<span>A group of people living in the same place or having a particular characteristic in common.</span>
N an experiment, suppose that the wings of fruit flies were clipped short for fifty generations. The fifty-first generation emerged with normal-length wings. This observation would tend to disprove the idea that evolution is based on
a. inheritance of natural variations
b. inheritance of acquired characteristics
c. natural selection
d. survival of the fittest
Inheritance of acquired characteristics. Thus, option "B" is correct.
<h3 /><h3>What is inheritance of acquired characteristics?</h3>
For fifty generations wings of fruit flies were clipped. Hence they acquired this trait in their lifetime and not genetically. If acquired characteristics were capable of passing on to next generation, 50 generations would have been enough to inculcate this clipped wing trait in fruit flies. Despite it, the fifty-first generation did not have clipped wings.
Hence evolution can not occur without genetic variation. A character simply acquired in a lifetime does not create a difference in germ cells and hence is not enough to be passed on to next generation or cause evolution
To learn more about genetic variation click here:
brainly.com/question/848479
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Answer:
Argon
Explanation:
Argon is actually used to determine the age. It's called potassium-argon dating b/c when volcano erupts the rocks in it contain potassium. The potassium slowly decays and produces argon. Scientists measure how much argon is in a given volcanic rock and do math backwards to figure out how many potassium atoms were needed to produce the given amount of argon. We already know the rate at which potassium decays so then they put two and two together to find the age.
Here's a link with some cool info on it:
https://divediscover.whoi.edu/archives/hottopics/volcano.html
Answer:
C) 2.0 kb
Explanation:
It is given that out of the 4 nucleotides A, T, C & G each one has equal probability to occur at any position on the DNA molecule which simply means that the probability of occurrence of any nucleotide at a position is 1/4.
Also, it is given that probability of occurrence of either A or T at 3rd position is equal which means that the probability at that particular position will be 2/4 = 1/2.
Now, GA(A/T)TC is the DNA sequence where Restriction enzyme HinfI cleaves so the total probability of an average HinfI cleavage fragment will be = 1/4 x 1/4 x 1/2 x 1/4 x 1/4 = 0.00195 = 0.2 i.e. 2 kb.
