Answer:
C. The DNA copied, a new cell membrane forms between the DNA copies, and the cell splits into two cells.
Explanation:
Each carbon atom in ethane forms 4<span> bonds: 1 to a carbon atom and 3 to each of the attached hydrogen atoms. A single covalent bond containing 2 shared electrons is also called a sigma (σ) bond.
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Answer:
See the answer below
Explanation:
Normal height - H
Short height - h
Heterozygous - Hh
A cross involving heterozygotes would be Hh x Hh
Hh x Hh
HH Hh Hh hh
Since H is dominant over h;
HH and Hh would produce normal height phenotypes.
hh will produce a short height phenotype.
Thus;
<em>Probability of normal height phenotype resulting from the cross = 3/4 or 75%</em>
<em>Probability of short height phenotype resulting from the cross = 1/4 or 25%</em>
Question is incomplete. Complete question is as follows:
You have decided to determine how many microbes are living on the lettuce in the salad bar at your favorite restaurant. You place 1 gram of lettuce and 99 mls of water in a blender and blend the mixture. This is sample A. You then transfer 1 ml of this dilution into to another that contains 9 mls of water. This becomes sample B. You next transfer 1 ml of sample B into a separate container that contains 9 mls of water. This is sample C. Next you transfer 1 ml each from samples B & C onto separate nutrient rich agar plates, swirl, let harden and incubate at 37C. When you examine the plates after 48 hours you find 110 colonies growing on plate C. How many microbes were living on that 1 gram of lettuce?
Answer:
1.1x10^6 microbes
Explanation:
When 1gm of lettuce is mixed with 99 ml of water, the microbe concentration is diluted 100 times (Sample A). 1 ml of sample A is mixed with 9ml of water to make sample B which further dilutes microbe concentration 10 times making the total dilution 1000 times. This process is repeated with sample B to make sample C increasing the dilution to 10000 times.
110 colonies are counted from this sample C. Each colony signifies a single microbe during plate counting method. So, 110 microbes were present on the plate. Original number of microbes = microbes counted * times of dilution =
110 * 10000 = 1100000 = 1.1x10^6 microbes
