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Aleks04 [339]
3 years ago
6

Which of the following outcomes would be most likely if the membrane of a cell were no longer selectively permeable? Harmful sub

stances would be able to enter the cell and cause mutations to the cell’s DNA. The cell would die as it would not be able to regulate which substances could enter and leave. A cell wall would form outside the membrane to protect the contents of the cell. The cell would grow as more nutrients would be able to easily enter the cell.
Biology
2 answers:
Mazyrski [523]3 years ago
8 0
The correct answer is option B. I just took the test and this was the correct answer!
Ray Of Light [21]3 years ago
7 0
If the membrane of a cell were no longer selectively permeable then "<span>The cell would die as it would not be able to regulate which substances could enter and leave", since a selectively permeable membrane allows both things to enter and leave for the cell's survival. </span>
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what are the two structures you would find at a passive continental margin that you would not find at an active continental marg
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3 years ago
Both alleles of a pair are the same, either both dominant or both recessive.
Sliva [168]

Answer:

Both Recessive

Explanation:

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List four benefits of cover crops, including soil health
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3 years ago
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Sickle cell anaemia is an inherited autosomal recessive
alekssr [168]

Answer:

1/8 (12.5%)

Explanation:

An autosomal recessive disease is an inherited disease in which an individual need to receive both defective alleles at the same gene <em>locus</em> to be expressed in the phenotype. In this case, both parents are carriers of the recessive mutant allele associated with the sickle cell anaemia trait, thereby both parents are heterozygous, ie., each parent has one copy of the normal allele 'H' and one copy of the defective mutant allele 'h' associated with this condition. In consequence, their first child has a 1/4 (25%) chance of having sickle-cell anaemia. Moreover, the chance of having a girl is 1/2 and the chance of having a boy is 1/2, thereby the final chance of having a girl sickle cell anaemia individual is 1/4 x 1/2 = 1/8 (12.5%).

- Parental cross for sickle cell anaemia trait = Hh x Hh >>

- F1 = 1/4 HH (normal); 1/2 Hh (normal); 1/4 hh (sickle cell anaemia) >>  

- Sex proportion of sickle cell anaemia individuals =  1/8 female sickle cell anaemia individuals + 1/8 male sickle cell anaemia individuals (1/8 + 1/8 = 1/4)

3 0
2 years ago
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