This is true.
An easy way to remember this is that Cis-acting elements are physically linked to a set of structural genes. So promoters and operators are physically linked to the gene they express. They are cis acting because they are on the same strand as the structural gene they control.

On the other side, trans means that they can come from somewhere else and act. A good example of this would be regulatory proteins. These could be expressed somewhere else (or on a different strand) and then bind and affect transcription.

4 0

3 years ago

Pete, a male mouse with a pigmented coat (C, the dominant allele), was mated with Theresa, an albino female (c, the recessive al

JulijaS [17]

The answer is option A

3 0

3 years ago

While gutting a fish, Nathan encountered an air-filled sac-like structure. What is the purpose of this structure in fish?

eimsori [14]

D. It provides buoyancy to a fish.

This air pocket is known as the swim bladder, and also called air bladder. This is a buoyancy organ possessed by most bony fish. The air bladder is located in the body cavity and is derived from an out pocketing of the digestive tube.

4 0

4 years ago

In 1985 a biologist counted 750 pine trees in a 250 hectare forest. Using similar counting techniques, the biologist counted 1,2

Alisiya [41]

Answer:

Suppose that we have a given function f(x)

The average rate of change of the function between two values x₁ and x₂ is given by:

$r = \frac{f(x_2) - f(x_1)}{x_2 - x_1}$

a) We want to find the average (rate) of change on the size of population from 1985 to 1995.

We have that:

f(1985) = 750

f(1995) = 1500

Then we have:

$r = \frac{1500 - 750}{1995 - 1985} = 750/10 = 75$

This means that the population of trees increases, in average, at a rate of 75 trees per year.

b) What is the density of trees each year that they were counted?

This will be equal to the quotient between the number of trees and the area.

1985: number of trees = 750 pines

area = 250 ha

Then the density is:

D(1985) = (750 pines)/(250 ha) = 3 pines/ha

So 1985, there were 3 pines per hectare.

1990: number of trees = 1250 pines

area = 250 ha

Then the density is:

D(1990) = (1250 pines)/(250 ha) = 5 pines/ha

1995: number of trees = 1500 pines

area = 250 ha

The density is:

D(1995) = (1500 pines)/(250 ha) = 6 pines/ha

3) now we want to get the average change between 1985 and 1995 in the density, this will be:

$r = \frac{D(1995) - D(1885)}{1995 - 1985} = \frac{6 pines/ha - 3pines/ha}{10} = 0.3 pines/ha$

So, on average, each year the number of pines per hectare increases by 0.3

7 0

3 years ago