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tensa zangetsu [6.8K]
3 years ago
9

why would having both polar and nonpolar properties in a protective boundary be advantageous for the cell?

Biology
1 answer:
Reika [66]3 years ago
5 0

Explanation:

The polar nature of the membrane’s surface can attract polar molecules, where they can later be transported through various mechanisms. Also, the non-polar  region of the membrane allows for the movement of small non-polar molecules across the membrane’s interior, while preventing the movement of polar molecules, thus maintaining the cell’s composition of solutes and other substances by limiting their movement.

Further explanation:

Lipids are composed of fatty acids which form the hydrophobic tail and glycerol which forms the hydrophilic head; glycerol is a 3-Carbon alcohol which is water soluble, while the fatty acid tail is a long chain hydrocarbon (hydrogens attached to a carbon backbone) with up to 36 carbons. Their polarity or arrangement can give these non-polar macromolecules hydrophilic and hydrophobic properties i.e. they are amphiphilic. Via diffusion, small water molecules can move across the phospholipid bilayer acts as a semi-permeable membrane into the extracellular fluid or the cytoplasm which are both hydrophilic and contain large concentrations of polar water molecules or other water-soluble compounds.

Similarly via osmosis, the water passes through the membrane due to the difference in osmotic pressure on either side of the phospholipid bilayer, this means that the water moves from regions of high osmotic pressure/concentration to regions of low pressure/ concentration to a steady state.

Transmembrane proteins are embedded within the membrane from the extracellular fluid to the cytoplasm, and are sometimes attached to glycoproteins (proteins attached to carbohydrates) which function as cell surface markers. Carrier proteins and channel proteins are the two major classes of membrane transport proteins; these allow large molecules called solutes (including essential biomolecules) to cross the membrane.

Learn more about membrane components at brainly.com/question/1971706

Learn more about plasma membrane transport at brainly.com/question/11410881

#LearnWithBrainly

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4) A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality)
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Answer and Explanation:

  • A homozygous groucho fly ( gro, bristles clumped above the eyes) is crossed with a homozygous rough fly (ro, eye abnormality).
  • The F1 females are testcrossed, producing these offspring: groucho 518 rough 471 groucho, rough 6 wild-type 5 1000 a) What is the linkage distance between the two genes? B) Plot the genes on a map c) If the genes were unlinked and the F1 females were mated with the F1 males, what would be the offspring in the F2 generation?

1st cross:

Parental) grogro ro+ro+ x  gro+gro+ roro

F1) gro+gro ro+ro

2nd cross:

Parental)  gro+gro ro+ro   x  grogro roro

Gametes) gro+ro+                       gro ro

                gro+ro                         gro ro

                gro ro+                        gro ro

                gro ro                          gro ro

Punnet square)  

                   gro+ro+             gro+ro              gro ro+            gro ro  

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

gro ro    gro+gro ro+ro   gro+gro roro    grogro ro+ro    grogro roro

F2)

0.518 grogro ro+ro (518 individuals)

0.471 gro+gro roro (471 individuals)

0.006 grogro roro (6 individuals)

0.005 gro+gro ro+ro (5 individuals)

Total number of individuals 1000

<u><em>Note</em></u>: These frequencies were calculated dividing the number of individuals belonging to each genotype by the total number of individuals in the F2.

To know if two genes are linked, we must observe the progeny distribution. <em>If individuals, whos </em><em>genes assort independently,</em><em> are test crossed, they produce a progeny with equal </em><em>phenotypic frequencies 1:1:1:1</em>. <em>If</em> we observe a <em>different distribution</em>, that is that <em>phenotypes appear in different proportions</em>, we can assume that<em> genes are linked in the double heterozygote parent</em>.  

In the exposed example we might verify which are the recombinant gametes produced by the F1 di-hybrid, and we can recognize them by looking at the phenotypes with lower frequencies in the progeny.  

By performing this cross we know that the phenotypes with lower frequencies in the progeny are groucho, rough and wild-type. So the recombinant gametes are <em>gro+ro+</em> and <em>gro ro</em>, while the parental gametes are <em>gro+ro</em> and <em>gro ro+.</em>

So, the genotype, in linked gene format, of the double heterozygote individual in the <u>F1</u> is gro+ro/gro ro+.

To calculate the recombination frequency we will make use of the next formula: P = Recombinant number / Total of individuals. The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product.

The recombination frequency is:

P = Recombinant number / Total of individuals

P = 6 + 5 / 1000

P = 11 / 1000

P = 0.011

The <u>genetic distance between genes,</u> is 0.011 x 100= 1.1 MU.

<u>Genetic Linkage Map:</u>

Parental Phenotypes)  

-----gro+------ro----              -----gro------ro+----

----- gro ------ro----               ---- gro------ ro ----

Recombinant phenotypes)

-----gro+------ro+----              -----gro------ro----

----- gro ------ ro----                -----gro------ro----

<u>If the genes were unlinked</u> and the F1 females were mated with the F1 males, the offspring in the F2 generation would have been

4/16 = 1/4 gro+gro ro+ro  

4/16 = 1/4 gro+gro roro  

4/16 = 1/4 grogro ro+ro    

4/16 = 1/4 grogro roro

Their phenotypic frequencies would be 1:1:1:1 related.                                                  

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