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sp2606 [1]
3 years ago
13

The serum elisa test is based on interaction between

Biology
1 answer:
umka2103 [35]3 years ago
6 0
The test is based on a microtiter plate that has a solid phase substrate (target protein, antigen) at a known concentration fixed to the plate that when exposed to an antibody that has an indicator attached (dye for color change or enzyme-labeled antibody) that can produce a color change.
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Albinism (the total lack of pigmentation) is recessive (and really rare!) in humans. If an albino female marries a normal but he
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Answer: not sure on the percentenge all i know is the albino gene is a recessive gene

Explanation:

5 0
3 years ago
Match each type of rock with the description of how it forms.
weeeeeb [17]

Answer:

metamorphic-existing rocks are exposed to extreme conditions

sedimentary rock-layers of sediment dry together

igneous rock-magma or lava cools and solidifies.

Explanation:

3 0
3 years ago
Describe what are harmattan winds.
kirza4 [7]

Answer: Harmattan, is a cool dry wind that blows from the northeast or east in the western Sahara and is strongest from late November to mid-March. It usually carries large amounts of dust, which it can transport hundreds of miles out

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3 years ago
A student has two solutions with the same acid-base indicator
castortr0y [4]

the indicator will have different colors in the different solutions

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3 years ago
Two linked genes, A and B, are separated by 18 cM. A man with genotype AB/ab marries a woman who is ab/ab. The man’s father was
kolbaska11 [484]

Answer:

a) 9%.  

b) 16.8%.

Explanation:

a).

We are provided with the information that Two linked genes, A and B, are separated by 18 cM (centiMorgan). i.e the recombinant frequency is 18%

Also , the man's genotype is AB/ab... This only result to one explanation, that The man will definitely produce 18% of recombinant gametes which entails

9% Ab & 9% aB

i.e 0.09 Ab & 0.09 aB

On the other-hand, The mother ab/ab have tendency to produce just one single type of gamete which is ab

∴

The probability that their first child will be Ab/ab will be

Pr ( Ab/ab) = (0.09) x (1)

= 0.09

= 9%.  

b).

If the father produces 18% of recombinant gametes which entails

9% Ab & 9% aB , this typically implies that the number of the non-recombinant gametes will be;

100%-18% = 82%  ( non-recombinant gametes)

i.e genotype AB/ab = 82%

AB =41%; ab = 41%

AB = 0.41 ; ab = 0.41

Now, the probability that their first two children will both be ab/ab:

Using Multiplication Rule to calculate the probability that their first two children (ab/ab); we have:

 (0.41)(1) ×(0.41)(1)

= 0.1681

= 16.8%.

5 0
4 years ago
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