Answer:
There will be a dominant and recessive trait that would be formed.
Explanation:
3 purple flowers would be dominant while 1 will be recessive.
Answer:
The enzyme is the pyruvate kinase
Explanation:
The designated enzyme can only grow in pyruvate or in ethanol. They are said to be in anaerobic conditions. If the end product of glycolysis is pyruvate and the enzyme that transforms phosphoenol pyruvate into pyruvate is called pyruvate kinase. In a reaction under anaerobic conditions, the pyruvate is transformed into ethanol. So if the pyruvate kinase enzyme is mutated, pyruvate cannot be transformed into ethanol and Saccharomyces cerevisiae could not survive in that medium.
