All categories

3 years ago

3. Which statement concerning an ecosystem is correct?

Biology

1 answer:

brilliants [131]3 years ago

6 0

Answer:

C) It involves interactions between biotic and

abiotic factors.

Explanation:

Ecosystems need energy imput, need both consumers and producers, and can exist on land, lakes, rivers and oceans.

You might be interested in

Any organism that has a circulatory system is a

Goshia [24]

Lungs have a pulmonary circulation

4 0

3 years ago

The point on earths surface directly above an earthquakes focus is called

vlada-n [284]

The point on earth's surface directly above an earthquakes focus (hypocenter) is called the epicenter.

8 0

3 years ago

When one atoms electron is given up to another atom both atoms become what?

Amiraneli [1.4K]

When an electron moves from one atom to another, both atoms become ions.

4 0

3 years ago

________ results in the production of spores in plants. meiosis fertilization germination mitosis spermatogenesis

babymother [125]

The answer is Meiosis.

8 0

3 years ago

In flies dumpy wings and ebony bodies are each mutations recessive to wild type, and you expect them to behave in a Mendelian fa

Dimas [21]

Answer:

See the answers below

Explanation:

Assuming that D (d) represents the allele for wing type and E (e) represents the allele for body type, crossing DdEe with DdEe will yield the following offspring according to the Punnet's square (see the attached image):

(a) 9/16 D_E_ wild type

3/16 D_ee wild wing, ebony body

3/16 ddE_ dumpy wing, wild body

1/16 ddee dumpy wing, ebony body

(b) Chi square $X^2$ = $\frac{(O - E)^2}{E}$ , where O = observed frequency and E = expected frequency.

Phenotype O E $X^2$

Wild type 473 9/16 x 830 = 466.875 $\frac{(473 - 466.875)^2}{466.875}$ = 0.08

wild w/ebony b 156 3/16 x 830 = 155.625 $\frac{(156 - 155.625)^2}{155.625}$ = 0.0009

Dumpy w/wild b 149 3/16 x 830 = 155.625 $\frac{(149 - 155.625)^2}{155.625}$ = 0.28

dumpy w/ebony b 52 1/16 x 830 = 51.875 $\frac{(52 - 51.875)^2}{51.875}$ = 0.0003

Total $X^2$ = 0.3612

Degree of freedom = 4 - 1 = 3

Tabulated value of $X^2$ (0.05)= 7.815

(c) Since the calculated Chi square value is less than the tabulated value, we conclude that the observed outcome agrees with the expected outcome and that the cross followed the standard Mendelian pattern.

6 0

3 years ago