Answer:
D: two jaguars per thousand hectares.
Explanation:
Population density refers to the computation of population by unit area. It is usually implemented in living organisms, generally in humans. It is a key geological term. In easy form, population density implies to the number of people existing in an area per square kilometer or hectare, etc.
So, two jaguars per thousand hectares is an example of population density as it involves the number of jaguars (population) per thousand hectares (per unit area).
Answer:
Explanation:
Option 4 bescause theres 4 extra cell and they are haploid.
Proteins function optimally at a specific temperature. So if you get too hot or too cold, biochemical reactions in your body start to function less well. If the situation becomes extreme enough, they can cease to function well enough to sustain life.
Warm-blooded animals have an advantage over cold-blooded ones in that their bodies automatically try to maintain the optimal termperature for things in their bodies to function. Cold-blooded animals depend on the environmental temperature to do this for them. That's why reptiles are very sluggish when they're cold, but will "wake up" when they get warm.
The cost to this benefit is that metabolically, warm-blooded animals require a lot more fuel to run their bodies. It's very energy-intensive to maintain a constant body temperature. Cold-blooded animals require far less fuel than warm-blooded ones relative to their size.
The way that proteins operate in a specific temperature is also true of the pH in your body which is also very tightly maintained.
Answer:
F (pp) = 0.04
F (Pp) = 0.32
F (PP) = 0.64
Explanation:
Studying population genetics, we used the Hardy-Weinberg equilibrium model. In the case of snails, the H-W principle predicts that the total number of cases in the population should follow the following equation: PP + 2.Pp + pp = 200.
If a population has 200 snails and 8 are affected by a recessive condition, we have 4% affected, or 0.04.
This means that 0.04 corresponds to p², and p corresponds to 0.2.
Therefore, P is equal to 0.8 because p + P = 1.
Knowing the allele frequencies, we go to the frequency of the genotypes.
F (pp) = p² = 0.04 = 8 snails
F (Pp) = 2.Pp = 0.32 = 64 snails
F (PP) = P² = 0.64 = 128 snails
