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Elza [17]
3 years ago
8

Which of the following statements are not true about Endoplasmic 1 Reticulum?

Biology
1 answer:
Serhud [2]3 years ago
6 0
Endoplasmic reticulum isn’t called the control center of the cell

The answer is B
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Question 9 of 10
arsen [322]

Answer:

B

Explanation:

a cell does NOT destroy energy

6 0
3 years ago
Read 2 more answers
Describe what happened when Mendel crossed purebred tall pea plants with short pea plants
Tomtit [17]
He found that the recessive trait such as the short plants disappeared, but only the dominent trait stayed which is tallness. When he allowed them to self pollinate, the recessive gene which is the shortness reappeared.
8 0
3 years ago
I - O selecionador pede para que a pessoa descreva como se comportaria em determinada situação ou trabalhe em cima de um case. P
sashaice [31]

Answer:

O seletor quer saber a ação da pessoa.

Explicação:

O seletor pede que a pessoa descreva como ela se comportaria em uma determinada situação ou trabalharia em um caso, porque o seletor quer saber sobre o comportamento e a inteligência da pessoa para lidar com a situação. o selecionador também quer treiná-lo porque essas situações vão vir na vida profissional dele então ele tem a informação e a experiência de como lidar com isso.

7 0
3 years ago
NEED ANSWER<br> IN PHOTOSYNTHESIS DO YOU LOSE OR GAIN ATONS
o-na [289]

Answer:

you gain

Explanation:

7 0
2 years ago
You discover a population of mustard plants and find it to be at Hardy–Weinberg equilibrium with respect to the R locus. Suppose
natulia [17]

Answer:

The fraction of heterozygous individuals in the population is 32/100 that equals 0.32 which is the genotipic proportion for these endividuals.

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 r-6 allelic frequency is 0,2. This means that if r-6=0.2, then the other allele frequency (R) is=0.8, and the sum of both the allelic frequencies equals one. This is:

p + q = 1

r-6 + R = 1  

0.2 + 0.8 = 1

Then, the genotypic proportion for the homozygous individuals RR is 0.8 ² = 0.64

The genotypic proportion for the homozygous individuals r-6r-6 is 0.2² = 0.04

And the genotypic proportion for heterozygous individuals Rr-6 is 2xRxr-6 = 2 x 0.8 x 0.2 = 0.32

5 0
3 years ago
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