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joja [24]
3 years ago
12

The flexible nature of a cell membrane results from its _________

Biology
2 answers:
MrMuchimi3 years ago
5 0
It results from the composition of the cell membrane. The cell membrane consists of a phospholipid bilayer, which have hydrophilic heads and hydrophobic tails. Hope this helps! :)
Juli2301 [7.4K]3 years ago
5 0
A phospholipid bilayer that is made up of fatty molecules that are polar
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Which physical properties do all-stars have? Check all that apply.
olga2289 [7]

Answer:

-Physical properties possessed by all stars:

  • They are made of gases such as hydrogen and helium.
  • They shine very brightly due to interaction of hydrogen and helium at appropriate pressure and temperature.
  • They contain iron in their cores which monitors the fusion reaction.
  • They make energy through fusion between the gases present in the stars which are hydrogen and helium.
  • They are not spherical in shape hence this physical property is false in terms physical properties of the stars.

Hope this helps. If it does, please mark me brainliest.

Explanation:

6 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 =<u> </u>\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

                                       <u>..................................................................................</u>

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

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

                                   <u>..........................................................................................</u>

                                <u>\frac{1}{4}AA + \frac{1}{2}Aa +\frac{1}{4} aa</u>

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
DNA is made up of monomers called nucleotides. How many does mono- represent?
Lunna [17]

Answer:

It is often said that life is “carbon-based.” This means that carbon atoms, bonded to other carbon atoms or other elements, form the fundamental components of many, if not most, of the molecules found uniquely in living things. Other elements play important roles in biological molecules, but carbon certainly qualifies as the “foundation” element for molecules in living things. It is the bonding properties of carbon atoms that are responsible for its important role.

Carbon Bonding

Explanation:

8 0
3 years ago
Gene mutations can be positive negative or neutral suppose that the normal gene in model 2 produces a polypeptide that was neces
IRINA_888 [86]
Mutations are the source of all new alleles. An allele is a variant form of a gene. Variations in alleles lead to variations in organisms. Positive mutations give an organism a better chance of survival. It means that the mutation may be passed on to the offspring. Negative mutations may lead to an early death.<span> Therefore, changes in alleles from one generation to another form the basis of evolution.</span>
6 0
3 years ago
The average baby will grow _____ inches in his or her first year.
Margaret [11]
<span>Infants grow about 3/4 inch per month during the first year, increasing their birth length by about 40 percent by their first birthday. Thus, the answer to the question is 3/4 inch during the first year upon giving birth of the mother.</span>
6 0
3 years ago
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