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ASAP PLEASE!!!!!! Which pair of atoms are ions?

Cerrena [4.2K]

A cation and an anion

7 0

3 years ago

Unlike a prokaryotic bacteria cell, a virus, like the one in the image, A) has a lipid cell wall. B) has no genetic material. C)

seraphim [82]

The appropriate response is C. Prokaryotes are unicellular life forms that need organelles or other interior film bound structures. In this manner, they don't have a core, be that as it may, rather, by and large have a solitary chromosome: a bit of roundabout, twofold stranded DNA situated in a territory of the phone called the nucleoid.

7 0

3 years ago

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MDCK and A549 cell cultures were grown in complete minimum essential medium (MEM) with 10% fetal bovine serum (FBS), 100 U/mL pe

UkoKoshka [18]

Answer:

Used the C1V1 = C2V2 method for calculations

Explanation:

Component Calculation Volume of Stock to be added

FBS: 100*V1 = 10*25 2.5 mL

Glutamine: 200*V1 = 2*25 0.25 mL

Penicillin: 10000V1 = 100*25 0.25 mL

Streptomycin: 10000V1 = 100*25 0.25 mL

NB: C1 = Stock Concentration; V1 = Volume of stock required; C2 = Final concentration required; V2 = Final volume required

3 0

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

3 0

3 years ago

Baby food manufacturers sometimes use proteases in their products. Proteases function by catalyzing the

Daniel [21]

Answer:

enzymes

Explanation:

Enzymes are organic catalysts which are protein in nature. The enzyme proteases found in baby food aid in break down of peptides into amino acids in the stomach. Extracellular enzymes are engineered outside the cells that produce them e.g., proteases. The proteases break down milk proteins into amino acids that are easily assimilated into the system reducing chances of babies developing allergies to milk. Particular proteases break down specific allergenic proteins that can cause allergic in hypoallergenic food .

6 0

3 years ago