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Tomtit [17]
4 years ago
10

Who was the first taxonomist to group animals by blood color and plants size and shape?

Biology
1 answer:
GaryK [48]4 years ago
6 0
<span>ARISTOTLE was the first </span>
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Jacob is studying cell organelles. He knows that light energy drives photosynthesis. Look closely at the diagram below.
Inessa05 [86]

Answer:

C- Plant cells are surrounded by a cell wall. Animal cells are not.

Explanation:

because only C is the correct answer

Hope this help..

5 0
3 years ago
Read 2 more answers
Industrial melanism refers to the dark pigmentation that evolved in some insects giving them protective coloration on vegetation
Natalka [10]

Answer:

  • The frequency of the dominant allele, p =  0.542
  • The proportion of black moths that are heterozygous 2pq = 0.496

Explanation:

According to Hardy-Weinberg, the allelic frequencies in a locus are represented as p and q, referring to the allelic dominant or recessive forms. The genotypic frequencies after one generation are p² (Homozygous dominant), 2pq (Heterozygous), q² (Homozygous recessive). 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 same way, the sum of genotypic frequencies equals 1, this is

p² + 2pq + q² = 1

Being

  • p the dominant allelic frequency,
  • q the recessive allelic frequency,
  • p² the homozygous dominant genotypic frequency
  • q² the homozygous recessive genotypic frequency
  • 2pq the heterozygous genotypic frequency

In the exposed example, 79% of the moths of the species Biston betularia were black due to the presence of a dominant gene for melanism.

If the genotypic frequency of back moths is 0.79, then, by performing the following equation we can get the not-black moths genotypic frequency:

p² + 2pq + q² = 1

where p² is the homozygous dominant genotypic frequency, q² the homozygous recessive genotypic frequency, and 2pq is the heterozygous genotypic frequency.

As 0.79 is the phenotypic frequency of black moths, then this frequency equals p²+2pq.

Clearing the equation:

p² + 2pq + q² = 1

0.79 + q² = 1

q² = 1 - 0.79

q² = 0.21

The genotypic frequency of non-black moths is 0.21. So, from here we can calculate the allelic frequency:

q² = 0.21

q= v 0.21

q = 0.458

If 0.46 is the allelic frequency of non-black moths, then by clearing the equation p + q = 1, we can get the p allelic frequency:

p + q = 1

p + 0.458 = 1

p = 1 - 0.458

p = 0.542

  • The genotypic frequency p² = (0.542)² = 0.294
  • The heterozygote genotypic frequency

        2 x p x q = 2 x 0.542 x 0.458 = 0.496

Finally, we can check this answer by clearing the following equation:

p² + 2pq + q² = 1

0.294 + 0.496 + 0.21 = 1                  

4 0
3 years ago
Help ASAP!!!
Ipatiy [6.2K]

Answer:

Its 0N

Explanation: 50 and 50 pulling from each side will be 0 its like 50-50 from each side wich equals 0

6 0
3 years ago
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Which statement best describes energy of an ecosystem?
vlabodo [156]

The energy of the ecosystem can be best described as option C. The amount of energy entering an ecosystem from the Sun is equal to the amount used by organisms plus the amount lost as heat.



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3 years ago
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Name and describe one of the three conditions Dr. Natterson-Horowitz mentions that have been well-studies by veterinarians, who
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Capture myopathy is a heart condition in animals specifically in tamarins. Dr Natterson-Horowitz studied the condition and discovered that it is similar to cardiomyopathy which is a heart condition in humans. She also studies some conditions that are parallel to both humans and animals.<span />
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