Lab Exam Final

Question 1

test for the presence of Starch

Answer 1

Potassium Iodine Reagent detects polysaccharide by turning blue/black

1 dropperful of unknown solution

3-4 drops of iodine

Turning violet = pos result

Question 2

test for the presence of Lipid

Answer 2

Put unknown solution on a piece of brown paper

Allow to dry

hold paper up to light

oily film = pos result

Question 3

test for the presence of Sugars

Answer 3

Benedicts Reagent detects Monosaccharides by turning red/orange and clumping at bottom

1 dropperful of unknown solution

2 droppersful of Benedict’s

Sit on hot plate for 5 minutes

clumping = pos result

Question 4

test for the presence of Protein

Answer 4

Biuret reagent detects protein by turning violet color

2 droppersful of unknown solution

1 dropperful of biuret

swirl gently

wait 2 minutes

violet color = pos result

Question 5

Subunits that make up carbohydrates

Answer 5

Monosaccharides

Question 6

Subunits of Proteins

Answer 6

Amino Acids

Question 7

how to focus the dissecting microscope

Answer 7

bring focus knob all the way down to the bottom

adjust oculars until there is a single field of view

center specimen

turn focus knob up until image is focused

Question 8

the eyepiece on a microscope has a magnification of 10X , what is the total magnification with a 10x objective? What is the total magnification with a 45 X objective?

Answer 8

100 / 450

Question 9

Why is it important to center a specimen on low power before attempting to focus on it at a high power?

Answer 9

By using low power, you have a wider field of view (you see more of the scene) so you know better which sub-part you want to ‘zoom into’.

Question 10

what purpose does mitosis serve in development and growth?

Answer 10

Mitosis helps organisms grow in size and repair damaged tissue.

Question 11

Besides the root tips, where else in plants would you expect mitosis take place in an onion plant?

Answer 11

Tip of the stem

Question 12

Explain how the substances that stimulate taste and smell receptors are similar.

Answer 12

Taste and smell are sensed by chemosensation. The substances are made up of complex mixtures of chemicals that stimulate special nerve cells in the nose, mouth, or throat.

Question 13

Photoreceptors are primarily responsible for dark adaptation. Explain how dark adaptation works.

Answer 13

Sensitivity to light increases in the dark because photosensitive pigments are reconstructed from vitamin A.

Question 14

Bivalves have little sensory ability and heavy shells, while cephalopods have well-developed senses and reduced shells. How are these characteristics adaptive for each group?

Answer 14

These characteristics are different adaptations of survival mechanisms. Because bivalves have little sensory ability they have adapted heavy shells to keep them safe from predators. Cephalopods on the other hand have come to rely on their well-developed senses to keep safe. Some have even been known to gather shells and fully wrap themselves for protection.

Question 15

Using as examples at least two bilaterally symmetrical organisms and two radially symmetrical organisms, discuss the adaptive value of bilateral versus radial symmetry.

Answer 15

Bilateral: Humans/Horse
Radially symmetrical: Jellyfish/Sea star



Bilateral symmetry is more adaptive than radial symmetry for a motile animal because the animal can move faster because it can move in one direction.

Bilateral symmetry favors locomotion by generating a streamlined body.

Question 16

Describe an experiment to test whether pill bugs are attracted to tannic
acid (a chemical found ln rotting wood)

Answer 16

By use of a choice chamber we can have a control condition in one chamber and the other chamber containing tannic acid. We can place 5 pill bugs into the control chamber and 5 into the acidic chamber and observe the choices they make over a period of time. This will allow us to determine whether they are attracted to, repelled by, or have a neutral response to the tannic acid.

Question 17

Porifera characteristics

Answer 17

(Sponge)

Spongocoel
Osculum
Mesohyl
Monoecious

Question 18

Cnidaria characteristics

Answer 18

(Jellyfish)

1) radial or biradial symmetry

2) reproduce both sexually and asexually

3) diploblastic

4) cnidocytes (“stinging cells”)
contain specialized organelle called the nematocysts

5) Alternate between two forms
Medusa- free floating (jelly fish)
Polyp – anemone

6) Three layers:
Epidermis – outer layer
Gastrodermis – inner layer
Mesoglea – jelly-like and non-living.

Question 19

Characteristics of Platyhelminthes

Answer 19

(Flatworm)

Bilaterally symmetrical

Question 20

Nematoda characteristics

Answer 20

(Round worm)

non segmented
Mouth
Pharynx
Intestine
Ventral nerve
Reproductive organs
Protective cuticle

Question 21

Arthropoda characteristics

Answer 21

(insects/crustaceans)

Largest phylum

functional segmentation of the body

the presence of jointed appendages

exoskeleton

Question 22

Mollusca attributes

Answer 22

(clams, barnacles, octupus, snail)

Asymmetrical

Gil – oxygen exchange

Radula – scrapes food particles

Question 23

Annelida traits

Answer 23

(segmented worm)

Bilaterally symmetrical

Circular and longitudinal muscles

Chaetae- chitinous hairlike extensions from the cuticle

Question 24

Four common features appear at some point in chordate development:

Answer 24

Notochord

dorsal hollow nerve cord

Pharyngeal slits

post-anal tail

note
(some of these traits are present only during embryonic development)

Question 25

Echinoderm traits

Answer 25

(sea star, sea urchins, sea cucumbers)

Radially symmetrical

water vascular system.- central ring that extends to each arm for gas, nutrient and waste exchange – regulated by the madreporite

“Tube feet,” which protrude through openings in the endoskeleton to allow for movement