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Aleks [24]
3 years ago
15

You have decided to determine how many microbes are living on the lettuce in the salad bar at your favorite restaurant. You plac

e 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
Biology
1 answer:
REY [17]3 years ago
8 0

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

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Answer:

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Explanation:

1. CGAGGTT → CGTT (Deletion) ---- This is called deletion because it involves the removal of 3 base pairs (AGG) from the DNA sequence.

2. ATTCGG → ATTCGGATTCGG (Duplication) ---- In this case, the particular sequence (ATTCGG) is copied again or duplicated.

3. CTTAAT → TAATTC (Inversion)----- This is called inversion mutation because the DNA sequence breaks off and is reattached but this time in a reverse order i.e. CTT becomes TTC, placing the last base first and the first base last.

4. CTTAAT → CTTAACGCT (Insertion) ------ This is called insertion mutation because it involves the addition of extra base pairs (CGC) into the sequence. The Insertion occurs between the last A and T nucleotide.

5. CGAT → CTAT (Substitution) ----- This is called substitution because Guanine base is replaced by Thymine in the DNA sequence. It is specifically called a transversion substitution because a purine (Guanine) is replaced by a pyrimidine (Thymine). It is called a point mutation because it involves a single base.

6. CCGGTT + TTAGGC = CCGTTA + GTTGGC (Translocation) ----- in this case, CCGGTT and TTAGGC are sequences on different chromosomes. Portions of sequence on the first chromosome (GTT) and second chromosome (TTA) breaks off and gets reattached/exchanged in each other i.e. the first chromosome gets TTA while the second gets GTT. This kind of mutation is called translocation.

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