All categories

3 years ago

Calculate the total magnification of the medium power ocular objective with a 10X magnification.

Biology

1 answer:

ziro4ka [17]3 years ago

8 0

Total magnification of the medium power ocular objective with a 10X magnification is 100X

Explanation:

Given: Medium power ocular objective=10X

magnification is 10X

Total magnification= power of ocular objective*magnification

10*10

The magnification of the ocular lens is multiplied with the power of the objective lens. We know that magnification of the objective lens for low, medium and high power are 4X, 10X, and 40X,

You might be interested in

A student wants to determine the mass of a large plastic cube. The student measures the volume of the cube as

arsen [322]

Answer:

see in explaination

Explanation:

Density (D) is a quantity defined as mass (m) per unit volume (v): D = m ÷ v. Since these three quantities are related, you can find one of them if you know the other two. This is more helpful than it sounds because density is often a known quantity that you can look up, assuming you know the composition of the material.

Calculations are straightforward for solid objects, but things get more complicated for liquids, where density changes with temperature, and for gases, for which density is dependent on temperature and pressure.

7 0

2 years ago

When you notice that someone has unusually blue eyes, you've noticed their

tatyana61 [14]

D. Phenotype because a phenotype is observable traits

4 0

2 years ago

Read 2 more answers

A lens that is thicker in the center than at the edges is called a ____________________ (concave, convex) lens.

denpristay [2]

Answer:

convex lens

Explanation:

because when parallel rays of light pass through a lens are brought together through a lens

5 0

2 years ago

Read 2 more answers

Termination of DNA synthesis in E. coli and humans differ significantly because of the genomic structures involved, namely a cir

Marina86 [1]

Answer:

Main protein in ending high fidelity in E. Coli is the Tus protein that binds to Ter sequences in order to prevent replication forks from passing through the end region. In the Ter sequences, the Tus protein blocks replication by establishing a close association with a particular G-C base pair.

The main protein in human cells is telomerase, which contains an RNA primer and is required to extend the synthesis of lagging strands in linear chromosomal telomeres.

3 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