All categories

3 years ago

All the genetic mutations may occur spontaneously in organisms is t he incidence of such mutations may be increased by what?

Biology

2 answers:

Mandarinka [93]3 years ago

6 0

Answer:

physical and chemical factors

Explanation:

Factors that alter the critical chemical and physical characteristics of ecological systems include temperature, pH electrochemical (redox) potential, and the transparency of air and water.

DaniilM [7]3 years ago

3 0

Answer:

Both physical and chemical factors

Explanation:

A mutation is any alteration in the DNA sequence of the genome of a particular cell/organism. Depending on the type and localization, a mutation can be neutral, deleterious, or even beneficial for the individual that possesses it. Moreover, the mutation rate refers to the probability for a mutation to appear in a cell/organism. Exposure to physical environmental factors (e.g. ultraviolet radiation, X-rays, gamma rays, etc) and chemical factors (e.g., bromine) can increase this likelihood.

You might be interested in

What does the phrase "Aspiring designers mean? HELP PLZ!!

anzhelika [568]

To have a great ambition or ultimate goal; desire strongly

8 0

2 years ago

Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

$C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}$

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

Best regards.

7 0

3 years ago

Why is feedback inhibition important in metabolic pathways?

GrogVix [38]

Answer:

To balance the production of certain products.

Explanation:

Feedback inhibition is when byproduct from metabolic reactions in cells accumulates and is in excess. The product goes and inhibits the enzyme that is responsible for speeding its chemical reaction, balancing the amount of product needed, with the amount already produced.

Therefore feedback inhibition is important in metabolic pathways because it balances the production of amino acids and nucleotides. It is there to ensure that the exact amount needed is produced.

4 0

3 years ago

There are 1000 micrometres (um) in a millimetre (mm).

Vesna [10]

Answer: The length of the cell in millimetres is 0.0015.

Explanation:

Given conversion :

$1000\mu m=1mm$

Thus $1\mu m=\frac{1}{1000}mm=0.001mm$

Given : Length of the cell = 1.5 micrometers $(\mu m)$

To find: Length of the cell in milllimeters (mm)

Length of the cell in milllimeters (mm) = $1.5\times 0.001mm=0.0015mm$

The length of the cell in millimetres is 0.0015.

4 0

3 years ago

if an organism occupies the secondary consumer trophic level in other food chains within a given food chain, is that organism al

nalin [4]

Answer:

The organisms that eat the primary consumers are called secondary consumers. Secondary consumers are generally meat-eaters—carnivores. The organisms that eat the secondary consumers are called tertiary consumers. These are carnivore-eating carnivores, like eagles or big fish.

Explanation:

7 0

2 years ago