Is the raccoon a producer, primary, secondary, or tertiary consumer?

Which stage of the human immunodeficiency virus (HIV) would a client with a CD4 T cell count of 325 cells/mm3 be classified

kirza4 [7]

Based on the CDC guidelines a person with a CD4+ cell count between 200-400 is regarded as being in <u>the second stage</u> of the HIV infection.

Acquired Immunodeficiency Syndrome

Acquired immunodeficiency syndrome (AIDS) is a condition caused by HIV (human immunodeficiency virus). AIDS is a chronic disorder and can become life-threatening for the infected person. The HIV viruses reproduce inside the host and while doing so they destroy the CD4+ immune cells which reduce the person's ability to fight against other diseases.

CD4+ cell count and AIDS

According to the CDC (U.S. Centers for Disease Control and Prevention), HIV infection is categorized into different stages based on the number of CD4+ cells.

Stage 1: - CD4+ cell count is more than 500 per microlitre. There are no symptoms or AIDS-related conditions.
Stage 2: - CD4+ cell count is more than 500 per microlitre. There are no symptoms or AIDS-related conditions.
Stage 3: - The CD4+ cell count falls below 200 and AIDS-related symptoms and conditions are present.

Learn more about HIV infection here:

brainly.com/question/13662165

#SPJ4

7 0

2 years ago

In hamsters, we can examine the genes for coat color and eye color which are linked:B = brown coat, b = white coat, G = green ey

Inessa [10]

Answer:

10.2 cM

Explanation:

Given,

pure breeding parent 1 : BBgg = brown coat and pink eyes

pure breeding parent 2 : bbGG = white coat and green eyes

BBGG X bbgg = F1 = BbGg ( All are brown with green eyes )

When F1 progeny is test crossed with white coat and pink eyes hamster:

BbGg X BbGg = F2 =

brown, pink eyes = 99 = parental

white, green eyes = 103 = parental

brown, green eyes = 12 = recombinant

white, pink eyes = 11 = recombinant

Recombination frequency =

(total number of recombinants / total progeny ) * 100

= [( 11 + 12 ) / ( 99 + 103 + 12 + 11 )] * 100 = (23/225)*100 = 0.102 * 100 = 10.2 %

1 % recombination frequency = 1 cM distance between genes

Hence, the distance between B and G gene is 10.2 cM

4 0

3 years ago

The heart damage associated with Lyme disease is

alexandr402 [8]

ANSWER:

The heart damage associated with Lyme disease is Lyme carditis.

~batmans wife

6 0

3 years ago

QUESTION 31 Chipmunk mitochondrial DNA was fingerprinted using the single primer amplified region technique (SPAR) and the PCR r

natima [27]

Answer:

It depends on the size of the amplified DNA fragments (for example for SPAR markers consisting of 200 bp in length >> 1600 / 200 = 8)

Explanation:

The PCR based single primer amplification reaction (SPAR) methods are tools for detecting genetic diversity by using DNA markers such as Simple Sequence Repeats (ISSRs), Minisatellite DNA regions (DAMDs) and Random Amplified Polymorphic DNAs (RAPDs), which can be used alone or combined with each other. In the last years, the SPAR methods have gained attention in the scientific community because they are cost-effective and highly effective for the detection of both intraspecific and interspecific genetic variation. In the SPAR methods, the size of the bands will depend on the length of the amplified DNA fragments.

6 0

3 years ago

During a workout, a football player pushes a blocking dummy a distance of 30 m. While pushing the dummy the same distance a seco

tresset_1 [31]

The player would need to use a greater force and more time.
Because realistically he would already be wore out from the first time. Therefore exerting more force would still wear you out and thus taking more time.

Take this from a former linebacker :)

7 0

3 years ago

Read 2 more answers