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Rasek [7]
3 years ago
13

Which reaction is responsible for the production of lipids when glycerol is joined to three fatty acids?

Biology
1 answer:
Anna35 [415]3 years ago
4 0

This reaction is called a condensation reaction. It is also known as  a dehydration reaction.

When three fatty acids become attached to a glycerol molecule, a triglyceride is formed.  The fatty acids become attached to the glycerol through a condensation reaction named so because 3 water molecules are formed from the 3 OH groups from the fatty acid chains and 3 H atoms from the glycerol.

The bond formed between the fatty acid chain and the glycerol is called an ester linkage.



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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
Which of the following must happen before the cell can go through mitosis? A. The cytoplasm must divide. B. The cell must copy i
alex41 [277]
B, because this occur before the others. 
4 0
4 years ago
Read 2 more answers
I realllyyyyy need help right now. it’s due it 30 mins!!
givi [52]

Answer:

LL

Explanation:

8 0
3 years ago
Read 2 more answers
Which of the following does NOT help increase conversion of short-term to long-term memories?
Contact [7]

Answer:

Memory consolidation

Explanation:

Short term memory is stored in the brain for relatively short period of time especially for few minutes. Long term memory is stored in brain for long period ( years or lifetime).

Short term memory can be converted into the long term memory by association, rehearsal and involved with emotional state. Memory consolidation is the process of learning new things and store them in previously stored memories. Hence, memory consolidation does not increase the conversion of short memory to long term memory.

Thus, the correct answer is option (1).

5 0
3 years ago
1. identify the organelle that regulates cell function and contains the dna
PilotLPTM [1.2K]

Answer:

the organelle that regulates cell function and contains the DNA is a nucleus

6 0
3 years ago
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