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Lelu [443]
2 years ago
11

When biological membranes are frozen and then fractured, they tend to break along the middle of the bilayer. The best explanatio

n for this is that
A)
the integral membrane proteins are not strong enough to hold the bilayer together.
B)
water that is present in the middle of the bilayer freezes and is easily fractured.
C)
hydrophilic interactions between the opposite membrane surfaces are destroyed on freezing.
D)
the carbon-carbon bonds of the phospholipid tails are easily broken.
E)
the hydrophobic interactions that hold the membrane together are weakest at this point.
Biology
1 answer:
user100 [1]2 years ago
4 0

<u>Answer</u>: E)   the hydrophobic interactions that hold the membrane together are weakest at this point.

The phospholipid bilayer is composed of a two-layered structure of phosphate and lipid molecules. In this cell membrane, the hydrophobic lipid ends are facing inward, whereas the hydrophilic phosphate ends are facing outward. As the hydrophobic interactions are weakest along the middle, freezing temperatures will cause them to break here.

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Gene A has two alleles (A1 and A2). A1A1 homozygotes are twice as likely to survive from birth to reproductive age as heterozygo
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The correct answer is C: <em>0.60 </em><em>A1</em> and <em>0.40 A2</em> will be the allele frequencies in those progeny when they reach reproductive age

Explanation:

<em>Genotype</em>                                <u><em>A1A1 </em></u><em>  </em><u><em>A1A2</em></u><em> </em><u><em>A2A2 </em></u>

<em>Relative aptitude, w</em>                  1           0.5    0.5

<em>Number of individuals</em>                 40            40     40

<em>Initial allelic frequency</em>      p0= (40+20)/120=0.5    q0= (40+20)/120=0.5

<em>Zygote frequency</em>                  p2= 0.25 2pq=0.25 q2=0.625

<em>Relative contribution</em>         0.25x1=0.25     0.5x0.5=0.25 0.25x0.5=0.125

<em>of each genotype</em>              

<em>Average aptitude W</em>              W= 0.025 + 0.25 + 0.125 = 0.625

<em>Population</em>                    AA= 0.25/0.625    AB=0.25/0.625  BB=0.125/0.625

<em>Genotype frequency</em>             AA= 0.4             AB=0.4          BB=0.2  

<em>New Allelic frequency</em>       p1=0.4+(0.4/2)=0.6   q1=0.2+(0.4/2)=0.4

  • <em>Total number of individuals</em>: 120
  • <em>Initial allelic frequency</em>:

                  (number of homozygote individuals + half number of  

                   heterozygote individuals) / Total number of individuals

  • <em>Relative contribution of each genotype</em>:

                           Zygote frequency x Relative aptitude

  • <em>Average aptitude W</em>: It is the sum of relative contribution of each genotype to the next generation.  

                    wA1A1 x p2 + WA1A2 x 2 x p x q + WA2A2 x q2

  • <em>Population Genotype frequency</em>:

                Relative contribution of each genotype / Average aptitude

  • <em>Allelic frequency</em>:

                      Homozygote population genotype frequency +  half  

                           heterozygote population genotype frequency

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