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Mrac [35]
3 years ago
12

Which of the following statements about proto-oncogenes is false? - Proto-oncogenes are normal genes with the potential to becom

e oncogenes. A mutation must occur in a cell's DNA for a proto-oncogene to become an oncogene. Many proto-oncogenes code for growth factors. If a proto-oncogene's gene product is produced in smaller quantities, a cell may become cancerous.
Biology
1 answer:
Anni [7]3 years ago
3 0

Answer:

If a proto-oncogene's gene product is produced in smaller quantities, a cell may become cancerous.

Explanation:

The transformation of a cell to a malign cell occurs by mutations accumulation in some specific genes. These genes are grouped into two groups or families:

  • Protooncogenes
  • Tumor suppressor genes

Protooncogenes direct the production of proteins of varying locations and functions that play a significant role in cell proliferation and differentiation. The proteins include cyclins, growth factors, receptors, etcetera. These genes are proteins codifiers and influence the cell cycle by favoring proliferation or inhibiting senescence and apoptosis.

Depending on the organism´s development stage, these genes can be physiologically active or reprimed.

Certain structural or functional changes in protooncogenes contribute to turning them into oncogenes. When protooncogenes mutate into oncogenes, they are capable of controlling cell multiplication and massively synthesize uncontrolled division triggers.

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Answer and Explanation:

<em><u>The number of observed individuals</u></em>:

  • AA 42
  • AG 24
  • GG 21

<u><em>Total number of individuals, N</em></u>= 87 = 42 + 24 + 21

<u><em>Allelic frequencies</em></u>:

  • f(p) = (2 x AA + AG)/ 2 x N

       f (p)= (2 x 42 + 24) /2 x 87

       f (p) = (84 + 24) / 174

       f (p)= 108 / 174

      f (p) = 0.62

  • f (q) = (2 x GG + AG)/2 x N

        f (q) = (2 x 21 + 24 )/2 x 87

        f (q) = (42 + 24)/ 174

        f (q) = 66/174

        f (q) = 0.38

p + q = 1

0.62 + 0.38 = 1

<em><u>The expected genotypic frequency:</u></em>

  • F (AA)= 0.62 ² = 0.3844
  • F (AG) = 2 x A x G = 2 x 0.62 x 0.38 = 0.4712
  • F (GG) = 0.38 ² = 0.1444

AA + AG + GG = 0.3844 + 0.4712 + 0.1444 = 1

<u><em>The number of expected individuals</em></u>:

AA= (0.62)² x 87 = 0.3844 x 87 = 33.44

AG= (0.4712) x 87 = 40.99

GG= (0.38)² x 87 = 12.563

<u><em>Total number of expected individuals</em></u> = 33.44 + 40.99 + 12.563 = 87

<u><em>Chi square</em></u>= sum (O-E)²/E

  • AA= (O-E)² /E

        AA=(42 - 33.44) ² / 33.44

        AA= 2.2

  • AB= (O-E)² /E      

        AB= (24 - 40.99)²/ 40.99

        AB=7.04

  • BB=(O-E)² /E

        BB= (21-12.563)²/12.563

        BB= 5.66

<u><em>Chi square</em></u>= sum ((O-E)²/E) = 2.2 + 7.04 + 5.66 = 14.9

<u><em>Degrees of freedom</em></u> = genotypes - alleles = 3 - 1 = 2

p value less than 0.05

There is enough evidence to reject the nule hypothesis. The genotype frequencies are not in equilibrium.

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