Answer:
Explanation:
Bacterial count in stock- 1.85x10^6 cfu/ml
Dilution methods
Take 100 uL or (0.1ml) from stock and add to 900ul (0.9ml) saline and mixed it- this makes 10^1dilution.
Now take 100ul from 10^1 dilution and add to next 900ul saline this is 10^2 dilution, similarly do upto 10^5 dilution.
Then take 100ul from 10^ 4 and 10^5 dilution seperately and plate on LB agar plate seperetely and count the colonies.
Cfu/ml formula= (No.of colonies x dilution factor)/0.1 ml
So suppose, 18 colonies formed on 10^4 dilution then total no. Of cells in stock will be 18x10^4/ 0.1= 18x10^5 cfu/ml.
If we dilute 10^4 or 10^5 that's leads to colony count of 18-19 colonies on 10^4 dilution while 2 colonies should come on plate of 10^5 dilution.
I think the correct answer from the choices listed above is the third option. The barrel of a gun is also known as bore. The barrel is the metal tube which the bullet is fired and the bore is the inside of the barrel. It is the one closest to the barrel so it must be the answer.
Answer:
a. Bands at 1kb, 6kb and 8kb
Explanation:
The EcoRI and BamHI are the restriction enzymes which cut the DNA sequence especially a plasmid at specific sites called the restriction sites.
The restriction enzymes produces bands of specific length therefore these restriction enzymes are used to estimate the approximate length of the DNA.
In the given question, the
1. EcoRI- produces two strands of 7 kb and 8 kb
2. BamHI- produces two strands of 1kb and 14kb
This shows that the length of DNA sequence is 15kb
But when the DNA strand are digested with both the enzymes simultaneously then it will produce three bands as:
i) 14 kb can be broken down in 2 bands of 6 kb and 8 kb
ii) 1 kb band is already produced by the Bam HI.
This shows that 1+6+8= 15 kb
Thus, Option-A is correct.
Answer:
small sprouts will compete for the remaining space that has adequate sunlight
