The stages of the cell cycle in order are interphase<span>, prophase, </span>metaphase<span>, anaphase and telophase. This process is known as </span>mitosis<span> and is used to generate new cells.</span>
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<h2>_____________________________________</h2><h2>Prokaryotes:</h2>
Prokaryotes are single cell organism. Prokaryotes were the most earliest organism and they lived and evolved all alone on earth for 2 billion years. They have continued to adapt and flourish on an evolving earth and in turn they helped the earth to change it. Prokaryotes, the simple creatures fed on carbon compounds that were accumulating in Earth's early oceans. Slowly, other organisms evolved that used the Sun's energy, along with compounds such as sulfides, to generate their own energy.
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h2 />
Enzymes catalyze reactions by decreasing the activation energy of a reaction.
A catalyst is a material that speeds up a chemical reaction without really being a reactant. Enzymes are the biological catalysts that drive biochemical reactions in living things. Most enzymes are proteins, however certain ribonucleic acid (RNA) molecules can also function as enzymes. Enzymes play a crucial role in reducing a reaction's activation energy, or the amount of energy required for the reaction to start.
Enzymes function by attaching to reactant molecules and holding them so that chemical bond-forming and bond-breaking reactions proceed more quickly. An enzyme will attach itself to one or more reactant molecules and start the process by catalyzing it. These compounds are the substrates for the enzyme. One substrate may undergo several product breakdowns in various reactions.
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Answer:
0.9342
Explanation:
The Hardy-Weinberg equation states that p² + 2pq + q² = 1,
where p is the frequency of the dominant 'normal' (n) allele and q is the frequency of the recessive 'albino' (a) allele in the population, while q² represents the frequency of the homo-zygous albino genotype (aa), p² represents the frequency of the homo-zygous normal genotype (nn) and 2pq represents the frequency of the heterozygous genotype (na).
In this case, the frequency of individuals in the population that have the genotype aa (q²) is equal to 26/6000 = 0.004333. In consequence, q is equal to √ 0.004333 = 0.0658.
Moreover, the allele frequency of the normal (n) allele p is equal to 1 - q = 1 - 0.0658 = 0.9342, so p² (nn) = (0.9342)² = 0.8727.
Finally, the frequency of the heterozygous genotype (na) is 2pq = 2 x 0.9342 x 0.0658 = 0.123.
C, presence of nucleic acid. not all living things are microscopic, many living things have nuclei in their cells, and not all living things need a host cell or body.