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ASHA 777 [7]
3 years ago
10

What is the role of an indicator? predict how the glucose test strips and iodine would react to the following foods:

Biology
2 answers:
USPshnik [31]3 years ago
4 0
The purpose of an indicator is to test  how acidic or basic something is.

polet [3.4K]3 years ago
3 0

Answer;

The role of an indicator is to establish the equivalence point when two substance has reached its neutralization process.

Explanation;

-The purpose of an indicator is to determine how acidic or basic the product or solution is. Indicator solutions work by actually reacting with the components of food that they test for, creating a new substance of a different color. This is a chemical reaction.If the substance that they test for is not present in the food, there will be no reaction, and therefore no color change. 

-The test strips will be more likely to be dark blue or black for  Grape, Tomato, Summer squash, Orange Bell pepper as they all contain starch. Color intensity may vary depending on the level of starch found in the samples. 

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How many fertilized eggs naturally don’t implant in the uterus, but they pass out of your body during your period?.
mr Goodwill [35]

Answer:

Between one-third and one-half of all fertilized eggs never fully implant. A pregnancy is considered to be established only after implantation is complete

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2 years ago
Alu element at the PV92 locus on Chromosome 16. In this population, 436 people have a heterozygous genotype and 102 people have
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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

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