Answer:
The molecules in the air can be displaced by a force creating sound waves. Vibrations generally travel faster in a solid compartment than as in gases.
In 1906, Harden and Young, in a series of classic studies on the fermentation of glucose to ethanol and CO 2 by extracts of brewer's yeast, made the observations inorganic phosphate was essential to fermentation; when the supply of phosphate was exhausted, fermentation ceased before all the glucose was used; during fermentation under these conditions, ethanol, CO 2 , and a sugar phosphate accumulated; when arsenate was substituted for phosphate, no sugar phosphate accumulated, but the fermentation proceeded until all the glucose was converted to ethanol and CO 2 . Which enzyme of glycolysis requires inorganic phosphate and, therefore, stops when no phosphate is available
Answer:
The enzyme is Glyceraldehyde-3-phosphate-dehydrogenase. (GAPDH). It is the enzyme that converts Glyceraldehyde-3-phosphate to D-glycerate 1,3-bisphosphate; the sixth Glycolytic pathway for breaking down glucose to ethanol, C02 in Glycolysis. This enzyme requires inorganic phosphate as substrate for the catalytic reaction to proceed. Since enzymatic reactions take place by forming enzyme-substrate complexes, absence of the inorganic phosphate substrate ; stops the conversion and progress of fermentation .
In dogs, black fur is dominant to white fur. Two heterozygous black dogs are mated. What would be the probability of the following combinations of offspring in a litter of six pups?
a) All with black fur.
b) Four with black fur and two with white fur.
c) At least two of the pups having white fur.
d) The firstborn pup with white fur, and among the remaining five pups, three with black fur and two with white fur.
This problem uses the concepts of Mendelian genetics and binomial probability to calculate the chances of various offspring combinations regarding black and white fur in dogs, assuming that the parent dogs are heterozygous for the trait.
Explanation:In this case, we are making use of the principles of Mendelian genetics to determine the probability of different outcomes when two heterozygous black dogs are mated.
a) As each pup is an independent event, the probability of an offspring being black is 0.75 (3 out of 4 possibilities). The likelihood of all six pups being black then is (0.75)^6 = about 0.18, or 18%.
b) For four black fur and two with white fur, this is a binomial probability case. Using the binomial coefficient, it works out to (6 choose 4)*(0.75^4)*(0.25)^2 = about 0.32, or 32%.
c) The probability of at least two of the pups having white fur is 1 minus the probabilities of having none or only one white puppy, which works out to 1- [(6 choose 0)*(0.75^6)*(0.25^0)+ (6 choose 1)*(0.75^6)*(0.25)] = about 0.89 or 89%.
d) For the firstborn pup with white fur, and among the remaining five pups, three with black and two with white, it would work out to (0.25)*(5 choose 3)*(0.75^3)*(0.25^2) = about 0.10 or 10%.
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