Explain why organisms that have an electron transport chain as well as fermentation pathways seldom ferment pyruvate, if an elec
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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>
Mutations are the changes produced in the nucleotide sequence of the genome.
There are four main types of mutations
Substitution
Insertion
Deletion
Duplication.
<h2>_____________________________________</h2><h3>DELETION:</h3>A small segment of chromosome mat be missing. This condition is known as deletion.
For example, Normal chromosome has A B C D E F G. If deletion mutation occurs then mutated chromosome has A D E F G and B C got deleted.
<h2>_____________________________________</h2><h3>DUPLICATION:</h3>In this condition, a part of chromosome present in exec ess to the normal chromosome.
For example, Normal Chromosome has A B C D E F G. If duplication mutation occurs, then mutated chromosome had A B C B C D E F G and B C is duplicated.
<h2>_____________________________________</h2><h3>SUBSTITUTION MUTATION:</h3>Substitution is a type of mutation where one base pair is replaced by a different base pair.
For example, in the sequence CAAGT, if C replaces G, it is a substitution mutation.
<h3>INSERTION MUTATION:</h3>In genetics, an insertion is the addition of one or more nucleotide base pairs into a DNA sequence.
For example, in the sequence CAAGT, if extra base G gets inserted after C, the new sequence would be CGAAGT.
<h2>_____________________________________</h2>Both substitution and insertion mutations change the position of nucleotide thus, the type of amino acid.
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>Answer:
a) Cell concentration when the dilution rate is one-half of the maximum is 0.598g cell/L
b) the substrate concentration when the dilution rate is 0.8
is 5.2g/l
c) the maximum dilution rate is : 0.41 h⁻¹
d) the cell productivity at 0.8
is 2.40g cell/L
Explanation:
Given data :
Mass doubling time of Pseudomonas sp. = 2.4 hr
Saturation constant = 1.3 g/L
Cell yield on acetate = 0.46g cell/g acetate
We are to find;
a. Cell concentration when the dilution rate is one-half of the maximum.
Here, cell yield =amount of cell produced / amount of substrate consumed.
[S] at 0.5D max is determined using the Monod's equation.
Using the formula :
, where D is the dilution rate,
[S] is substrate concentration; &
Ks is the saturation constant.
By replacing the values, we get :
[S]=1.3g/L
The cell concentration at 0.5Dmax= cell yield x substrate consumed at 0.5Dmax.
=0.46×1.3
= 0.598g cell/L
b)
Substrate concentration when the dilution rate is 0.8 is calculated as:
0.8=[S]/1.3+[S]
1.04+0.8[S]=[S]
[S]= 5.2g/L
Therefore , the substrate concentration when the dilution rate is 0.8
is 5.2g/l
c)
Maximum dilution rate is calculated using the expression
=1/2.4
=0.41 h⁻¹
So, the maximum dilution rate is : 0.41 h⁻¹
d)
The cell productivity at 0.8 can be calculated by multiplying the amount of the cell yield with the amount of the substrate consumed at 0.8
Cell yield =
Cell productivity at 0.8 = 0.46 × 5.2
=2.40g cell/L
Therefore, the cell productivity at 0.8
is 2.40g cell/L