The stage of the design process that the designer conducts a beta test is stage 5.
Nucleus could be considered the place where the rollercoasters are controlled and stopped and started
Explanation:
These questions are related to the extraction of DNA using detergents and ethanol.
1. Consistency and texture- the extracted DNA is white, spongy and moist or slime in nature.
2. Salt is used during the extraction process as the DNA is negatively charged molecule and to neutralize the negative charge which will allow the formation of precipitate.
3. The DNA is soluble in aqueous solution as the DNA is a negatively charged molecule therefore the negative charge makes it a polar molecule and it can form the interactions in the aqueous solution.
4. RNA could be present in the aqueous solution along with the DNA as both are the nucleic acid.
5. The base-pairing rule was suggested by the Chargaff who proposed that Adenine binds thymine and guanine binds cytosine.
5. The binding of adenine to thymine involves two hydrogen bonds and between guanine to cytosine involves three hydrogen bonds.
Answer:
D. The methyl group of acetyl CoA becomes radio-labeled
Explanation:
During the steps in glycolysis, the carbon at position 1, becomes C-1 in dihydroxyacetone phosphate during the cleavage of fructose-1,6-bisphosphate to dihydroxyacetone phosphate and glyceraldehyde-3-phosphate. Subsequently on isomerization of dihydroxyacetone phosphate to glyceraldehyde-3-phosphate, C-1 of dihydroxyacetone phosphate becomes C-3 of glyceraldehyde-3-phosphate.
Furthermore, in pyruvate, the end product of glycolysis, C-3 is converted to a methyl group which then becomes the methyl group in the acetyl-CoA molecule produced from the oxidative decarboxylation of pyruvate.
Since the radioactive 14-C of radio-labeled glucose occupies position 1, it will become the methyl group of acetyl-CoA.
The phenotype would b all short hair; the genotype would be all heterozygous dominant.
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