Answer:
a. It is a dihybrid cross
b. 0%
c. 0%
d. 0%
e. 100%
f. 0%
g. 0%
Note: Answers are given assuming that hhRD = hhDD
Explanation:
a. The cross, HHdd x hhDD is a dihybrid cross involving two traits: fruit colour and fruit shape
2. Gametes produced in the cross are given below:
for HHdd= Hd and Hd
For hhDD = hD and hD
Offspring produced in the cross:
All HhDd, which represents white and the flattened disc-shaped fruit white since They are both dominant characters.
b. Percentage of the offspring from this cross expected to have the HHDD genotype = 0%
c. Percentage of the offspring from this cros expected to have the hhDD genotype = 0%
d. Percentage of the offspring from this cross expected to have the HhDd genotype = 0%
e. Percentage of the offspring from this cross are expected to produce white and disc-shaped fruits (HHDD or HhDD or HhDd) = 100%
f. Percentage of the offspring from this cross expected to produce white and spherical fruits (HHdd or Hhdd) = 0%
g. Percentage of the offspring from this cross expected to produce yellow and disc-shape (hhDD or hhDd) = 0%
Elements in the first row (hydrogen and helium) will have outer electrons in the first energy level. Their principal quantum number is 1. Elements in the second row (lithium through neon) will have valence electrons in the second energy level with a principal quantum number of 2.
Answer: If a cell is placed in a hypertonic solution, water will leave the cell, and the cell will shrink. In an isotonic environment, there is no net water movement, so there is no change in the size of the cell. When a cell is placed in a hypotonic environment, water will enter the cell, and the cell will swell.
Explanation:
B) After determining the optimum pH, they could vary the temperature of the environment to see if catalase is temperature specific
Enzymes are proteins which catalyze reactions by acting on substrates in order to speed up reactions- like the breakdown of large polysaccharides by amylase. Here, the enzyme catalase facilitates the breakdown of hydrogen peroxide into oxygen and hydrogen. Catalase specificity is affected by pH, temperature and the presence of inhibitors.
In temperatures beyond its optimal range, catalase may undergo changes to its physical structure called denaturation; when denatured, enzymes lose their ability to bind specifically to their substrate -i.e. substrate binding specificity is lost. H2O2 would no longer be able to bind to the active site, and thus would not be broken down.
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Plant cells have a cell wall that adds an additional layer of protection from the external environment.