Answer:
1. Minimum viable population is the estimate of the fewest number of organisms a population needs to avoid extinction.
2. The measurement will most likely stay the same if the number of offspring each female in the population produces increases.
3. If the death rate increases, this measurement will most likely increase.
Explanation:
Minimum Viable Population:
- Minimum viable population is an ecological threshold that signifies the minimum number of individuals required to prevent the species from going extinct.
- MVP determines population conservation benchmarks.
- MVP also helps in keeping a population genetically diverse as a limited number of individuals present limited mating opportunities and therefore limited genetic diversity.
Organisms eat plants, which is comprised of carbohydrates; those carbohydrates come from Carbon<span> dioxide in the atmosphere. Thus, an abiotic factor (</span>carbon<span>dioxide) helps create a </span>biotic<span> factor (the plants made out of carbohydrates).</span>
Q1. The answer is removing metabolic wastes from the body.
Excretion is the process through which metabolic wastes are removed from the body. Skin, lungs, and greatly kidney, which are the part of the excretory system, are responsible for excretion of metabolic waste in vertebrates. Invertebrates have special systems (insects, for example, have Malpighian tubules) or use skin to excrete metabolic wastes while single-celled organisms use the whole surface of the cell.
Q2. The answer is some animals live in dry or salty environments.
Kidneys are important organs in maintaining water balance. Some animals that live in dry and salty environments must preserve water in order to maintain homeostasis. They drink and eat food with more salt in it. If they lose that precious water in such conditions, the amount of different salts in the organism will increase and it will affect a normal functioning of the organism.
Q3. The answer is simple diffusion across the skin.
Ammonia is very toxic substance and a lot of water is needed for its neutralisation and excretion. Therefore, animals that live in water excrete ammonia directly in the water. Many freshwater invertebrates eliminate ammonia through skin. In animals that do not live in the water, kidneys and liver help conversion of ammonia into urea which is then excreted.,
Q4. They both actively pump salt across their gills.
Both saltwater and freshwater fishes use gills to eliminate nitrogenous wastes while kidneys have a little role in the elimination of this kind of the waste. Salt that is lost is replaced by active transport of salt ions into the body by the gills.
Q5. The answer is They both convert nitrogenous wastes to uric acid.
A garden spider and a sparrow are terrestrial organisms. They do not live in the water and do not excrete metabolic wastes in the water. It is known that ammonia is toxic nitrogenous substance and a lot of water is needed for its excretion. For water organisms this is not a problem, they are surrounded by water, but terrestrial organisms, such as the garden spider and the sparrow, have no such amount of water in the environment, so their kidneys and liver must convert ammonia into urine which can then easily be excreted.
Answer:
Explanation:
According to the exercise we can infer that for the Alu insert:
p: positive allelic frequency
q: negative allelic frequency
maintaining that the population is in equilibrium we can carry out the following formula
p + q = 1 and pp + 2pq + qq = 1
looking for the genotype frequency we clear and obtain the following data
genotype frequency of 2pq = 436/1000 = 0.436
The genotype frequency of qq = 102/1000 = 0.102
This is how we look now:
number of positive people for Alu = 1000- (436 + 102) = 1000- 538 = 46
In this way it is resolved that:
genotypic frequency of pp = 462/1000 = 0.462
p = 0.462 + (0.436 / 2) = 0.462 + 0.218 = 0.680
q = 0.102+ (0.436 / 2) = 0.102 + 0.218 = 0.320
According to the exercise carried out it is deduced:
The value of p in the population is = 0.68
We conclude that our prognosis showing a homozygous positive genotype is: 0.462
