My Classroom Material 119 - BIOLOGY JUNCTION

Genetics

Mendelian Genetics
All Materials © Cmassengale


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)

Graph Examples

Examples of Graphs

LINE GRAPHS

Line Graph title

A line graph is most useful in displaying data or information that changes continuously over time. The example below shows the changes in the temperature over a week in January. Notice that the title of the graph is “Average Daily Temperature for January 1-7 in degrees Fahrenheit”.

To the left is a table that shows the date in one column and the corresponding temperature in the second column. The line graph on the right shows the degrees of temperature going up the vertical axis (up and down numbers on the left of the graph) and the days of the week on the horizontal axis (going sideways from left to right). The points for the temperature for each day are connected by a line – thus the graph is a line graph.

Average Daily Temperature for January 1-7 in Degrees Fahrenheit

Date	Temperature
1	10
2	25
3	30
4	42
5	23
6	25
7	40

Bar Graph Animated title

BAR GRAPHS

Bar graphs are an excellent way to show results that are one time, that aren’t continuous – especially samplings such as surveys, inventories, etc. Below is a typical survey asking students about their favorite after school activity. Notice that in this graph each column is labeled – it is also possible to label the category to the left of the bar. In this case, the numbers for each category are across the bottom of the chart.

A bar chart is marked off with a series of lines called grid lines. These lines typically mark off a numerical point in the series of numbers on the axis or line. In this case, each grid line going up and down marks a multiple of 20 as the graph is divided. More gridlines can make it easier to be exact with the amounts being shown on the bar graph, but too many can make it confusing. Notice that for data that does not fall evenly on a multiple of 20, the bar is in between two grid lines. Bar graphs are useful to get an overall idea of trends in responses – which categories get many versus few responses.

Favorite Student After School Activity

Activity	Number
Visit W/Friends	175
Talk on Phone	168
Play Sports	120
Earn Money	120
Use Computers	65

Circle Pie Graph Title

CIRCLE/PIE GRAPHS

Circle or pie graphs are particularly good illustrations when considering how many parts of a whole are inception. In the table below both the number of hours in a whole day devoted to certain activities is listed as well as the percent of time for each of these activities. The pie chart is then divided very much as a baker’s pie would be into slices that represent the proportional amounts of time spent on each activity.

To the right of the pie chart is a legend that tells which color stands for which category. In addition, the percents are also near the pie slice that stands for that particular amount of time spent.

Percent of Hours of a Day Spent on Activities

ACTIVITY	HOURS	PERCENT OF DAY
Sleep	6	25
School	6	25
Job	4	17
Entertainment	4	17
Meals	2	8
Homework	2	8

Pie Graph of Day's Activities

Loss of Biodiversity Activity

Loss of Biodiversity

Students will make a PowerPoint presentation on the topic of loss of biodiverisity in one of the following areas:

Fauna of Arkansas
North American Vertebrates
North American Invertebrates
North American Plants
Flora of Arkansas
Aquatic Habitats of Arkansas
Florida Everglades
Alaskan Tundra
United States Deserts
Along the Mississippi River
North American Waterfowl
North American Raptors
North American Reptiles
North American Amphibians
North American Mammals

The PowerPoint presentation will be presented to the class and must include 25 slides, 15 of which must include graphics such as images from your web search (save on disk as .jpeg), pictures from books or magazines that you have scanned and inserted into your program, or photographs taken with a digital camera. You should also include three of the following as part of your slide presentation:

Maps
Graphs
Lists
Photograph of a person you interviewed

Your PowerPoint presentation must be accompanied by a written script that corresponds to the numbered order of your slides. The following must be included in your PowerPoint presentation and script:

Name/Description of your chosen area (include a picture if available)
Explanation of the physical environment of the area — climate, water, temperature, etc.
Examples of threatened organisms ( include pictures)
Reasons for organisms endangerment
How the loss of these organisms is affecting other organisms &/or the environment
Conservation measures being taken to prevent the loss of biodiversity in this area

Mammal Orders

Mammal Orders

Locate the orders of mammals and then list the common names of the animals in each order

ARTIODACTYLA	CARNIVORA	CETACEA
CHIROPTERA	EDENTATA	INSECTIVORA
LAGOMORPHA	MARSUPIALIA	MONOTREMATA
PERISSODACTYLA	PRIMATES	PROBOSCIDEA
RODENTIA	SIRENIA

Solution

Lab 8 Ap Sample Population Genetics

Lab 8 Population Genetics

Introduction:
G. H. Harding and W. Weinberg both came up with the idea that evolution could be viewed as changes in the frequency of alleles in a population. They used the letter “p” to represent and “A” allele and the letter “q” to represent the “a” allele. So, in a population of 100 individuals and 40% of the alleles are “A”, then “p” is .40, “q” would equal .60. The frequency at which the alleles show up is known as their allele frequency. The frequency of the possible combinations of the alleles can be figured using this equation:

p2 + 2pq + q2 = 1.0

Hardy and Weinberg also came up with 5 conditions in which a population’s frequencies would remain constant from generation to generation. One of the conditions is that the population has to be large. Another condition is the mating has to be random. Another condition that is very important is that there isn’t any mutation of the alleles. Another condition is there isn’t any migration going on. The last condition is there can’t be any natural selection.
Using these methods, allele frequency and evolution can be measured in a population. Evolution is difficult to work with in a natural population, so the class will serve as a model to represent a population under different conditions.

Hypothesis:
The purpose of this lab is to learn how to measure allele frequencies in a population under certain conditions and observe evolution and natural selection take its role.

Materials:
Materials used in this lab involved: PTC tasting paper, calculators, and note cards.

Methods:
Exercise 8A: Everyone in the class needs to get a piece of PTC paper and put it into their mouths. If you can taste a bitter flavor, then you are considered a taster, if you can’t taste a bitter flavor, then you are considered a non-taster. In order to calculate the percent of tasters in the class, divide the number of tasters by the total number of students in the class. Calculate the percent of the non-tasters the same way and record the numbers into the table. To figure out the allele frequency of q, calculate the square root of the percent of non-tasters. To calculate the frequency of p, subtract the frequency of q from 1. Record your answers in the graph.

Exercise 8B CASE I: This part of the lab simulates a population breeding under the conditions of Hardy-Weinberg. The entire class will participate in this case. Each student needs to randomly pick a partner genotype or sex doesn’t matter. When the students have a partner, then they must all get 4 cards. Each card either has an “A” or an “a” written on it, the students need to be sure to get 2 cards with matching letters. When that’s all sorted out, the students will randomly pick a card out of their hand and lay it down. This simulates the alleles of the first offspring. Note whether or not the cards show a heterozygote, homozygous dominant, or homozygous recessive in a table. The student pair will repeat the procedure once more. Once each pair of students has drawn cards twice and noted the genotypes in a table, they must then assume the genotypes of their offspring. Say one genotype was Aa and another was AA, then one student will have all 4 cards being A, a, A, a and the other student will have all his/her cards be A, A, A, A. Once the exchanging has been done, the students then must rotate, randomly, to another partner. All in all, each student will rotate 5 times with different partners each time representing 5 generations of offspring. All the genotypes must be counted after then ends of each generation and put into a graph.

After the generations have been “born” and the genotypes recorded, frequencies can be calculated. To calculate “p”, multiply the total number of offspring with the genotype AA by 2. Then multiply the total number of offspring with the genotype Aa by 1. Add up the 2 values and divide that number by the total number of genotypes multiplied by 2, representing the total number of alleles in the population. To calculate “q”, multiply the total number of offspring with genotypes aa by 2 and add the total number of offspring with the genotypes Aa. Next, divide the sum of the 2 values by the total number of alleles in the population, in this case, the total number of genotypes multiplied by 2. There you have it!

CASE II: This case is exactly like CASE I except that the aa genotypes won’t be allowed. This simulates that the environment will favor some genotypes over others, in this case, aa. If the 2 students draw aa, then they must try again until they get a surviving offspring. Once the students have reached 5 generations, calculate the frequencies for p and q. Remember to record the genotypes after each generation.

CASE III: This case simulates sickle-cell anemia in human genomes. Just like in CASE II, the homozygous recessive, aa, alleles never survive. Individuals that are heterozygous to the disease are slightly more resistant than individuals who are homozygous dominant. With this in mind, if the pair draws up AA, then they must flip a coin to see whether or not the offspring will survive or not, heads survives, tails dies. The student pair must always try to get 2 offspring, no matter how long it takes. Remember to record the genotypes after each generation. Once everything is finished, calculate the frequencies for p and q.

CASE IV: The case is just like CASE I except after each generation is produced, there will be no random selection of new mates. The purpose of this case is to simulate isolated populations and genetic drift. Each group must no interact in any way in order to represent isolated populations. Record the genotypes after each generation and calculate the frequencies for p and q as usual.

Results:
Exercise 8A:

		Phenotypes				Allele Frequency Based on the H-W Equation
		Tasters (p2 + 2pq)		Non-tasters (q2)		P	q

Class Population		#	%	#	%	.53	.47
	7	77.78	2	22.22
North American Population		0.55		0.45		.329	.671

1. What is the percentage of heterozygous tasters (2pq) in your class? 49.82%

2. What percentage of the North American population is heterozygous for the taster trait? 44.15%

Exercise 8B CASE I:

	AA	Aa	aa
F1	1	5	2
F2	2	4	2
F3	1	6	1
F4	1	5	2
F5	1	5	2
Total	6	25	9

Frequencies:
P= .46

Q= .54

AA=.21 Aa=.50 aa=.29

1. What does the Hardy-Weinberg equation predict for the new p and q? That the frequency of AA alleles is 46% while the frequency of the aa alleles is 54%.

2. Do the results you obtained in this simulation agree? If not, why not? The results that I obtained agree because it’s normal for the heterzygotes to be large in number because they can carry on both alleles.

3. What major assumption(s) were not strictly followed in this simulation? All of the factors in the environment that could change the results.

CASE II:

	AA	Aa	aa
F1	2	6	0
F2	6	2	0
F3	5	3	0
F4	5	3	0
F5	4	4	0
Total	22	18	0

Frequencies:

P= .78

Q= .23

AA=.61 Aa=.36 aa=.05

1. How do the new frequencies of p and q compare to the initial frequencies in Case I? Well, p has gone up a lot and q has gone down a lot.

2. How has the allelic frequency of the population changed? It has gotten smaller.

3. Predict what would happen to the frequencies of p and q if you simulated another five generations. The p value would continue to increase, and the q value would decrease.

4. In a large population, would it be possible to completely eliminate a deleterious recessive allele? Explain. No because if the recessive alleles are being carried on by the heterozygous alleles, then they can’t be eliminated.

CASE III:

	AA	Aa	aa
F1	3	5	0
F2	1	7	0
F3	2	6	0
F4	2	6	0
F5	4	4	0
Total	12	28	0

Frequencies:

P= .65

Q= .35

AA=.42 Aa=.46 aa=.12

1. Explain how the changes in p and q frequencies in Case II compare with Case I and Case III. The changes in Case II astounded the results in Case I, but in Case III, the results are similar because the aa genotype is wiped out.

2. Do you think the recessive allele will be completely eliminated in either Case II or Case III? No because if the recessive alleles are being carried on by the heterozygous alleles, then they can’t be eliminated.

3. What is the importance of heterozygotes in maintaining genetic variation in populations? They contain both the dominant and the recessive allele, which make a variation right off the bat.

CASE IV:

	AA	Aa	aa
F5	7	18	15

Frequencies:

P=.40

Q=.60

AA=.16 Aa= .48 aa=.36

1. What do your results indicate about the importance of population size as an evolutionary force? When a population is very large, there is more diversity. When there is a small population with only a couple of offspring reproducing at each generation, and then eventually, there will be no diversity at all.

Error Analysis:
Not many things could have affected the results to this lab unless someone didn’t compute the correct information into their calculators.

Conclusions:
Based on this lab, I can conclude that allelic frequencies appear normal in a perfect environment like Case I. In the rest of the cases however, as we related more to the real world, changes occurred in the frequencies. In Case II the homozygous dominant alleles became the dominant genotype when the aa genotype died because as the recessive alleles became scarcer, the heterozygous alleles also suffered because they are half recessive. In Case III, the heterozygotes being more resistant to sickle-cell had an affect on the results, they were dominant. In Case IV, evidence of the populations becoming fixed is showing as the dominant AA alleles slowly disappear.

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Trait – plant height

Trait – plant height