What polysaccharide provides rigidity and strength in plants? Brainliest and 10 points Glycogen Starch Cellulose Chitin

In an experiment, what is the factor that is tested? control specimen hypothesis variable

e-lub [12.9K]

The factor being tested is the variable. There are two types of variables, independent and dependent. The independent variable is the factor being changed by the scientist or the person controlling the experiment. The dependent variable is the thing that gets changed due to the changing of the independent variable. The dependent variable cannot be changed physically by the person doing the experiment.

6 0

3 years ago

Which question can be answered by science?

tekilochka [14]

Answer: B. Does this medicine reduce fevers?

Explanation: This is because you can run experiments and use the scientific method to answer your question.

5 0

2 years ago

Read 2 more answers

This diagram shows the menstrual cycle.

LiRa [457]

Answer:

According to the hormone diagram of the menstrual cycle, the woman is not pregnant due to the behavior of progesterone and estrogens, whose levels do not increase, in addition to the absence of human chorionic gonadotropin.

Explanation:

The graph shows the behavior of hormones during a woman's menstrual cycle in the absence of pregnancy.

During a woman's normal cycle, estrogen, luteinizing hormone (LH) and follicle stimulating hormone (FSH) tend to increase prior to ovulation, reach their peak values at ovulation, and then decline, as shown in the graph. Progesterone, on the other hand, increases after ovulation and decreases if the woman does not become pregnant.

In the case of a pregnant woman:

Estrogens continue to increase after ovulation, produced by the ovaries and placenta.
Progesterone also increases its levels, as it is a hormone produced by the ovaries and placenta.
Hormone human chorionic gonadotropin (HCG) appears and increases during pregnancy, due to the secretory activity of the placenta.

The diagram represents the normal cycle of a woman who is not pregnant.

4 0

3 years ago

Ganymede is one of the many moons of Jupiter. It is nearly spherical in shape. It is larger than the planet Mercury and slightly

Aneli [31]

3. It moves in an orbit around Jupiter

5 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