A) explain the hardy-weinberg principle of equilibrium theory. (4 pts.)
b.apply the hardy-weinberg principle to this scenario. in plants, violet flower color (v) is dominant over white (v). if p=.8 and q = 0.2 in a population of 500 plants, how many individuals would you expect to be homozygous dominant ( vv), heterozygous (vv), and homozygous recessive (vv)? how many plants would you expect to have violet flowers, and how many would have white flowers? (6 pts.)

Microbiological experiments are conducted to analyse the growth pattern and capabilities to consume a substrate and undergo multiplcation in number and mass for multiple microbes. Generally, agar or brotg remains the preliminary substrates for analysis of microbial growth. The usual practices include dilution of the stock solution, plating on the plate and observations after a growth period of atleast 24 hours. The number of colonies appearing on the plate are observed and counted manually or automatically. Using appropriate mathematical formula, the actual number of microbial units can be calculated in thw stock solution.

Thus, based upon number of colonies appearing on the plate, the growth can be distinguished as following:

1. No growth: this refers to no appearance of any growth on plate, neither diluted nor stock solution.

2. Poor growth: Qualitatively, poor growth refers to appearance of only 10 or less number of colonies in the stock solution. Such stocks are further concentrated and re-analysed for growth.

3. Good growth: Good microbial growth refers to appearance of 15 or more colonies in the stock solution. Such solutions are considered as moderately concentrated in nature.

Thus, based upon the above mentioned qualitative analyses, the nature of growth can be determined easily.

A non random influence affecting who survives or reproduces is selective pressure. A selective pressure is considered to be a phenomenon in which behaviors of the living organisms are being driven to evolution and the natural selection in which they are also divided in to two types.

Excessive concentrations of phosphorus can lead to eutrophication in aquatic ecosystems, causing harm to aquatic organisms.

Excessive concentrations of phosphorus are considered a pollutant because they can lead to eutrophication. When phosphorus enters aquatic ecosystems from sources such as fertilizer runoff and sewage, it promotes excessive growth of microorganisms. This growth depletes oxygen levels in the water, leading to the death of aquatic organisms like shellfish and finfish.

https://brainly.com/question/13232104

#SPJ6

Answer:

The answer is D. Glycogen

Explanation:

Glycogen is an energy reserve polysaccharide, which is the main source of blood glucose. Its main function is to be an unlimited energy source for the organism. As is known, glucose is necessary to produce energy in the living being and to perform all the processes that occur in an organism. Glycogen is abundantly found in the liver and in smaller amounts in the muscles.

Human waste, excreted from the digestive system, _____.

is considered sterile

contains about 50 percent bacterial cells

has passed through the liver

The question asks for a point on a graph which indicates an outbreak of moose ticks and a rapid decline in the moose population in the context of a predator-prey system study. The point of interest should show a notable drop in the moose population, after a decrease due to a harsh winter and before any significant downward trend in the wolf numbers.

Without the accompanying graph, it's difficult to definitively point out which point represents the outbreak of moose ticks and the consequent drastic decline in the moose population. However, given the context, you'd want to look for a point in the graph where there's a significant drop in the moose population. This point should fall after the first major decrease representing the severe winter and before any major decreases in the wolf population, if such data is represented in the graph. An understanding of the predator-prey system and effects of disease outbreaks on such systems can aid interpreting the graph.

https://brainly.com/question/36444438

#SPJ12

Karyotyping is unable to diagnose microdeletions and microinsertions due to its limitation to detect changes on a micro-scale. While it can identify larger genetic alterations, the minute changes caused by microdeletions and microinsertions are too small to be accurately observed.

The reason why it is difficult to diagnose microdeletions and microinsertions using karyotyping is that the methodology of karyotyping typically involves visualizing the chromosomes in a cell, and the size of the deletions or insertions that can be easily observed via this technique are typically rather large, such as whole chromosomes or significant portions of them. While karyotyping is effective in identifying gross chromosomal alterations such as those in Down syndrome (extra chromosome 21) or Turner syndrome (only one X chromosome in women), it does not offer a high resolution for smaller changes on a micro-scale. Microdeletions and microinsertions involve changes of a very small number of base pairs, too small to be accurately captured by karyotyping. Therefore, these minute changes might get unnoticed in a karyotype.

https://brainly.com/question/18024374

#SPJ3

Final answer:

The MPF protein complex turns itself off by the targeted degradation of its cyclin subunit towards the end of mitosis, allowing the cell to progress to the next stage of the cell cycle.

Explanation:

The MPF protein complex turns itself off by initiating the degradation of its cyclin subunit. MPF, also known as Mitosis Promoting Factor, is composed of two subunits: cyclin and cyclin-dependent kinase (cdk). While levels of cdk remain relatively constant, the levels of cyclin vary, accumulating during the G₁ phase and peaking at mitosis before they are targeted for degradation. This degradation process involves proteasomes that break down cyclin, leading to the inactivation of MPF towards the end of mitosis. The breaking down of cyclin is a critical step in ensuring that MPF activity ceases, allowing the cell to exit mitosis and enter the G₁ phase of the cell cycle.

Answer:

increase in erosion

Explanation

i just tuck the test 3.17

The serratus anterior functions as an antagonist to the rhomboid muscles by performing protraction of the scapula, which is the opposite action of the rhomboids' retraction of the scapula.

The serratus anterior is a large and flat muscle that originates on the ribs and inserts onto the medial border of the scapula. Its primary function is to pull the scapula forward and around the rib cage, a motion known as protraction. This movement is opposite to that of the rhomboid muscles, which retract the scapula by pulling it towards the vertebral column.

Antagonistic muscles have opposite actions. When the serratus anterior contracts, it works against the rhomboids. If the rhomboids are contracting to bring the scapula medially (retraction), the serratus anterior will be relaxed. Conversely, when the serratus anterior contracts to move the scapula laterally (protraction), the rhomboids will be relaxed. This opposition is what makes the serratus anterior an antagonist to the rhomboids.

Answer:

The answer is the interface

Explanation:

Interface is defined as the surface between two portions of matter or the space in which they have a common boundary. For example, the interface allows to know the separation surface between two liquids. In addition to this, the interface allows molecules or particles to cross the membrane, for example, what happens in the osmosis process.