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3 years ago

What are the similarities between biosphere and ecosystem

1 answer:

6 0

The difference between them is that an ecosystem is a community of organisms and their environment. And a biosphere is all the living organisms.

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What was the actual proportion of the basis for equal amounts of DNA

Stells [14]

Answer:

Equal amounts of DNA contain equal proportions of nitrogenous bases.

Explanation:

According to Chargaff's rule, the nitrogenous base pairs (Purines and pyrimidines) have an equal ratio (1:1) in the DNA of all the organisms. More precisely, the amount of adenine is equal to thymine and the amount of guanine is equal to cytosine in the DNA of all the organism. A-T base pair has two H-bonds while G-C base pair has three H-bonds.

7 0

4 years ago

Systolic blood pressure is the amount of pressure that blood exerts __________

andrey2020 [161]

Answer:

while the heart beats

Explanation:

Blood pressure of the heart is the pressure by which heart pumps blood through the arteries. Since the blood flow through the heart and arteries is pulsatile, there are two heart blood pressures: systolic and diastolic blood pressure. Normal systolic pressure is about 120 mmHg.

Diastolic blood pressure is the pressure of the blood between heartbeats (when heart is not active, relaxed). Normal diastolic blood pressure is about 80 mmHg.

3 0

3 years ago

Noble gases have a full outer energy level and are in active these Atoms are considered

777dan777 [17]

Explanation:

Noble gases have a full outer energy level and are inactive. These atoms are considered inert.

Hope this helps

Good Luck

7 0

4 years ago

Which of the following are examples of what scientists can learn from studying fossils?

FrozenT [24]

The answer is a and c

7 0

2 years ago

Read 2 more answers

Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some

Sphinxa [80]

Answer and Explanation:

The number of observed individuals:

AA 42

AG 24

GG 21

Total number of individuals, N= 87 = 42 + 24 + 21

Allelic frequencies:

f(p) = (2 x AA + AG)/ 2 x N

f (p)= (2 x 42 + 24) /2 x 87

f (p) = (84 + 24) / 174

f (p)= 108 / 174

f (p) = 0.62

f (q) = (2 x GG + AG)/2 x N

f (q) = (2 x 21 + 24 )/2 x 87

f (q) = (42 + 24)/ 174

f (q) = 66/174

f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

The expected genotypic frequency:

F (AA)= 0.62 ² = 0.3844

F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712

F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

The number of expected individuals:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

Total number of expected individuals = 33.44 + 40.99 + 12.563 = 87

Chi square= sum (O-E)²/E

AA= (O-E)² /E

AA=(42 - 33.44) ² / 33.44

AA= 2.2

AB= (O-E)² /E

AB= (24 - 40.99)²/ 40.99

AB=7.04

BB=(O-E)² /E

BB= (21-12.563)²/12.563

BB= 5.66

Chi square= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

Degrees of freedom = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

5 0

4 years ago