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Vikentia [17]
4 years ago
6

What would be the magnification of a specimen viewed with a compound light microscope that has an objective power of 40x and an

ocular lens power of 10x?
Biology
2 answers:
lorasvet [3.4K]4 years ago
5 0
400 would be the lens power because you multiply them together
Mila [183]4 years ago
3 0

Answer:

The correct answer is 400 times.

Explanation:

The procedure of amplifying the ostensible size, not the physical size of something is known as magnification. This amplification is enumerated by a calculated number also known as magnification. Generally magnification is associated with scaling up of images or visuals in order to witness the details more clearly, with the use of printing techniques, microscopes, or digital processing.  

In the given case, in order to find the magnification, there is a need to multiply the objective power and the power of the ocular lens, that is, M = O × OL

It is mentioned, that Objective power = 40 times and Ocular lens power = 10 times,  

Therefore, M = 40 × 10 = 400. Hence, the magnification of the sample is 400 times.  

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4. A population of water snakes is found on an island in Lake Erie. Some of the snakes are banded and some are unbanded; the ban
Colt1911 [192]

Answer:

233 snakes are heterozygous for the banding allele

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the alleles. The genotypic frequencies after one generation are p² (Homozygous for allele p), 2pq (Heterozygous), q² (Homozygous for the allele q). Populations in H-W equilibrium will get the same allelic frequencies generation after generation. The sum of these allelic frequencies equals 1, this is p + q = 1.

In the exposed example,

  • The banding phenotype is autosomal recessive, bb
  • The frequency of banded snakes on the island is 0.4
  • There are 500 total snakes on the island

How many snakes are heterozygous for the banding allele?

The frequency of banded snakes refers to the genotypic frequency for the trait, which is bb= q2= 0.4.

If q2= 0.4, then q = √0.4 = 0.63

The allelic frequency for b is 0.63.  

This means that the allelic frequency for B or p is 0.37, which we deduce by clearing the equation p + q = 1

                          p + 0.63 = 1

                          p = 1 - 0.63

                          p = 0.37

The allelic frequency of B is 0,37, and the allelic frequency for b is 0,63. The population heterozygote frequency for this allele is 2 x p x q = 2 x 0,37 x 0,63 = 0.466. The percentage of the population that is heterozygous for this allele is 46%.

As the population size is 500 individuals, then we can calculate how many of these snakes are heterozygous. This is: 0.466 x 500 = 233

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lesya [120]

Answer:

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Explanation:

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sleet_krkn [62]

Explanation:

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