Survivorship patterns within a species are pre-set by life history characteristics and can never change in response to variable environmental conditions.a. True
b. False

Answers

Answer 1
Answer:

Answer:

B. False

Explanation:

Life history characteristics and survivorship patterns within a species are pre-set but may be changed according to the changes in environmental conditions needed for growth and reproduction. Allocation of energy to reproduction or other processes varies according to food availability, temperature, rainfall, etc.  

Abundant resource availability favors the production of a large number of smaller offspring while limited resources make a population to produce a small number of larger progeny to increase the survival rate.  

Example: Clutch size in birds is adjusted to obtain the maximum survival of progeny.  


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The Isotope calcium-41 decays Into potassium-41, with a half-life of 1.03 x 105 years. There is a sample of calcium-41 containing 5 x 10⁹ atoms.How many atoms of calcium-41 and potassium-41 will there be after 4.12 x 105 years?
OA 3.125 x 108 atoms of calcium-41 and 4.375 x 10⁹ atoms of potassium-41
OB. 6.25 x 108 atoms of calcium-41 and 4.6875 x 10° atoms of potassium-41
OC. 6.25 x 108 atoms of calcium-41 and 4.375 x 109 atoms of potassium-41
OD. 3.125 x 108 atoms of calcium-41 and 4.6875 x 10° atoms of potassium-41
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Answers

The correct answer for this question will be option D.

Calcium-41 = 3.12×10^8 atoms

Potassium-41 = 4.69×10^9 atoms

As stated in the question,

Half life of isotope of Ca-14 which decays into K-14 = 1.03 × 10^5 years

Therefore, after 1.03 × 10^5 years (1 half life)

Ca-41 will be 50%

and, K-41 will be 50%

After 2.06 × 10^5 years (2 half lives)

Ca-41 will be 25%

K-41 will be 75%

After 3.09 × 10^5 years (3 half lives)

Ca-41 will be 12.5%

K-41 will be 87.5%

After 4.12 × 10^5 years (4 half lives)

Ca-41 will be 6.25%

K-41 will be 93.75%

After 4.12 × 10^5 years,

Ca-41 = 6.25/100 × 5×10^9

= 3.12 × 10^8 atoms

K-41 = 93.75/100 × 5×10^9

= 4.69 × 10^9 atoms

Therefore, option D is correct.

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A population of snakes has 900 individuals. They have a death rate of 30 per year and a birth rate of 20 per year. Assuming 6 snakes migrate into the population and 3 migrate out of the population per year, how many individuals would there be after the first year? (To answer, write numerals in the box below.)

Answers

Answer:

893 snakes after first year

Explanation:

total population= 900

death = 30 per year    left=  900-30= 870

birth each year= 20 per year      so 870 +20= 890 snakes

6 snakes migrate into the population = 890 + 6= 896 snakes

3 migrate out of the population per year = 896-3= 893 snakes

Final answer:

After accounting for annual birth rate, death rate, and net migration, the snake population after one year would be 893 individuals.

Explanation:

To find out how many individuals are in the population after the first year, we start with the initial count of 900 snakes. We then subtract the death rate, or number of snakes dying per year (30), add the birth rate, or number of snakes being born per year (20), then also add the net migration, or difference between snakes moving into and out of the area (6 entering - 3 exiting = 3 net influx).

So the calculation would look like this: 900 (initial population) - 30 (death rate) + 20 (birth rate) + 3 (net migration) = 893

Therefore, after one year, there would be 893 snakes in this population.

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An ecologist decides to study the effect of pathogens of the pollinator of a particular plant species on its reproductive success. Which level of ecological study does this variable represents?

Answers

Answer:

pp

Explanation:

Ten grams of hamburger were added to 90 ml of sterile buffer.This was mixed well in a blender. One-tenth of a ml of this slurry
was added to 9.9 ml of sterile buffer. After thorough mixing, this
suspension was further diluted by a 1/100 dilution followed by a
1/10 dilution. One-tenth of a ml of this final dilution was plated
on agar plates. After incubation, 52 colonies were present. How
many colony-forming units were present in the total 10 gram sample
of hamburger?

Answers

Answer:

5.2 × 10 ⁹ cfus

Explanation:

Using the dilution factors

0.1 ml of the final dilution has 52 colonies

1 ml will have approximately 520 colonies

10 ml of the final sample will have 5200 colonies

at 1 / 100 dilution

1 ml of the sample will have 5200 colonies

100 ml of the sample will have 520000 colonies

1 ml of the 0.1 ml + 9.9 ml has 520000 colonies

10 ml will have   5200000

at the second stage of the dilution

0.1 ml of the slurry had 5200000 colonies

1 ml will have 52000000 colonies

10 ml will have 520000000 colonies

100 ml of the initial sample ( 10 grams + 90 ml ) = 5200000000 colonies =

5.2 × 10 ⁹ cfu

List three differences between DNA and RNA:

Answers

DNA exists naturally as a double helix structure
RNA consists of a single strand that folds into various shapes.
DNA was then heralded to be the blueprint of life because it is chemically more stable than RNA
DNA has two strands
RNA has two complementary strands

Final answer:

DNA and RNA have differences in structure, bases, and function.

Explanation:

There are three main differences between DNA and RNA:

  1. Structure: DNA is a double-stranded helix while RNA is a single-stranded molecule.
  2. Bases: DNA contains the bases adenine (A), thymine (T), cytosine (C), and guanine (G), while RNA contains adenine (A), uracil (U), cytosine (C), and guanine (G).
  3. Function: DNA stores genetic information and is involved in heredity, while RNA plays a role in protein synthesis and gene expression.

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ATP is produced during the light reactions of photosynthesis. This ATP is used to_____________in the Calvin Cycle.A. split water into oxygen has and hydrogen
B. produce NADPH
C. reduce carbon dioxide into glucose
D. None of these are correct

Answers

Answer:

C

Explanation:

The energy source is ATP, and the reducing agent is NADPH, which adds high-energy electrons to produce sugar.