Both of the cells are eukaryotic, both have:
Cell Membrane
Nucleus
Endoplasmic reticulum
Mitochondria
Lysosome
Golgi Apparatus
Peroxisomes
Cytosol
Cytoskeleton
Vocules
And that it's my notebook contain these only
:D
Answer:
The correct answer would be "NADH delivers its electrons to complex I and FADH₂ deliver its electrons to complex II" in cellular respiration.
There are mainly four complexes associated with electron transport chain of cellular respiration.
Complex I or NADH: ubiquinone oxidoreductase is the complex at which NADH is oxidized to form NAD⁺. The free electrons are transported with the help of ubiquinone.
Complex II or succinate dehydrogenase is the complex associated with oxidation of FADH₂ to FAD⁺. It also transports the free electrons with the help of the ubiquinone pool.
Complex III or cytochrome bc1 complex transport free electrons from ubiquinone to the cytochrome C which is a water-soluble electron carrier.
Complex IV or cytochrome c oxidase transport the free electrons to oxygen to form water.
Taxol is a chemotherapeutic drug that will attach to the spindle fibers and prevent their function thereby killing the cancer cell. The Anaphase process is avoided from occurring by this drug.
<h3>What is Taxol?</h3>
Taxol (chemical name: paclitaxel) is a anti-cancer chemotherapy drug used to treat cancer, blocks cancer cell growth by stopping cell division, resulting in cell death. It is used in treatment of breast cancer, ovarian cancer, non-small cell lung cancer, pancreatic cancer, and AIDS-related Kaposi sarcoma.
Taxol interferes with anaphase process by inhibiting the shortening of the microtubules attached to the sister chromatids, so the chromatids do not separate and hence, will not move to the ends of the cells. Ultimately the cell division does not take place.
To learn more about Taxol, click the given link brainly.com/question/15860122
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Answer:
0.25%
Explanation:
20 people start the new population. So there are 20 genes or 40 alleles for the recessive disorder phenylketonuria. 2 out of 40 alleles are recessive for the condition hence frequency of the allele = 2/40 = 0.05
Frequency of the allele does not change when the population increases so it is in Hardy-Weinberg equilibrium. According to it, if q is the frequency of recessive allele, q² = frequency of the recessive condition
Here, q = 0.05 So,
q² = (0.05)² = 0.0025
In percentage, it is 100 * 0.0025 = 0.25%
Hence, incidence of phenylketonuria in the new population is 0.25%
