Category: 2nd Semester

Seed Germination & Detergents

 

Detergent & Seed Germination

Introduction:

Seeds come in different sizes, shapes, and colors. Some are edible and some are not. Some seeds germinate readily while others need specific conditions to be met before they will germinate. Within every seed lives a tiny plant or embryo.The outer covering of a seed is called the seed coat. Seed coasts help protect the embryo from injury and also from drying out. Seed coats can be quite thin and soft as in beans or very thick and hard as in locust or coconut seeds. Endosperm, which is a temporary food supply, is packed around the embryo in the form of special leaves called cotyledons or seed leaves. These generally are the first parts visible when the seed germinates. Plants are classified based upon the number of seed leaves (cotyledons) in the seed. Plants such as grasses and grass relatives can be monocots, containing one cotyledon. Dicots are plants that have two cotyledons.

 Seeds remain dormant or inactive until conditions are right for germination. All seeds need water, oxygen, and proper temperature in order to germinate. Some seeds require proper light also. Some germinate better in full light while other require darkness to germinate.When a seed is exposed to the proper conditions, water and oxygen are taken in through the seed coat. The embryo’s cells start to enlarge and the seed coat breaks open and root or radicle emerges first, followed by the shoot or plumule which contains the leaves and stem.

Many factors contribute to poor germination. Over-watering results in a lack of proper oxygen levels. Planting seeds to too deep results in the seed using up all of its stored energy before reaching the soil surface, and dry conditions result in the lack of sufficient moisture to start and sustain the germination process.

Objective:

The students will be able to describe how some environmental factors affect seed germination.

Materials:

Masking tape, Scissors, 3 ziplock bags, Marker, Forceps, Paper Towels, Metric Ruler, 3 colored pencils, 25 seeds, distilled water, 50 ml graduated, 1% detergent solution, 10% detergent solution, graph paper

Procedure:

  1. Label the 3 zip lock bags: Control, 1% Solution and, 10% Solution.
  2. Cut 6 square pieces of paper toweling to fit each bag.
  3. Place 2 squares in each bag.
  4. Distribute 6 seeds on each side of the paper towel between the plastic and towel.
  5. In the control bag add 25 ml of distilled water completely moistening the paper towel.
  6. In the 1% solution bag add 25 ml of 1% detergent solution making sure to completely moisten the towel.
  7. Do the same to the 10% solution bag by adding 25 ml of 10% detergent solution.
  8. Make sure all bags are sealed tightly.
  9. Place the bags in a dark warm place designated by the instructor.
  10. Write a hypothesis predicting the results of the experiment.
  11. Examine the bags daily for 5 days. Record any changes that might have occurred. If the roots is visible the seed is considered germinated.
  12. Record your date in the table below.
  13. Do not allow your towels to dry out. Moisten each bag with the appropriate solutions in equal amounts.
  14. Measure the root growth of each seed daily from the time it appeared.
  15. Graph the data from the table using the colored pencils to represent each of the zip lock bags.

Number of Seeds Germinated

 

Day Control 1% Detergent Solution 10% Detergent Solution
1
2
3
4
5

 

Average Growth of Germinating Seeds(mm)

Day Control 1% Detergent Solution 10% Detergent Solution
1
2
3
4
5

Graph Title: ________________________________________

Analysis:

1. How many of the seeds germinated after 5 days in distilled water? ________. In 1% solution? _______ in 10% solution? ________.

2. Was there a difference in the number of seeds germinated?

3. In which of the three bags did seeds germinate faster?

4. What was the purpose of the control?

5. Did the detergent strength have an effect on the seed’s germination? If so What was it?

6. Was your hypothesis correct? Why or why not?

7. If it was not, what will you do now?

 

Seed Plants Bi

For the Angiosperms the two variation of this basic design are seen in the two Classes (Monocots versus Dicots) (see fig. 23-2).

 

MONOCOTS		 DICOTS
Flower structure arranged in group of three arranged in groups of four or five
Leaves narrow with parallel veins wider with branching netlike veins
Vascular tissue scattered vascular bundles Ring of vascular bundles
Roots Many smaller roots One main taproot
Seed One cotyledon Two cotyledons

Second Semester Study Guide Bl

 

Second Semester Review 

 

Are animals autotrophs or heterotrophs? Explain.
What type of symmetry does a sea anemone have?
At which end of an animal is the tail located?
What supportive rod along the back do all chordates have at some time during their life?
How do sponges differ from all other animals?
How does a sponge obtain its food?
What hard, needle like structures are found in the walls of sponges?
Do all animal cells have cell walls? Explain.
In what phylum are squid & octopus found?
Name 4 animals that are classified as cnidarians.
What is the function of collar cells in sponges?
Are animals unicellular or multicellular organisms?
At which end of an animal is the head located?
What kind of symmetry do insects have?
Flatworms use what method to asexually reproduce?
Why do flatworms NOT need circulatory & respiratory systems?
What group of worms has a pseudocoelom?
What is the function of the radula in mollusks?
Which class of mollusks uses “jet propulsion” to move?
Describe torsion in gastropod mollusks.
Give several examples of appendages in arthropods.
Describe the body of all arthropods.
Describe the appendages of all arthropods.
In what group are clamworms found?
Name 3 main classes of mollusks.
What muscles open & close bivalve mollusks?
What makes up the exoskeleton of arthropods?
Name 4 members of the class Crustacea.
What group of animals has 3 body regions & 6 legs?
How do insects benefit agricultural crops?
Name 4 characteristics of all chordates.
In what order are amphibians without tails found?
From what structure in fish did jaws probably arise?
List 4 examples of echinoderms.
What structure in fish filters wastes from blood?
The urinary bladder & kidneys in fish make up what system?
Where are shark eggs fertilized?
What does “Agnatha” mean?
What does “Chondrichthyes” mean?
Describe caecilians.
Name 4 things used by sharks to detect their prey.
What type of symmetry do echinoderms have?
Why do most amphibians have thin, moist skin?
What does “amphibian” mean?
Describe development in placental mammals.
Where is the diaphragm found in mammals?
Name a reptilian characteristic found in birds.
What covers the body of birds?
What covers the body of reptiles?
Describe a reptile’s skin.
Are reptiles ectotherms or endotherms? Explain.
Where are the chorion & amnion found?
How many chambers does a bird’s heart have?
What adaptation of reptiles allowed them to live & reproduce on land?
How many chambers does the heart of most reptiles have?
How many chambers does the heart of mammals have?
Which group of vertebrates has a diaphragm & what is its function?
Are mammals endotherms or ectotherms?
Name 3 groups of ectothermic vertebrates.
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Salamander Key

 

 

Dichotomous Key to Salamanders

 

Introduction:

A dichotomous key is constructed of a series of couplets, each consisting of two separate statements. For example: couplet 1. Seeds round soybeans
1. Seeds oblong 2 (this statement indicates that you go to couplet “2”)

couplet 2. Seeds white northern beans
2. Seeds black black beans

By reading the two statements of each couplet, you progress through the key from typically broad characteristics to narrower characteristics until only a single choice remains. As long as the correct statement of each couplet is chosen, and the unknown organism is included in the key, a confident identification is usually achieved. Many types of organisms can be identified using a dichotomous key. In this lab, you will identify salamanders.

Materials:

pictures of various salamanders, dichotomous key, metric ruler, pencil

Procedure:

  1. Use the dichotomous key provided to identify the salamanders in Figure 1.
  2. Write the pathway you took to get to the name of the salamander next to the drawing.
  3. Write the correct name for the salamander on the line below each picture.

Figure 1 – Types of salamanders

Key to the Salamanders:

 

1 a  Hind limbs absent Siren
b  Hind limbs present Go to 2
2 a  External gills present in adults Mud puppy
b  External gills absent in adults Go to 3
3 a  Large size (over 7 cm long) Go to 4
b  Small size (under 7 cm long) Go to 5
4 a  Body background black, large white spots irregular in shape and size completely covering body & tail Tiger salamander
b  Body background black, small, round, white spots in a row along each side fro eye to tip of tail Spotted Salamander
5 a  Body background black with white spots Go to 6
b  Body background light color with dark spots and or lines on body Go to 7
6 a  Small white spots on a black background in a row along each side from head to tip of tail Jefferson salamander
b  Small white spots on a scattered throughout a black background from head to tip of tail Slimy salamander
7 a  Large irregular black spots on a light background extending from head to tip of tail Marbled salamander
b No large irregular black spots on a light background Go to 8
8 a Round spots scattered along back and sides of body, tail flattened like a tadpole Newt
b Without round spots and tail not flattened like a tadpole Go to 9
9 a Two dark lines bordering a broad, light mid-dorsal stripe with a narrow median dark line extending from the head onto the tail Two-lined salamander
b Without two dark lines running the length of the body Go to 10
10 a A light stripe running the length of the body and bordered by dark pigment extending downward on the sides Red-backed salamander
b A light stripe extending the length of the body, a marked constriction at the base of the tail Four-toed salamander

 

Sample 6a Transformation Lab

 

 

Lab 6A – Bacterial Transformation & Ampicillin Resistance

 

 

Introduction:
Bacterial transformation occurs when a bacterial cell takes up foreign DNA and incorporates it into its own DNA. This transformation usually occurs within plasmids, which are small circular DNA molecules separate from its chromosome. There can be 10 to 200 copies of the same plasmid within a cell. These plasmids may replicate when the chromosome does, or they may replicate independently. Each plasmid contains from 1,000 to 200,000 base pairs. Certain plasmids, called R plasmids, carry the gene for resistance to antibiotics such as ampicillin, which is used in this lab.

Plasmids function in transformation in two different ways. They can transfer genes that occur naturally within them, or they can act as vectors for introducing foreign DNA. Restriction enzymes can be used to cut foreign DNA and insert it into the plasmid vectors. The bacteria used in this lab were Escherichia coli (E. coli). It was ideal for this transformation study because it can be easily grown in Luria broth or on agar, and it has a relatively small genome of about five million base pairs.

Transformation is not the only method of DNA transfer within bacteria. Conjugation is a DNA transfer that occurs between two bacterial cells. A bridge is formed between the two cells and genetic information is traded. In transduction, a virus is used to transfer foreign DNA into a bacterial cell.

Hypothesis:
The transformed E. coli with the ampicillin resistance gene will be able to grow in the ampicillin plates, but the non-transformed E. coli will not.

Materials:
The materials needed for this lab were 2 sterile test tubes, 500 μL of ice cold 0.05M CaCl2, E. coli bacteria cultures, a sterile inoculating loop, a sterile micropipette, 10 μL of pAMP solution, a timer, ice, a water bath, 500 μL of Luria broth, a spreading rod, 4 plates: 2 ampicillin+ and 2 ampicillin – , and an incubator.

 

Methods:
One sterile tube was labeled “+” and the other “-“. A sterile micropipette was used to transfer 250 μL of ice cold 0.05M CaCl2 to each tube. A large colony of E. coli was transferred with an inoculating loop to each tube. The suspension was then mixed by repeatedly drawing and emptying a sterile micropipette. 10μL of pAMP solution was added to the cell suspension in the tube marked “+” and mixed by tapping the tube. Both tubes were immediately put on ice for 15 minutes and then soaked in a 42° C water bath for 90 seconds. The tubes were then returned to ice for another 2 minutes.

After the heat shock, 250 μL of Luria broth were added to each tube. The tubes were mixed by tapping. Two plates of ampicillin + agar were labeled LB/AMP+ and LB/AMP-. The two plates of ampicillin- agar were labeled LB+ and LB-. 100 μL of the cell suspension in the “+” tube were placed on the LB+ and the LB/AMP+ plates. 100μL of the cell suspension in the “-” tube were added to the LB- and the LB/AMP- plates. These were spread with a spreading rod that was sterilized by passing it over a flame after each use. The plates were allowed to sit for several minutes and then incubated over night inverted at 37° C.

 

Results:

 

 

 

LB+

(Positive Control)

  

LB-

(Positive Control)

  

LB/AMP+ (Experimental)

  

LB/AMP- (Experimental)
 

Bacterial Growth

 lawn lawn 3 colonies No growth

 

Questions:
1. Compare and contrast the number of colonies on each of the following pairs of plates. What does each pair of results tell you about the experiment?
LB+ and LB- Both of these plates had a lawn of bacteria. This proves that the bacteria are capable of growing on the agar and that there was nothing preventing growth beside the ampicillin.

LB/AMP- and LB/AMP+ The LB/AMP- had no growth, but the LB/AMP+ had small growth. This shows that the bacteria was transformed and developed a resistance to ampicillin.

LB/AMP+ and LB+ The LB/AMP+ had less growth than the LB+. This shows that the transformation was not completely effective and only transformed some of the most competent bacterial cells.

 

2.  Total mass of pAMP used = 0.05 μg

 

Total volume of cell suspension = 510 μL

 

Fraction of cell suspension spread on the plates = 0.196

 

Mass of pAMP in cell suspension = 0.0098

 

Number of colonies per μg of plasmid = 0.0294

3. What factors might influence the transformation efficiency? Explain the effect of each you mention.
Transformation efficiency could be affected by the size of the colony added to the solution. In a larger colony the efficiency would increase because there would be more receptive cells. Another factor would b the amount of pAMP added. The more pAMP added, the higher the efficiency. The amount of Luria broth added could also affect efficiency. If the amount of Luria broth was increased, the efficiency would decrease.

 

Error Analysis:
This lab had several steps, each giving the potential for error. All of the measurements had to be precise and accurate, and the heat shock timing was also a very complicated procedure. Error in this lab could have been caused by the concentration of the CaCl2 due to the fact that most of it was frozen.

 

Discussion and Conclusion:
The bacteria treated with the pAMP solution developed a resistance to ampicillin and were able to grow on the ampicillin+ plate. Those that were not treated with the pAMP were not able to grow on this medium. The plates with no ampicillin served as a control to show how the bacteria would look in normal conditions. Transformation is never fully effective, Only cells that are competent enough are able to take up the foreign DNA. Therefore, the ampicillin + plates showed less growth than the control plate.

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