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Mekhanik [1.2K]
2 years ago
8

Multiple arteries joined in order to serve a single capillary network are called:______.

Biology
1 answer:
ra1l [238]2 years ago
6 0

Answer:

they are called collaterals

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How would a mistake in transcribing the DNA message affect your ability to catch the killer?
mel-nik [20]

Answer: When the DNA collected on the crime scene is being analysed in the forensic lab, the results will be inconclusive, because error in DNA transcription, particularly in mRNA encoding leads to inheritable phenotype change by reprogramming the transcriptional network, without changing the DNA. Transcription errors are brief with no long-term consequences. Due to mRNA being short-lived and the erroneous proteins are degraded. A decrease in transcription accuracy triggers cellular identity change.

Explanation: This is called epimutation, a heritable modification that results in the change of gene expression, but not the DNA sequence. Epimutation is not associated with DNA mutation, but is associated with the loss or gain of DNA methylation or other heritable changes of the chromatin.

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3 years ago
Which of the following best describes the cycle of matter between humans and the environment?
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Answer:

B: humans inhale oxygen from plants which is used in cellular respiration to produce water

Explanation:

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2 years ago
Most black bears (Ursus americanus) are black or brown in color. However, occasional white bears of this species appear in some
Sphinxa [80]

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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