0:3:1
Explanation:
Tasters have the dominant allele for the itter taste of PTC, they can be homozygous TT and heterozygous Tt.
Non-tasters are supposed as autosomal recessive characters
they are homozygous recessive tt when expressed
Given in the equation two heterozygotes parents would produce offspring with the trait in the following ratio.
Tt (parent 1)
Tt (parent 2)
If a punnet square is made
T t
T TT Tt
t Tt tt
The phenotypic character is in the ratio of 0:3:1
There will be 25% chances of recessive trait of non-tasters in the progeny of heterozygous parents.
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The lungs are lined with epithelial cells:)
An enzyme is an organic catalyst.
<h3>
What is an enzyme?</h3>
Enzymes are proteins that help speed up metabolism or the chemical reactions in our bodies.
They build some substances and break others down.
All living things have enzymes. Our bodies naturally produce enzymes.
But enzymes are also in manufactured products and food.
Examples of specific enzymes:
Amylase: In the saliva, amylase helps change starches into sugars.
Maltase: This also occurs in the saliva, and breaks the sugar maltose into glucose.
Trypsin: These enzymes break proteins down into amino acids in the small intestine.
To learn about enzymes, refer
https://brainly.in/question/15327487
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Answer:
When the patch occupancy rate (c) equals the patch extinction rate (e), patch occupancy (P) is 0
Explanation:
According to Levin's model (1969):
<em>dP/dt = c - e</em>
where P represents the proportion of occupied patches.
<em>c</em><em> </em>and <em>e </em>are the local immigration and extinction probabilities per patch.
Thus, the rate of change of P, written as dP/dt, tells you whether P will increase, decrease or stay the same:
- if dP/dt >0, then P is increasing with time
- if dP/dt <0, then P is decreasing with time
- if dP/dt = 0, then P is remaining the same with time.
The rate dP/dt is calculated by the difference between colonization or occupancy rate (<em>c</em>) and extinction rate (<em>e</em>).
c is then calculated as the number of successful colonizations of unoccupied patches as a proportion of all available patches, while e is the proportion of patches becoming empty. Notice that P can range between 0 and 1.
As a result, if the patch occupancy rate (c) equals the patch extinction rate (e), then patch occupancy P equals to 0.
