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mamaluj [8]
2 years ago
13

When testing a person's skin for allergens using patch testing, tiny amounts of allergens are placed on stainless steel discs an

d applied to the back. How many hours later is the skin under the discs assessed for reaction?
Biology
2 answers:
Airida [17]2 years ago
7 0

Answer:

48 hours

Explanation:

Patch testing is a method to test the allergic reactions called contact dermatitis caused by the allergens found in the environment the patient is surrounded by like home allergens or work allergens.

To perform the test, the allergens are placed on the metal discs which are pasted on the back of the person with hypoallergenic tape.

Then the discs are allowed to be pasted on the back of the person for about 2 days or 48 hours after which the skin is examined for the allergy reactions caused by an allergen in the form of inflammatory response or hypersensitivity.

Thus, 48 hours is the correct answer.

Jet001 [13]2 years ago
7 0

Answer:

The correct answer is - 48 hours.

Explanation:

Skin patch testing is the process or the testing method that is used to detect allergic dermatitis an individual person could be affected of from the atmosphere around him. In this test the patch with substances to test, in a small discs to skin of a person.

The test substances to the skin under the disc that are then left in place for 48 hours. The skin is then assessed a further 48 hours later for any allergic reaction.

Thus, the correct answer is - 48 hours.

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Drag each tool below into the box corresponding to the step in the dmaic cycle where it is most likely to be used.
aev [14]
I found the exercise on the internet and the tools are organised below into their corresponding step of the DMAIC cycle.
             

Define (a<span>rticulate the problem, objectives, resources, project, etc):</span>

-Quality Circles

-Flow charts

Measure (collection of data, establish base lines for improvement):

-Pareto Charts

-Check Sheets

Analyse (identify, validate):

-Cause-and-effect diagrams

Improve (test and implement solutions):

-Brainstorming 

Control (monitoring improvements and ensure continuing success):

-Control Charts for new Process

7 0
2 years ago
Suppose the sub-hypothesis that wing waving alone reduces predation by jumping spiders was supported by the Zonosemata experimen
Sliva [168]

Answer:

The correct answer is: <em>Zonosemata flies with housefly wings are attacked less frequently</em>

Explanation:

The Zonosemata experiment is a scientific experiment that was conducted to investigate how zonosemata flies are able to protect themselves from predation by jumping spiders. In this study, housefly wings were transplanted onto zonosemata flies. This made zonosemata flies protected from attack by jumping spiders. On the other hand, when zonosemata wings were transplanted onto houseflies, they were easy targets for jumping spiders. Therefore, if the sub-hypothesis that wing waving alone reduces predation by jumping spiders is proven, the logical result would be that:  Zonosemata flies with housefly wings are attacked less frequently.

4 0
2 years ago
Gene A has two alleles (A1 and A2). A1A1 homozygotes are twice as likely to survive from birth to reproductive age as heterozygo
Usimov [2.4K]

Answer:

The correct answer is C: <em>0.60 </em><em>A1</em> and <em>0.40 A2</em> will be the allele frequencies in those progeny when they reach reproductive age

Explanation:

<em>Genotype</em>                                <u><em>A1A1 </em></u><em>  </em><u><em>A1A2</em></u><em> </em><u><em>A2A2 </em></u>

<em>Relative aptitude, w</em>                  1           0.5    0.5

<em>Number of individuals</em>                 40            40     40

<em>Initial allelic frequency</em>      p0= (40+20)/120=0.5    q0= (40+20)/120=0.5

<em>Zygote frequency</em>                  p2= 0.25 2pq=0.25 q2=0.625

<em>Relative contribution</em>         0.25x1=0.25     0.5x0.5=0.25 0.25x0.5=0.125

<em>of each genotype</em>              

<em>Average aptitude W</em>              W= 0.025 + 0.25 + 0.125 = 0.625

<em>Population</em>                    AA= 0.25/0.625    AB=0.25/0.625  BB=0.125/0.625

<em>Genotype frequency</em>             AA= 0.4             AB=0.4          BB=0.2  

<em>New Allelic frequency</em>       p1=0.4+(0.4/2)=0.6   q1=0.2+(0.4/2)=0.4

  • <em>Total number of individuals</em>: 120
  • <em>Initial allelic frequency</em>:

                  (number of homozygote individuals + half number of  

                   heterozygote individuals) / Total number of individuals

  • <em>Relative contribution of each genotype</em>:

                           Zygote frequency x Relative aptitude

  • <em>Average aptitude W</em>: It is the sum of relative contribution of each genotype to the next generation.  

                    wA1A1 x p2 + WA1A2 x 2 x p x q + WA2A2 x q2

  • <em>Population Genotype frequency</em>:

                Relative contribution of each genotype / Average aptitude

  • <em>Allelic frequency</em>:

                      Homozygote population genotype frequency +  half  

                           heterozygote population genotype frequency

8 0
2 years ago
Alleles A and a are located on a pair of metacentric chromosomes. Alleles B and b are located on a pair of acrocentric chromosom
Leto [7]

Answer:

See the second attached image showing the gametes from the parent with the location of the centromere indicating the type of chromosome

Explanation:

According to the position of the centromere, chromosomes can be of 4 types namely:

  1. Metacentric
  2. Acrocentric
  3. Telocentric
  4. Sub-metacentric

A metacentric chromosome is a chromosome that has no short or long arm. The arms are equal in length with the centromere joining the two sister chromatids located at the center.

Acrocentric chromosomes have unequal arm lengths with the centromere skewed towards one end of the chromosome.

Telocentric chromosomes have their centromeres at one end of the chromosome.

Sub-metacentric chromosomes have unequal arm lengths but the centromere is not as skewed to one end of the chromosome length as found in acrocentric chromosome.

The only possible gamete from <em>aa bb</em> parent is <em>ab</em>. Recall that the allele <em>a </em>is located on a metacentric chromosome while allele <em>b </em>is located on acrocentric chromosome.

4 0
2 years ago
Next, you perform a cross with a mouse with black fur and a mouse with white fur. which result(s) for the offspring would indica
Komok [63]

Answer:

8 mice with black fur, 1 mouse with white fur

5 mice with black fur, 6 mice with white fur

6 mice with black fur, 2 mice with white fur

Explanation:

I just got it right

6 0
2 years ago
Read 2 more answers
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