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inysia [295]
3 years ago
12

Suppose an enzyme that is involved in cellular respiration is assembled by a ribosome. where will this enzyme need to be transpo

rted in order to perform its function?
Biology
1 answer:
Natasha2012 [34]3 years ago
7 0
Needs to be transported to the mitochondria or the cytosol.

<span>
</span>There are four stages of cellular respiration: glycolysis, pyruvate oxidation, Krebs cycle and Oxidative phosphorylation. Glycolysis takes place in the cytosol of a cell,while the <span>pyruvate oxidation and the following steps take place </span> in the matrix of the mitochondria. If the enzyme is involved in cellular respiration, it need to be in one of these places.

<span />
You might be interested in
While a relatively inefficient mode of transmission, _____________ accounts for 90% of HIV infections worldwide.
Marat540 [252]

Answer:

The correct answer is - sexual transmission.

Explanation:

HIV infections can be transmitted by the various mode. Sexual transmission is one of the major mode of transmission which is comprise almost 90 % of HIV infection, according to the reports. These are relatively inefficient mode, unprotected or unnatural sex is the major mode for the HIV transmission worldwide. Cases are now comparatively less than when it was discovered due to awareness.

Thus, the correct answer is - sexual transmission.

3 0
3 years ago
Inheritance patterns cannot always be explained by Mendel's models of inheritance. If a pair of homologous chromosomes fails to
soldier1979 [14.2K]

The question is incomplete as it does not have the options which are:

A) n+1; n+1; n-1; n-1

B) n+1; n-1; n; n

C) n+1; n-1; n-1; n-1

D) n+1; n+1; n; n

Answer:

Option-1

Explanation:

The laws of inheritance were concluded from the result of Mendel's experiment which is based on the fact that gametes are formed. Later research suggested that gametes are formed by the process of meiosis which takes place in two phases and recombination is a characteristic of Meiosis.

If during anaphase I of meiosis I, the alleles fails to separate that is nondisjunction takes place at anaphase I, Then the resulting daughter cells will have unequal distribution of chromosomes.

One daughter cell will receive 1 extra copy of the chromosome while another daughter cell will receive 1 less chromosome therefore ploidy level will be n+1 and n-1.

During meiosis II, 2 more daughter cells will be formed with the same ploidy level therefore in last the meiosis will result in 2 (n+1) and 2 (n-1) cell.

Thus, Option-1 is the correct answer.

4 0
3 years ago
Rank the following in order from most general to most specific: 1. gametic isolation 2. reproductive isolating mechanism 3. sper
mihalych1998 [28]
<h2>Reproductive Method </h2>

Explanation:

<em>The rank in order from the most specific which is following .</em>

<em>(1) Reproductive isolating mechanism</em>

<em>(2) Sperm-egg incompatibility in sea urchins</em>

<em>(3) Gametic isolation </em>

<em>(4)Prezygotic isolating mechanism</em>

<em>(1) Reproductive isolating mechanism-</em> The components of regenerative confinement are an assortment of transformative instruments, practices and <em>physiological procedures basic for speciation.</em> They keep individuals from various species from delivering posterity, or guarantee that any posterity are sterile.

(<em>2) Sperm-egg contradiction in ocean urchins-</em> Bindin is a gamete acknowledgment protein known to control species-explicit <em>sperm-egg grip</em> and layer combination in ocean urchins.

<em> (3)Gametic isolation - Prezygotic hindrances </em>keep preparation from occurring. Gametic disengagement is a sort of prezygotic hindrance where the<em> gametes (egg and sperm) </em>come into contact, yet no preparation happens. Gametes might be not able to remember each other in various species  

<em> (4) Prezygotic isolating mechanism- </em>while postzygotic segregation forestalls the arrangement of rich posterity. Prezygotic systems incorporate environment segregation, mating seasons, "mechanical" disconnection, gamete detachment and conduct seclusion. 

4 0
3 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.

5 0
3 years ago
The sun is a significant factor that helps to support life. Which of the following earht-sun relationship is not true.
Pavel [41]

Answer:

D

Explanation:

I know FOR SURE that A, B, and C are correct, so D isn't

7 0
3 years ago
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