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kati45 [8]
3 years ago
11

Explain one medical reason why balanced diet may vary

Biology
1 answer:
balandron [24]3 years ago
8 0
One medical reason is diabetes and those who need food or else they faint. Diabetes need a balanced diet in SUGAR, and just like me, there are people, (I don't know what the sickness is called), who need a lot of food to eat and water to drink or else they start to get dizzy and fainnt. So that's why balanced diets vary.  
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Answer:

Dissolved or gaseous substances have to pass through the cell membrane to get into or out of a cell. Diffusion occurs when particles spread. ... They move from a region where they are in high concentration to a region where they are in low concentration.

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3 years ago
If a heterozygous plant with green seeds self-fertilizes, what is the probability that 6 seeds in a single pod of the progeny pl
nalin [4]

Here is the complete question

The seeds in bush bean pods are each the product of an independent fertilization event. Green seed color is dominant to white seed color in bush beans.

If a heterozygous plant with green seeds self-fertilizes, what is the probability that 6 seeds in a single pod of the progeny plant will consist of:

A) 4 green and 2 white seeds?

B) all white seeds?

C) at least 5 white seeds?

Answer:

A) 0.3

B) 0.0002

C) 0.001

Explanation:

Let the allele for Green plant be = P

Let the allele for white plant be =  p

Let the heterozygous plant be Pp, if self fertilization occurs we have:

                   P                    p

P                 PP                 Pp                                                              

                                                           

p                 Pp                 pp                  

From the punnet square above, we have the following  cross;

PP, Pp, Pp, pp

Since P is dominant to p; we have a phenotypic ratio of 3:1 between the dominant phenotype ( PP, Pp, Pp ) and recessive phenotype ( pp )

NOW, the probability for the green plant(i.e PP or Pp) will be: Pr(G)=\frac{3}{4}

the probability for the white plant (pp) will be: Pr (W)= \frac{1}{4}

A)

what is the probability that 6 seeds in a single pod of the progeny plant will consist of 4 green and 2 white seeds?

Pr(4G2W) = (\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{1}{4})*(\frac{1}{4})

The probability includes the use of permutation:

Permutation(P)=\frac{6!}{(4!*2!)} * (\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{1}{4})*(\frac{1}{4})

= 15*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{3}{4})*(\frac{1}{4})*(\frac{1}{4})

= 0.2966

≅ 0.3

B)

What is the probability that 6 seeds in a single pod of the progeny plant will consist of all white seeds?

Since probability for white seed = \frac{1}{4}

The probability that 6 seeds will consist of all white seeds will be = (\frac{1}{4})^6

=  2.44 × 10⁻⁴

= 0.000244

≅ 0.0002

C)

What is the probability that 6 seeds in a single pod of the progeny plant will consist of at least 5 white seeds?

For at least 5 white seeds, the probability will consist of  5 white seeds and 6 white seeds.

For 5 white seeds; we have:

5 White  and 1 Green  Since there are 6 permutations.

The probability For 5 white seeds will therefore be :

Pr(5 seeds)=(\frac{1}{4})^5*(\frac{3}{4})^1 *6

= 7.32 × 10⁻⁴

= 0.000732

For 6 white seeds; we have:

Only one permutation for 6 whites since there will no be any presence of any green seed.

Pr(6seeds)=(\frac{1}{4})^6

= 2.44 × 10⁻⁴

= 0.000244

∴

The total probability for at least 5 white seeds =  0.000732 + 0.000244

= 9.76  × 10⁻⁴

= 0.000976

≅ 0.001

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