinants / total progeny ) * 100

= [( 11 + 12 ) / ( 99 + 103 + 12 + 11 )] * 100 = (23/225)*100 = 0.102 * 100 = 10.2 %

1 % recombination frequency = 1 cM distance between genes

Hence, the distance between B and G gene is 10.2 cM

4 0

3 years ago