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3 years ago

Which would best explain why one red blood cell hemolyzes more quickly than another?

1 answer:

lina2011 [118]3 years ago

4 0

The statement that would best explain why one red blood cell hemolyzes more quickly than another is if the cell that hemolyzes more quickly acquires NaCl at a faster rate.

Water moves by osmosis in and out of cells from the region of high water potential or low solute concentration to the region of low water potential or high solute concentration.

A cell with a higher solute concentration than the surrounding solution will keep acquiring water from its surrounding until the cell becomes turgid and bursts or lyses due to over-turgidity.

If the reverse occurs, such a cell will lose water and become flaccid.

Thus, if a red blood cell acquires NaCl, a solute, at a faster rate, such a cell will also acquire water and become turgid/lyses at a faster rate.

More on hemolysis can be found here: brainly.com/question/6598052

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Select the correct answer.

Klio2033 [76]

The best answer to go with is c you’re welcome help it help

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3 years ago

Identify whether the following statements are true or false.

Vanyuwa [196]

Answer:

They are all true besides "Organisms of the same species are equipped with the same survival skills."-that one is false.

Explanation:

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3 years ago

Read 2 more answers

If color blindness is an X-linked recessive trait, what genotype in a father and a mother would be predicted to produce a 1:1 ra

Likurg_2 [28]

The genotype would be X'Y for the father (he would be colorblind)
The genotype would be X'X for the mother (she would be a carrier)
The offspring would be X'X' _ X'X_ X'Y_ XY
The ratio would be 1:1 normal to colorblind
Hope this was helpful(;

3 0

3 years ago

After infection of a cell, a viral particle must transport itself to the nucleus in order to produce viral proteins. What is the

kykrilka [37]

Answer:

Double-stranded DNA

Explanation:

If a virus requires to transport its genome in the nucleus to produce viral protein then the viral genome content must be DNA. This DNA of the virus will use the RNA polymerase of the host cell and will first convert into mRNA in the nucleus.

Then the mRNA of the virus will come out of the nucleus because the protein synthesis takes place outside the nucleus and in the cytoplasm. So in the cytoplasm by using host translational machinery the viral mRNA will code for viral proteins. So the correct answer is double-stranded DNA.

3 0

3 years ago

2. Dominant trait: cleft chin (C) Mother’s gametes: Cc

andre [41]

.2. Offspring Genotypes will be Cc or cc.

Offspring phenotypes : Cleft chin or no cleft chin.

% chance child will have cleft chin: 50%

3. % chance child will have arched feet: 25%

4. % chance child will have blonde hair: 50%

5. % chance child will have normal vision: 25%

Explanation:

CASE 1 :

Dominant trait: cleft chin (C)

Recessive trait: lacks cleft chin (c)

Father’s gametes: cc

Mother’s gametes: Cc

There are two possible combination of Gametes ,

C fom mother and c from father= Cc

c from mother and c from father = cc

Gametes of Cc Parents= $\frac{1}{2}C + \frac{1}{2} c$ ........(i)

Gametes of cc parents = $\frac{1}{2}c + \frac{1}{2}c$ .........(ii)

Combining (i) and (ii) we get,

$\frac{1}{2} Cc + \frac{1}{2} cc$

There fore offspring Genotypes will be Cc or cc

Offspring phenotypes :

Genotype Cc then phenotype= Cleft chin

Genotype cc then phenotype = Lacks cleft chin.

percentage chance child will have cleft chin = $\frac{0.5}{1}$ ×100

Therefore the chance is 50%.

CASE 2 :

Dominant trait: flat feet (A)

Recessive trait: arched feet (a)

Mother’s gametes: Heterozygous (Aa)

Father’s gametes: Heterozygous (Aa)

There are four possible combination of genotypes are =AA , Aa, Aa and aa

i.e. A from mother, A from father= AA

A from mother, a from father =Aa

a from mother, A from Father = Aa

a from mother, a from father = aa

Gametes of Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

Gametes of other Aa parent = $\frac{1}{2} A + \frac{1}{2} a$

..................................................................................

$\frac{1}{4} AA + \frac{1}{4} Aa$

+ $\frac{1}{4} Aa$ + $\frac{1}{4} aa$

..........................................................................................

$\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa$

Offspring Genotypes will be: AA or Aa or aa

Offsprings phenotype will be:

Genotype AA then phenotype will be Flat feet

Genotype Aa then phenotype will be flat feet

Genotype aa then Phenotype will be arched feet.

Percentage chance child will have arched feet = $\frac{0.25}{1}$ × 100 = 25%

CASE 3:

Dominant trait: Brown hair (B)

Recessive trait: Blonde hair (b)

Mother’s gametes: Homozygous recessive (bb)

Father’s gametes: Heterozygous (Bb)

This case is very similar to the case 1 as one parent is homozygous recessive and other parent is heterozygous.

Resulting in half Bb and halve bb combination.

Genotypes will be Bb or bb

Phenotypes will be :

Genotype Bb then phenotype Brown hair

Phenotype bb then Phenotype bb.

% chance child will have blonde hair: 50%

CASE 4:

Dominant trait: farsightedness (F)

Recessive trait: normal vision (f)

Mother’s gametes: Heterozygous (Ff)

Father’s gametes: Heterozygous (Ff)

This Case is similar to case 2

it will result in one-fourth FF , half Ff and one-fouth ff combination.

Therefore Genotypes will be: FF, Ff and ff

Phenotypes:

Genotype FF then phenotype farsightedness

Genotype Ff then phenotype farsightedness

Genotype ff then phenotype normal vision.

% chance child will have normal vision: 25%

3 0

3 years ago