Biology Junction Team - BIOLOGY JUNCTION

Molecular Biology

Molecular Biology

Introduction:
The bacterium Escherichia coli or E. coli is an ideal organism for the molecular geneticist to manipulate and has been used extensively in recombinant DNA research. It is a common inhabitant of the human colon and can easily be grown in suspension culture in a nutrient medium such as Luria broth, or in a petri dish of Luria broth mixed with agar (LB agar) or nutrient agar.

The single circular chromosome of E. coli contains about five million DNA base pairs, only 1/600th the haploid amount of DNA in a human cell. In addition, the E.coli cell may contain small circular DNA molecules (1,000 to 200,000 base pairs) called plasmids, which also carry genetic information. The plasmids are extra chromosomal; they exist separately from the chromosome. Some plasmids replicate only when the bacterial chromosome replicates, and usually exists only as single copies within the bacterial cell. Others replicate autonomously and often occur in as many as 10 to 200 copies within a single bacterial cell. Certain plasmids, called R plasmids, carry genes for resistance to antibiotics such as ampicillin, kanamycin, or tetracycline.

In nature, genes can be transferred between bacteria in three ways: conjugation, transduction, and transformation. Conjugation is a mating process during which genetic material is transferred from one bacterium to another of a different mating type. Transduction requires the presence of a virus to act as a vector to transfer small pieces of DNA from one bacterium to another. Bacterial transformation involves transfer of genetic information into a cell by direct uptake of the DNA. During gene transfer, the uptake and expression of foreign DNA by recipient bacterium can result in the conferring a particular trait to a recipient lacking the trait.

Plasmids can transfer genes that occur naturally within them, or plasmids can act as carriers for introducing foreign DNA from other bacteria, plasmids, or even eukaryotes into recipient bacterial cells. Restriction endonucleases can be used to cut and insert pieces of foreign DNA into the plasmid vectors (figure 6.1).

Figure 6.1 Bacterial Transformation using a Restriction Endonuclease

Exercise 6A: Bacterial Transformation-Ampicillin Resistance*
Background Information:
You will insert a plasmid that contains a gene for the resistance to ampicillin , an antibiotic that is lethal to many bacteria, into competent E.coli cells. Transformed bacteria can be selected based on their resistance to ampicillin by spreading the transformed cells on nutrient medium containing ampicillin. Any cell that grown on this mediums has been transformed.

Procedure:
1. Mark one 15 mL tube “+”; this tube will have the plasmid added to it. Mark another tube “-” ; this tube will have no plasmid added.

2. Use a sterile pipette to add 250 micro liters (uL) of ice cold 0.05M CaCl2 to each tube.

3. Transfer a large (3 mm) colony of E.coli from a starter plate to each of the tubes using a sterile inoculating loop. Try and get the same amount of bacteria into each tube. Be careful not to transfer any agar.

4. Vigorously tap the loop against the wall of the tube to dislodge the cell mass.

5. Mix the suspension by repeatedly drawing in and emptying a sterile micro pipette with the suspension.

6. Add 10 uL of pAMP solution (0.005 ug/uL) directly into the cell suspension in tube “+”. Mix by tapping the tube with your finger. This solution contains the antibiotic resistance plasmid.

7. Keep both tubes on ice for 15 minutes.

8. While the tubes are on ice, obtain two LB agar plates and two LB/Amp agar(LB agar containing ampicillin) plates. Label each plate on the bottom as follows: one LB agar plate “LB+” and the other “LB-“. Label one LB/Amp plate “LB/Amp+” and the other “LB/Amp-.”

9. A brief pulse of heat facilitates entry of foreign DNA into the E. coli cells. Heat shock cells in both the “+” and “-” tubes by holding the tubes in a 42 degree C water bath for 90 seconds. It is essential that cells be given a sharp and distinct shock, so take the tubes directly from the ice to the 42 degree C water bath.

10. Immediately return cells to ice for two minutes.

11. Use sterile micro pipette to add 250 uL of Luria broth to each tube. Mix by tapping with your finger and set at room temperature. Any transformed cells are now resistant to ampicillin because they possess the gene whose product renders the antibiotic ineffective.

12. Place 100 uL of “+” cells on the “LB+” plate and on the “LB/Amp+” plate. Place 100 uL of “-” cells on the “LB-” plate and on the “LB/Amp-” plate.

13. Immediately spread the cells using a sterile spreading rod. ( Remove the spreading rod from alcohol and briefly pass it through a flame. Cool by touching it to the agar on a part of the dish away from the bacteria. Spread the cells and once again immerse the rod in alcohol and flame it.) Repeat the procedure for each plate.

14. Allow plates to set for several minutes. Tape your plates together and incubate inverted overnight at 37 degrees C.

* Exercise 6A is adapted with permission from DNA Science: A First Course in Recombinant-DNA Technology by David A Micklos, DNA Learning center of Cold Spring Harbor Laboratory , and Greg A. Freyer, Columbia University College of Physicians and Surgeons, Copyright 1990 Cold Spring Harbor Laboratory Press and Carolina Biological Supply Company. It is based on a protocol published by Douglas Hanahan, University of California, San Francisco

Analysis of Results:
1. Observe the colonies through the bottom of the culture plate. Do not open the plates. Count the number of individual colonies; use a permanent marker to mark each colony as it is counted. If cell growth is too dense to count individual colonies, record “lawn.”

LB + ( Positive Control ) ___________________ LB – ( Positive Control ) ______________________

LB/Amp + ( Experimental ) _________________ LB/Amp – ( Negative Control ) _________________

2. 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/

a. LB+ and LB- _______________________________________________

b. LB/Amp- and LB/Amp+ ____________________________________________________________

c. LB/Amp + and LB+ _____________________________________________________________

3. Transformation efficiency is expressed as the number of antibiotic-resistant colonies per microgram of pAMP. Because transformation is limited to only those cells that are competent, increasing the amount of plasmid used does not necessarily increase the probability that a cell will be transformed. A sample of competent cells is usually saturated with small amounts of plasmid and excess DNA may actually interfere with the transformation process.

a. Determine the total mass of pAMP used. _____________________

( you used 10 uL of pAMP at a concentration of 0.005ug/uL.)

Total Mass = volume x concentration.

b. Calculate the total volume of cell suspension prepared. _______________________

c. Now calculate the fraction of the total cell suspension that was spread on the plate.

( Number of uL spread/total volume) _____________________________________

d. Determine the mass of pAMP in cell suspension. __________________________

(Total mass of pAMP X fraction spread.)

e. Determine the number of colonies per ug of plasmid. Express in scientific notation.

( Number of colonies observed/mass pAMP spread ( from calculation in step (d) = transformation efficiency.)

4. This is the transformation efficiency. What factors might influence transformation efficiency? Explain the effect of each you mention.

___________________________________________________________________

Exercise 6B: Restriction Enzyme Cleavage of DNA and Electrophoresis
Background Information:
Restriction enzymes or restriction endonucleases are essential tools in recombinant DNA methodology. Several hundred have been isolated from a variety of prokaryotic organisms. Restriction endonucleases are named according to a specific system of nomenclature. The letters refer to the organism from which the enzyme was isolated. The first letter of the name stands for the genus name of the organism. The next two letters represent the second word or the species name. The fourth letter (if there is one) represents the strain of the organism. Roman numerals indicate whether the particular enzyme was the first isolated, the second, or so on.

Examples:

EcoRI E = genus Escherichia

co= species coli

R = strain RY13

I = first endonuclease isolated

HaeII H = genus Haemophilus

ae= species aegyptus

I I = second endonuclease isolated

Restriction endonucleases recognize specific DNA sequences in double stranded DNA (usually a four to six base pair sequence of nucleotides) and digest the DNA at these sites. The result is the production of fragments of DNA of DNA of various lengths. Some restriction enzymes cut cleanly through helix at the same position on both strands to produce fragments with blunt ends ( figure 6.2a ). Other endonucleases cleave each strand off center at specific nucleotides to produce fragments with “overhangs” or sticky ends (figure 6.2b). By using the same restriction enzyme to “cut” DNA from two different organisms, complementary “overhangs” or sticky ends will be produced and allow the DNA from two sources to be “recombined.”

Figure 6.2a

Hae III

Cleavage by HaeIII produces blunt ends

5’…GGCC…3′

3’…CCGG…5′

Figure 6.2b

EcoR I

Cleavage by EcoRI produces sticky ends

5’…GAATTC…3′

3’…CTTAAG…5′

In this exercise, samples of DNA obtained from the bacteriophage lambda have been incubated with different restriction enzymes. The resulting fragments of DNA will be separated by using gel electrophoresis. One sample has been digested with the restriction endonuclease EcoRI, one with the restriction endonuclease HindIII, and the third sample is uncut. The DNA samples will be loaded into wells of an agarose gel and separated by the process of electrophoresis. After migration of the DNA through an electrical field, the gel will be stained with methylene blue, a dye which binds to DNA.

When any molecule enters an electric field, the mobility or speed at which it will move is influenced by the charge of the molecule, the strength of the electrical field, the size and shape of the molecule, and the density of the medium (gel) through which it is migrating. When all molecules are positioned at a uniform starting site on a gel and a gel is placed in a chamber containing a buffer solution and electricity is applied, the molecules will migrate and appear as bands. Nucleic acids, like DNA and RNA, move because of the charged phosphate group in the backbone of the DNA molecule. Because the phosphates are negatively charged at neutral pH, the DNA will migrate through the gel toward the positive electrode.

In this exercise, we will use an agarose gel. In agarose, the migration rate of linear fragments of DNA is inversely proportional to their size; the smaller the DNA molecules, the faster it migrates through the gel.

General Procedure:
A: Preparing the Gel
1. Prepare the agarose gel for electrophoresis according to the directions given by you teacher or in the kit.

2. Obtain the phage lambda DNA digested with EcoRI endonuclease. The DNA is mixed with a gel-loading solution containing a tracking dye, bromophenol blue, that will make it possible to “track” the processes of its migration in the agarose gel.

3. Obtain the phage lambda DNA digested with HindIII endonuclease. The DNA fragments are of a known size and will serve as a “standard” for measuring the size of the EcoRI fragments from step 2. It also contains the tracking dye.

4. Obtain the undigested phage lambda DNA to use as a control. It also contains the tracking dye.

B: Loading the Gel

Helpful Hints for Loading Gel

1. Put a small amount of gel-loading solution into the end of a micropipette. Do not allow the solution to move up into the pipette, or bubbles will be introduced into the well of the agarose gel during loading. loading.

2. Hold the tip of the pipette above the gel and gently dispense the solution. The loading dye is denser than the buffer and will move into the well. ( Do not place the tip of the pipette into the well or you might puncture the gel).

1. Pour enough buffer gently over the gel to cover it.

2. Load 5-10 uL of undigested lambda phage DNA (control) into a well.

3. Load 5-10 uL of the HindIII digest into a second cell.

4. Load 5-10 uL of the EcoRI digest into a third well.

See the figure below for a side view of a typical gel box.

Figure 6.3 Gel Box.

C: Electrophoresis:
1. Place the top on the electrophoresis chamber and carefully connect the electrical leads to an approved power supply (black to black and red to red). Set the voltage to the appropriate level for your apparatus. When the current is flowing, you should see bubbles on the electrodes.

2. Allow electrophoresis to proceed until the tracking dye has moved nearly to the end of the gel.

3. After electrophoresis is complete, turn off the power, disconnect the leads, and remove the cover of the electrophoresis chamber.

D: Staining and Visualization:
Note: Wear Gloves!

1. Carefully remove the gel bed from the chamber and gently transfer the gel to a staining tray for staining. Use the metal spatula under the gel during the transfer. Do not stain in the electrophoresis chamber.

E: Determining Fragment Size:
1. After observing the gel, carefully wrap it in plastic wrap and smooth out all the wrinkles.

2. Using a marking pen, trace the outlines of the sample wells and the location of the bands.

3. Remove the plastic wrap and flatten it out on a white piece of paper on the laboratory bench. Save the gel in a zip lock bag. Add several drops of buffer, store at 4degrees C. You can make your measurements directly from the marks on the plastic wrap.

Analysis and Results:
Background Information
The size of the fragments produced by a specific endonuclease can be determined by using standard fragments of known size. When you plot the data on semilog graph paper, the size of the fragments is expressed in the log of the number of base pairs they contain. This allows data to be plotted on a straight line. The migration distance of the unknown fragments, plotted on the x-axis, will allow their size to be determined on the standard curve.

Graphing:
A. Standard Curve for HindIII
1. Measure the migration distance in cm) for each HindIII band on your gel. Measure from the bottom of the sample well to the bottom of the band. Measurement of the longest standard fragment does not need to be measured (23,120 base pairs). Record these measurements on table 6.1.

2. Plot the measured migration distance for each band of the standard HindIII digest against the actual base pair (bp) fragment sizes given in Table 6.1 using the semilog graph paper. Draw the best fit line to your points. This will serve as a standard curve.

B. Interpolated Calculations for EcoRI:
From your standard curve for HindIII, made from known fragment sizes, you can calculate fragment sizes resulting from a digest with EcoRI. The procedure is as follows:

1. Measure the migration distance in cm for each EcoRI band. Record the data in Table 6.1

2. Determine the sizes of fragments of lambda phage DNA digested with EcoRI. Locate on the x axis the distance migrated by the first EcoRI fragment. Using a ruler, draw a vertical line from this point to its intersection with a best fit data line. Now extend a horizontal line from intersection point to the Y axis. This point gives the base pair size for this EcoRI fragment. Repeat this procedure and determine the remaining EcoRI fragments. Enter your interpolated data in Table 6.1, in the interpolated bp column.

3. Your teacher will provide you with the actual bp data. Compare your results to these actual sizes. Note: This interpolation technique is not exact. You should expect as much as 10% to 15% error.

Table 6.1: Distance HindIII produced fragments migrate in agarose gel (cm)

HindIII
Actual bp	Measured Distance (cm)
23,130
9,416
6,557
4,361
2,322 +
*570 +**
125 *
+ may form a single band	* may not be detected

Table 6.2: Distance EcoRI produced fragments migrate in agarose gel (cm)

	EcoRI
	Measured Distance (cm)	Interpolated bp	Actual bp
Band 1
Band 2
Band 3
Band 4
Band 5
Band 6

4. For which fragment size was your graph most accurate? For which fragment size was it least accurate/ What does this tell you about the resolving ability of agarose-gel electrophoresis?

_____________________________________________________________________

Analysis:
1. Discuss how each of the following factors would affect the results of electrophoresis:

a. Voltage used _____________________________________________________________________

_____________________________________________________________________

b. Running time_________________________________________________________________

_____________________________________________________________________

c. Amount of DNA used_________________________________________________________________

_____________________________________________________________________

d. Reversal of polarity______________________________________________________________

____________________________________________________________________

2. Two small restriction fragments of nearly the same base-pair size appear as a single band, even when the sample is run to the very end of the gel. What could be DNA to resolve the fragments? Why would it work?

_____________________________________________________________________

____________________________________________________________________

Questions:
1. What is a plasmid? How are plasmids used in genetic engineering?

____________________________________________________________________

2. What are restriction enzymes? How do they work? What are recognition sites?

____________________________________________________________________

3. What is the source of restriction enzymes? What is there function in nature?

____________________________________________________________________

4. Describe the function of electricity and agarose gel in electrophoresis.

____________________________________________________________________

5. If restriction enzyme digest resulted in DNA fragments of the following sizes: 4, 000 base pairs, 2,500 base pairs, 2,000 base pairs, 400 base pairs, sketch the resulting separation by electrophoresis. Show starting point, positive and negative electrodes, and the resulting bands.

6. What are the functions of loading dye in electrophoresis? How can DNA be prepared for visualization?

____________________________________________________________________

7. Use the graph you will prepared from your lab data to predict how far in centimeters a fragment of 8,000 bp would migrate.

____________________________________________________________________

8. How can a mutation that alters a recognition site be detected by gel electrophoresis?

____________________________________________________________________

Graph paper:

Mollusk & Annelid Study Guide B1

Mollusk & Annelid Study Guide

Be able to answer these questions:

Name the kingdom for mollusks & annelids.
What is the oldest part of a bivalves shell called?
What muscles open & close a bivalves shell?
What tissue surrounds & protects the soft body of mollusks?
What is the larval stage of mollusks called & describe it?
What structure enables a squid to move by jet propulsion?
What are the external segments of an earthworm’s body called?
Name the internal shell of a squid.
How do earthworms breathe?
Earthworms are hermaphrodites. What does this mean?
What type of symmetry do annelids & mollusks have?
What are the respiratory organs of aquatic mollusks & annelids called?
What are aortic arches & what organism has them?
What type of circulatory system do mollusks & annelids have?
What is the tongue-like structure called that snails use to scrap algae?
What is the “lip” of an earthworm called & how is it used?
Name the muscular organ used by mollusks for movement.
Name several examples of bivalve mollusks.
Give an example of a univalve mollusk.
Give an example of a marine, shelled cephalopod mollusk.
What is the area of a mollusk’s body called that contains most of the body organs?

Know the class for each of the following mollusks & annelids:

clams & scallops
snails & slugs
clam worms
chitons
squid & octopus
leeches
earthworms

Be able to label these internal parts of a clam:

heart
gills
anus
adductor muscles
incurrent siphon

Be able to recognize pictures of these mollusks & annelids:

clam
snail
lugworm
leech
earthworm
chiton
squid
octopus

Notes

Study Guides

Mitosis PPT Questions

Cell Cycle and Mitosis
ppt Questions

Cell Cycle

1.Prokaryotic organisms include ___________, while plants and animals are ____________.

2. Describe prokaryotes.

3. How do bacteria asexually reproduce?

4. Name the 3 main steps of binary fission in bacteria.

5. Name a bacterial cell that reproduces by binary fission.

6. Describe eukaryotes.

7. How do eukaryotes asexually reproduce cells?

8. The stages in the growth and reproduction of a cell are called the __________ ___________.

9. List the 5 stages in the cell cycle.

10. What does G1 stage stand for?

11. Name two things that happen to a cell during G1?

12. What is the S stage of the cell cycle?

13. _________ instructions are copied in the S phase as ___________ are duplicated.

14. _______ stands for second growth stage.

15. G2 is the time between ____________ and ___________.

16. Cells continue to _________ during G2 and to make __________ that will be needed for mitosis or cell division.

17. Mitosis or cell division is known as the ________ stage.

18. How does a cell use its energy during the M phase?

19. Does a cell continue growing & making proteins in the M phase?

20. Mitosis is also called _______________ which means division of the ____________.

21. ____________ is called the resting stage and makes up the longest part of a cell’s life cycle.

22. What happens to cells during interphase?

23. Are chromosomes visible during interphase?

Mitosis

24. Name the 4 stages of mitosis.

25. Name 2 things that happen to a cell during prophase.

26. Can chromosomes be seen during prophase?

27. Sketch a eukaryotic chromosome and label the centromere and kinetochore fiber that attaches to it.

28. How many pairs of chromosomes are found in humans?

29. List 3 things that occur during metaphase.

30. Where are chromosomes located during metaphase of a cell?

31. What stage occurs after metaphase?

32. List 2 things that happen to cells during anaphase.

33. Sketch and label the mitotic spindle and attached chromosomes.

34. What is the last stage of mitosis?

35. Where are the two sets of chromosomes located at Telophase?

36. What two things reform during Telophase?

37. Chromosomes ___________ during Telophase so they are no longer visible.

38. In plants, what begins to form that will separate the two cells?

39. How are the two cells separated from each other in animals?

40. _____________ or division of the cytoplasm follows ___________, division of the nucleus, and forms ____________ daughter cells.

41. How do the two, new daughter cells compare to each other?

42. Label the following stages of mitosis.

Printable Copy

Genetics


1862	1868	1880

Genetic Terminology:

Trait – any characteristic that can be passed from parent to offspring
Heredity – passing of traits from parent to offspring
Genetics – study of heredity
Alleles – two forms of a gene (dominant & recessive)
Dominant – stronger of two genes expressed in the hybrid; represented by a capital letter (R)
Recessive – gene that shows up less often in a cross; represented by a lowercase letter (r)
Genotype – gene combination for a trait (e.g. RR, Rr, rr)
Phenotype – the physical feature resulting from a genotype (e.g. tall, short)
Homozygous genotype – gene combination involving 2 dominant or 2 recessive genes (e.g. RR or rr); also called pure
Heterozygous genotype – gene combination of one dominant & one recessive allele (e.g. Rr); also called hybrid
Monohybrid cross – cross involving a single trait
Dihybrid cross – cross involving two traits
Punnett Square – used to solve genetics problems

Blending Concept of Inheritance:

Accepted before Mendel’s experiments
Theory stated that offspring would have traits intermediate between those of its parents such as red & white flowers producing pink
The appearance of red or white flowers again was consider instability in genetic material
Blending theory was of no help to Charles Darwin’s theory of evolution
Blending theory did not account for variation and could not explain species diversity
Particulate theory of Inheritance, proposed by Mendel, accounted for variation in a population generation after generation
Mendel’s work was unrecognized until 1900

Gregor Mendel:

Austrian monk
Studied science & math at the University of Vienna
Formulated the laws of heredity in the early 1860’s
Did a statistical study of traits in garden peas over an eight year period

drawing of a flower cross-section showing both male and female sexual structures

Why peas, Pisum sativum?

Can be grown in a small area
Produce lots of offspring
Produce pure plants when allowed to self-pollinate several generations
Can be artificially cross-pollinate

Picture of Pisum sativum
GARDEN PEA

Mendel’s Experiments:

Mendel studied simple traits from 22 varieties of pea plants (seed color & shape, pod color & shape, etc.)
Mendel traced the inheritance of individual traits & kept careful records of numbers of offspring
He used his math principles of probability to interpret results
Mendel studied pea traits, each of which had a dominant & a recessive form (alleles)
The dominant (shows up most often) gene or allele is represented with a capital letter, & the recessive gene with a lower case of that same letter (e.g. B, b)
Mendel’s traits included:

         a. Seed shape — Round (R) or Wrinkled (r)
            b. Seed Color —- Yellow (Y) or Green (y)
            c. Pod Shape — Smooth (S) or wrinkled (s)
            d. Pod Color — Green (G) or Yellow (g)
            e. Seed Coat Color — Gray (G) or White (g)
            f. Flower position — Axial (A) or Terminal (a)
            g. Plant Height — Tall (T) or Short (t)
            h. Flower color — Purple (P) or white (p)

Mendel produced pure strains by allowing the plants to self-pollinate for several generations
These strains were called the Parental generation or P1 strain

Mendel cross-pollinated two strains and tracked each trait through two
generations (e.g. TT x tt )

Trait – plant height
Alleles – T tall, t short
P1 cross TT x tt			genotype — Tt
	t	t	phenotype — Tall
T	Tt	Tt	genotypic ratio –all alike
T	Tt	Tt	phenotypic ratio- all alike

The offspring of this cross were all hybrids showing only the dominant trait & were called the First Filial or F1 generation
Mendel then crossed two of his F1 plants and tracked their traits; known as an F1 cross

Trait – plant height
Alleles – T tall, t short
F1 cross Tt x Tt			genotype — TT, Tt, tt
	T	t	phenotype — Tall & short
T	TT	Tt	genotypic ratio —1:2:1
t	Tt	tt	phenotypic ratio- 3:1

When 2 hybrids were crossed, 75% (3/4) of the offspring showed the dominant trait & 25% (1/4) showed the recessive trait; always a 3:1 ratio
The offspring of this cross were called the F2 generation
Mendel then crossed a pure & a hybrid from his F2 generation; known as an F2 or test cross

Trait – Plant Height
Alleles – T tall, t short
F2 cross TT x Tt			F2 cross tt x Tt
	T	t		T	t
T	TT	Tt	t	Tt	tt
T	TT	Tt	t	Tt	tt

genotype – TT, Tt			genotype – tt, Tt
phenotype – Tall			phenotype – Tall & short
genotypic ratio – 1:1			genotypic ratio – 1:1
phenotypic ratio – all alike			phenotypic ratio – 1:1

50% (1/2) of the offspring in a test cross showed the same genotype of one parent & the other 50% showed the genotype of the other parent; always a 1:1 ratio

Problems: Work the P1, F1, and both F2 crosses for all of the other pea plant traits & be sure to include genotypes, phenotypes, genotypic & phenotypic ratios.

Mendel also crossed plants that differed in two characteristics (Dihybrid Crosses)
such as seed shape & seed color
In the P1 cross, RRYY x rryy, all of the F1 offspring showed only the dominant form for both traits; all hybrids, RrYy

Traits: Seed Shape & Seed Color
Alleles: R round Y yellow r wrinkled y green
P1 Cross: RRYY x r r yy
	ry	Genotype:	RrYy
RY	RrYy	Phenotype:	Round yellow seed
Genotypic ratio:	All alike
		Phenotypic ratio:	All Alike

When Mendel crossed 2 hybrid plants (F1 cross), he got the following results

Traits: Seed Shape & Seed Color
Alleles: R round Y yellow r wrinkled y green
F1 Cross: RrYy x RrYy
	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	r rYY	r rYy
ry	RrYy	Rryy	r rYy	r ryy

Genotypes	Genotypic Ratios	Phenotypes	Phenotypic Ratios
RRYY	1	Round yellow seed	9
RRYy	2
RrYY	2
RrYy	4
RRyy	1	Round green seed	3
Rryy	2
r rYY	1	Wrinkled yellow seed	3
r rYy	2
r ryy	1	Wrinkled green seed	1

Problems: Choose two other pea plant traits and work the P1 and F1 dihybrid crosses. Be sure to show the trait, alleles, genotypes, phenotypes, and all ratios.

Results of Mendel’s Experiments:

Inheritable factors or genes are responsible for all heritable characteristics
Phenotype is based on Genotype
Each trait is based on two genes, one from the mother and the other from the father
True-breeding individuals are homozygous ( both alleles) are the same
Law of Dominance states that when different alleles for a characteristic are inherited (heterozygous), the trait of only one (the dominant one) will be expressed. The recessive trait’s phenotype only appears in true-breeding (homozygous) individuals

Trait: Pod Color
Genotypes:	Phenotype:
GG	Green Pod
Gg	Green Pod
gg	Yellow Pod

Law of Segregation states that each genetic trait is produced by a pair of alleles which separate (segregate) during reproduction

Rr
R	r

Law of Independent Assortment states that each factor (gene) is distributed (assorted) randomly and independently of one another in the formation of gametes

RrYy

RY	Ry	rY	ry

Other Patterns of Inheritance:

Incomplete dominance occurs in the heterozygous or hybrid genotype where the 2 alleles blend to give a different phenotype
Flower color in snapdragons shows incomplete dominance whenever a red flower is crossed with a white flower to produce pink flowers

In some populations, multiple alleles (3 or more) may determine a trait such as in ABO Blood type
Alleles A & B are dominant, while O is recessive

Genotype	Phenotype
IOIO	Type O
IAIO	Type A
IAIA	Type A
IBIO	Type B
IBIB	Type B
IAIB	Type AB

Polygenic inheritance occurs whenever many variations in the resulting phenotypes such as in hair, skin, & eye color
The expression of a gene is also influenced by environmental factors (example: seasonal change in fur color)

Meiosis Labeling

Meiosis

On each of the images, label the phase of meiosis

1. _______________	2. _______________	3. _______________
4. _______________	5. _______________	6. _______________
7. _______________	8._______________	9._______________
10. _______________	11. A cell with a diploid number of 20 undergoes meiosis. This will produce ________ daughter cells, each with ________ chromosomes. 12. Synapsis occurs during this phase: _______________________ 13 How many different possible combinations are there for a cell that has 10 chromosomes (5 pairs): _____________ 14. Tetrads line up along the equator during this phase: ______________ 15. At the end of meiosis I, ________ daughter cells are created. These daughter cells are [ diploid \| haploid ]. 16. Meiosis occurs in what type of cells: ____________________________

Now label the photographs.
17. _______________	18. _______________	19. _____________
20. _______________	21. _______________	22. _____________
23. _______________	24. _______________	24. _____________
25. _______________

Trait – plant height

Trait – plant height