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3 years ago

What impact might the increasing worldwide use of the internet have on the final step of the experimental method?

1 answer:

8 0

The experimental method is usually mentioned in relation to Psychology. This method focuses on the manipulation of the independent variable and making observations with relation to the dependent variable. The use of the internet in the final steps of the experimental method can lead to experimental procedures that are ignored or not examined thoroughly.

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Besides glucose what other kinds of molecules can be used to produce ATP in cellular respiration?

Makovka662 [10]

Lipids, carbs, and protein

4 0

3 years ago

Read 2 more answers

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

In semiconservative dna replication, each new double helix formed will have

Gnesinka [82]

In semiconservative DNA replication, each new double helix that will form will have 1 polynucleotide strand that is from the old DNA molecule and is an Old or Parent strand, and will have a polynucleotide strand from the newly synthesized one, the new DNA strand.

7 0

3 years ago

Which of the following is an inherited behavior?

Marina CMI [18]

Many of these options can be inherited based on the context. The only one i can see that is absolutely inherited is saying "please" and "thank you", because those are not built into a person. They are taught to a child so they can show respect when given something. The porcuipine using its quills when scared is the only instictive trait.

4 0

3 years ago

B=black and b=brown. A Bb guinea pig crosses with a Bb guinea pig, and four offspring are produced. All of the offspring are bla

Sladkaya [172]

If two Bb guinea pigs cross, there are four possible outcomes:

both pass the B: the son is BB
one of the pass the B and the other pass the b: the son is Bb
both pass the b: the son is bb

Since B (being black) is dominant, the son will be black in both BB and Bb cases. In order for a son to be brown, he must be a pure bb specimen.

So, if all of the offspring are black, it means that all four guinea pigs are either BB or Bb, which in turn means that at least one of the parents passed the B gene.

8 0

3 years ago