Mendelian Genetics.

Mendelian Genetics

Mendel’s Experiments Gregor Mendel
Joined the Augustinian Monastery at the age of 21 Taught in a secondary school, was fascinated with science and nature (physics, evolution, botany and natural sciences) Attended the University of Vienna to study physics and biology Returned to the monastery and began his experiments with the common garden plant (Pisum sativum)

Mendel’s Experimental, Quantitative Approach
Advantages of pea plants for genetic study: Many varieties with distinct heritable features, or characters (such as flower color); character variants (such as purple or white flowers) are called traits Mating of plants can be controlled Each pea plant has sperm-producing organs (stamens) and egg-producing organs (carpels) Cross-pollination (fertilization between different plants) can be achieved by dusting one plant with pollen from another

TECHNIQUE Parental generation (P) Stamens Carpel RESULTS First filial
Fig. 14-2 TECHNIQUE 1 2 Parental generation (P) Stamens Carpel 3 4 RESULTS First filial generation offspring (F1) 5

Mendel’s Experiments Mendel’s Experimental Design
Crossed only peas with desired traits, two methods Self-fertilization – pollen from anther falls onto the stigma of the same flower before it opens Cross-fertilization – pollen from one plant fertilizes another Mendel opened the keel and removed the anther before self-fertilization could occur Selected 7 discrete, nonoverlapping characteristics Flower color – purple or white Nonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.

Table 14-1

Grew the plants for two years to identify homogenous, pure-breeding characteristics Started by crossing two pure-breeding plants (the parent generation, P), one purple, one white The offspring (first filial generation, F1), were referred to as hybrids (mixture of both parents) Monohybrids – hybrid of only one characteristic All the F1 plants were purple – purple flower color is a dominant trait Thus the white flower color trait is recessive Nonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.

He allowed the F1 to self-fertilize The resulting second filial generation (F2) showed both dwarf and tall characteristics 705 purple, 224 white, a ratio of 3:1 Mendel did not recognize that the traits were controlled by genes (term coined in 1909) He did propose that each trait relied on two related, but different, determinants. Alleles represent different forms of a gene Nonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.

Phenotype – observable characteristics Allele for purple flower plants (P) is dominant over white flower plants (p) Genotype - combination of alleles an organism possesses Homozygous purple– PP (Both alleles are the same) Heterozygous purple– Pp (Two alleles are different) Nonoverlapping characteristics: heights may vary so he chose the dwarf variety to compare to the tall. Tall simply refers to the nondwarf type.

EXPERIMENT P Generation (true-breeding parents) Purple flowers White
Fig EXPERIMENT P Generation (true-breeding parents)  Purple flowers White flowers F1 Generation (hybrids) All plants had purple flowers F2 Generation 705 purple-flowered plants 224 white-flowered plants

Fig. 14-5-3 P Generation Appearance: Purple flowers White flowers
Genetic makeup: PP pp Gametes: P p F1 Generation Appearance: Purple flowers Genetic makeup: Pp Gametes: 1/2 P 1/2 p Sperm Figure 14.5 Mendel’s law of segregation F2 Generation P p P PP Pp Eggs p Pp pp 3 1

Segregation Law of Segregation
During gamete formation the alleles will separate randomly Fertilization is the fusion of two gametes, reestablishing the two copies of a gene Allele for purple flowers Homologous pair of chromosomes Locus for flower-color gene Allele for white flowers

Segregation Law of segregation explains Mendel’s experiments with four related concepts: 1. Alternate versions of a gene accounts for variations in inherited characters 2. For each character, an organism inherits two alleles, one from each parent

Segregation Law of segregation explains Mendel’s experiments with four related concepts: 3. If two alleles at a locus differ, then one, the dominant allele, determines the organisms appearance, the other, the recessive allele, has no noticeable effect on the organisms appearance. 4. The two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.

How are each of the 4 parts of Mendel’s law of segregation portrayed in his experiment with crossing peas plants with different flower colors?

Segregation Mendel’s segregation model accounts for the 3:1 ratio he observed in the F2 generation of his numerous crosses The possible combinations of sperm and egg can be shown using a Punnett square, a diagram for predicting the results of a genetic cross between individuals of known genetic makeup A capital letter represents a dominant allele, and a lowercase letter represents a recessive allele

Segregation Testing the Law of Segregation
Along with the 3:1 phenotype ratio, there should also be a 1:2:1 genotypic ratio 1 PP, 2 Pp, 1 pp This can be tested by self-fertilizing the F2 to create an F3 The resulting whites should be homozygous The purple F2 plants should be 1/3 homozygous and 2/3 heterozygous The homozygous should produce only purple flower plants The heterozygous should produce 3 purple flower plants for each 1 white flower plant (3:1)

Phenotype Genotype PP Purple 1 (homozygous) 3 Purple Pp (heterozygous)
Fig. 14-6 Phenotype Genotype PP Purple 1 (homozygous) 3 Purple Pp (heterozygous) 2 Purple Pp (heterozygous) Figure 14.6 Phenotype versus genotype pp 1 White 1 (homozygous) Ratio 3:1 Ratio 1:2:1

Segregation Testing the Law of Segregation
Another way to test is by using a testcross – cross any organism with a homozygous recessive If the organism in question is homozygous dominant, then all progeny will have the dominant phenotype If the organism is heterozygous, then the progeny will be 50% phenotypically dominant, and 50% phenotypically recessive

Practicing a Test Cross
Draw punnett square crossing homozygous white flower pea plant with heterozygous purple flower pea plant. What are the possible offspring? Can you determine the genotypes of the offspring? Draw punnett square crossing homozygous white flower pea plant with homozygous purple flower pea plant.

TECHNIQUE RESULTS Dominant phenotype, unknown genotype: PP or Pp?
Fig. 14-7 TECHNIQUE  Dominant phenotype, unknown genotype: PP or Pp? Recessive phenotype, known genotype: pp Predictions If PP If Pp or Sperm Sperm p p p p P P Pp Pp Pp Pp Eggs Eggs Figure 14.7 The testcross P p Pp Pp pp pp RESULTS or All offspring purple 1/2 offspring purple and 1/2 offspring white

Independent Assortment
Mendel analyzed the inheritance of two different traits Homozygous round yellow seeds (YY) crossed with homozygous wrinkled(rr) green seeds The F1 (dihybrid) were all round yellow seeds Dihybrid-individuals that are heterozygous for two characters When the F1 was self-fertilized, the resulting F2 had all four combinations of characteristics The ratio is very close to 9:3:3:1

EXPERIMENT RESULTS Fig. 14-8 P Generation F1 Generation Hypothesis of
YYRR yyrr Gametes YR  yr F1 Generation YyRr Hypothesis of dependent assortment Hypothesis of independent assortment Predictions Sperm or Predicted offspring of F2 generation 1/4 YR 1/4 Yr 1/4 yR 1/4 yr Sperm 1/2 YR 1/2 yr 1/4 YR YYRR YYRr YyRR YyRr 1/2 YR YYRR YyRr 1/4 Yr Eggs YYRr YYrr YyRr Yyrr Eggs Figure 14.8 Do the alleles for one character assort into gametes dependently or independently of the alleles for a different character? 1/2 yr YyRr yyrr 1/4 yR YyRR YyRr yyRR yyRr 3/4 1/4 1/4 yr Phenotypic ratio 3:1 YyRr Yyrr yyRr yyrr 9/16 3/16 3/16 1/16 Phenotypic ratio 9:3:3:1 RESULTS 315 108 101 32 Phenotypic ratio approximately 9:3:3:1

Punnett squares are used to visualize possible gamete fusions Assumption: Four types of gametes from each dihybrid parent will be produced in equal numbers WG Wg wG wg WWGG WWGg WwGG WwGg WWgg Wwgg wwGG wwGg wwgg

Law of Independent Assortment Alleles for one gene can segregate independently of alleles for other genes Each phenotypic class is made of several different genotypes Except the homozygous recessive The genotypic ratio is 1:2:1:2:4:2:1:2:1 WG Wg wG wg WWGG WWGg WwGG WwGg WWgg Wwgg wwGG wwGg wwgg

Testing the Law of Independent Assortment This law can be tested by doing a dihybrid testcross Review: monohybrid heterozygous testcross resulted in a 1:1 phenotypic ratio Testcross WwGg with wwgg The result is a 1:1:1:1 phenotypic ratio

Strictly speaking, this law applies only to genes on different, nonhomologous chromosomes Genes located near each other on the same chromosome tend to be inherited together

Laws of Probability Mendel’s laws of segregation and independent assortment reflect the rules of probability

Probability Types of Probability
Probability (P) = Number of times an event is observed (a) / the total number of possible cases (n) P=a/n Examples: Probability of rolling a 4 on a six-sided dice P=1/6 Probability of drawing a 7 of clubs from a card deck P=1/52

Probability Types of Probability
An event that is certain has a probability of 1 P=1/1 An event that is impossible has a probability of 0 An event has a probability of P, the likelihood of the alternative event is Q=1-P Probability of rolling a 4 on a six-sided dice P=1/6 Probability of rolling anything else Q=1-1/6= 5/6 The probability of all possible events must equal 1 P+Q=1

Probability Types of Probability Mutually exclusive outcomes
Event in which the occurrence of one possibility excludes all other possibilities Rolling a dice, only one side can face up Independent outcomes Events that do not influence one another Rolling two dice, the face value of one does not influence the other Two rules of probability that affect genetics Sum rule and product rule

Probability Sum Rule When events are mutually exclusive
The probability that one of several mutually exclusive events will occur is the sum of the probabilities Probability of a dice showing either a 4 or 6 P=1/6 + 1/6 = 2/6 = 1/3 The probability increases as the number of possible outcomes increase Probability of a dice showing any number P=1/1 Not used for traits expressed on a continuum (human heights)

Probability Product Rule When one event is independent of other events
The probability that two events will both occur is the product of their separate probabilities Probability of throwing a die two times and getting a 4 and then a 6 P=1/6 x 1/6 = 1/32 The probability will decrease as you increase the number of independent events Much like the lottery, the more numbers you need the less likely you are to win

Probability Using Probabilities
In a monohybrid cross, Dd X Dd (1:2:1), what is the probability of getting either a homozygous dominant or heterozygous? P= ¼ + ½ = ¾ In a dihybrid cross WwGg X WwGg, what is the probability of getting a round and green seed? P=3/4 x 1/4 = 3/16

Solving Complex Genetics Problems
We can apply the rules of probability To predict the outcome of crosses involving multiple characters A dihybrid or other multicharacter cross Is equivalent to two or more independent monohybrid crosses occurring simultaneously In calculating the chances for various genotypes from such crosses Each character first is considered separately and then the individual probabilities are multiplied together

Probability Branch-Line Approach to Calculate Probabilities
Punnett squares require 16 squares for a dihybrid cross, 64 squares for a trihybrid Punnett squares are useful, but hard to use with more complex crosses The branch-line approach is based on the law of independent assortment In branch-line each trait is examined independently

AaBb x AaBb To determine the probability of this dihybdrid cross calculate the probability of each trait and apply the product rule A phenotype 3/4 (either AA or Aa) B phenotype ¾ (either BB or Bb) AB phenotype 9/16 (A-B- genotype) b phenotype 1/4 (bb genotype) Ab phenotype 3/16 (A-bb genotype) Here you can see the 9:3:3:1 ratio a phenotype 1/4 (aa genotype) B phenotype ¾ (either BB or Bb) aB phenotype 3/16 (aaB- genotype) b phenotype 1/4 (bb genotype) ab phenotype 1/16 (aabb genotype)

Use branch-line to determine the probabilities of this trihybrid cross: AaBbCc x AabbCc A phenotype 3/4 a phenotype 1/4 B phenotype 1/2 b phenotype 1/2 C phenotype 3/4 c phenotype 1/4 ABC phenotype 9/32 ABc phenotype 3/32 AbC phenotype 9/32 Abc phenotype 3/32 aBC phenotype 3/32 aBc phenotype 1/32 abC phenotype 3/32 abc phenotype 1/32 The key is to look at each cross and what the probabilities are going to be. Aa x Aa has ¾ probability to express the A allele and ¼ the a allele. Then look at the next cross, Bb x bb. This yields a ½ B and ½ b phenotype expression. Do the same with the Cc cross.

The branch-line approach can also be used to determine genotypes The key is to look at each cross and what the probabilities are going to be. Aa x Aa has ¾ probability to express the A allele and ¼ the a allele. Then look at the next cross, Bb x bb. This yields a ½ B and ½ b phenotype expression. Do the same with the Cc cross.

Statistics When dealing with probabilistic events, there is a chance the data will cause us to support a bad hypothesis Mendel’s F2 heterozygous pea plants yielded 788 tall and 277 dwarf. 2.84:1 not exactly 3:1 Is 2.84:1 close enough to represent 3:1?

Statistics Hypothesis Testing
Statistics are used by scientists to summarize data and test their hypothesis by comparing data predicted results To determine if the data is consistent with the hypothesis we generate a null hypothesis The null hypothesis assumes the difference between the observed and expected results are due to chance

Statistics Chi-Square (O – E)2 χ2=∑ E
This test is used when data is distributed among discrete categories (tall and dwarf plants) Formula for the chi-square χ is the Greek letter chi, O is the observed number for a category, E is the expected number for the category, and ∑ means to sum the calculations for all categories χ2=∑ (O – E)2 E

Statistics Chi-Square (787 – 798)2 = 0.15 798 (277 – 266)2 = 0.45 266
Example: Mendel observed 787 tall plants and 277 dwarf. Total of The expected results for a 3:1 ratio are 798 tall and 266 dwarf. Start with figuring the chi-square for tall plants Then figure it for the dwarf plants Now sum the two results (787 – 798)2 798 = 0.15 (277 – 266)2 266 = 0.45 = 0.60

Statistics Chi-Square (787 – 532)2 = 122.23 532 (277 – 532)2 = 122.23
Example: Use the same data, but assume we were testing for a 1:1 ratio. How does the chi-square change? Mendel observed 787 tall plants and 277 dwarf. Total of 1064. Start with figuring the chi-square for tall plants Then figure it for the dwarf plants Now sum the two results (787 – 532)2 532 = (277 – 532)2 532 = =

Statistics Chi-Square
To properly use chi-square values we need to convert them to a probability value (p) As the number of categories increases so will the chi-square value (because we sum each category) To help solve this we use degrees of freedom (# of categories minus 1) Example: two categories would have 1 degree 0f freedom Consult the chi-square table

Statistics Chi-Square Table Probabilities 0.99 0.95 0.80 0.50 0.20
Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01 1 0.000 0.004 0.064 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3 0.115 0.352 1.005 2.366 4.642 7.815 11.345 4 0.297 0.711 1.649 3.357 5.989 9.488 13.277 5 0.554 1.145 2.343 4.351 7.289 11.070 15.086 6 0.872 1.635 3.070 5.348 8.558 12.592 16.812

Statistics Chi-Square Table We will look a p value of 0.05
We have only 1 degree of freedom Probabilities Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01 1 0.000 0.004 0.064 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3 0.115 0.352 1.005 2.366 4.642 7.815 11.345 4 0.297 0.711 1.649 3.357 5.989 9.488 13.277 5 0.554 1.145 2.343 4.351 7.289 11.070 15.086 6 0.872 1.635 3.070 5.348 8.558 12.592 16.812

Statistics Chi-Square Table
There is a 0.05 probability of getting a χ2 value of or larger by chance alone, given the hypothesis is correct Probabilities Degrees of freedom 0.99 0.95 0.80 0.50 0.20 0.05 0.01 1 0.000 0.004 0.064 0.455 1.642 3.841 6.635 2 0.020 0.103 0.446 1.386 3.219 5.991 9.210 3 0.115 0.352 1.005 2.366 4.642 7.815 11.345 4 0.297 0.711 1.649 3.357 5.989 9.488 13.277 5 0.554 1.145 2.343 4.351 7.289 11.070 15.086 6 0.872 1.635 3.070 5.348 8.558 12.592 16.812

Inheritance Patterns Inheritance patterns are often more complex than predicted by simple Mendelian genetics The relationship between genotype and phenotype is rarely simple

Extending Mendelian Genetics for a Single Gene
The inheritance of characters by a single gene may deviate from simple Mendelian patterns The Spectrum of Dominance Complete dominance Occurs when the phenotypes of the heterozygote and dominant homozygote are identical Codominance Two dominant alleles affect the phenotype in separate, distinguishable ways Ex. The human blood group MN is an example of codominance

The Spectrum of Dominance
Incomplete dominance The phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties P Generation F1 Generation F2 Generation Red CRCR Gametes CR CW  White CWCW Pink CRCW Sperm Cw 1⁄2 Eggs CR CR CR CW CW CW

The Relation Between Dominance and Phenotype
Dominant and recessive alleles do not really interact; it is in the pathway from genotype to phenotype that dominance and recessiveness come into play Lead to synthesis of different proteins that produce a phenotype Examples: Mendel’s Pea Shape, Tay-Sachs disease Dominant alleles are not always the most common in a population Example: polydactyly (+5 digits)

Multiple Alleles Most genes exist in populations
Table 14.2 Most genes exist in populations In more than two allelic forms The ABO blood group in humans Is determined by multiple alleles

Blood Types Glycoproteins on surface determine blood type; Important in transfusions/transplants IA and IB are codominant ii (type O) is recessive to A or B Type O = universal donor Type AB= universal recipient Differences in Rh factor (Mom Rh- and baby Rh+) can result in erythroblastosis fetalis

Pleiotropy Pleiotropy-a gene has multiple phenotypic effects
Ex. Hereditary diseases such as cystic fibrosis & sickle cell disease

Extending Mendelian Genetics for Two or More Genes
Some traits may be determined by two or more genes epistasis- a gene at one locus alters the phenotypic expression of a gene at a second locus polygenic inheritance- an additive effect of two or more genes on a single phenotype

EPISTASIS EX: Coat color in mice B = Black b = brown
C = color deposited in coat c = color NOT deposited cc-mouse looks white even though it has color genes Image from Biology; Campbell and Reece; Pearson Prentice Hall publishing as Benjamin Cummings © 2005

An example of epistasis
BC bC Bc bc 1⁄4 BBCc BbCc BBcc Bbcc bbcc bbCc BbCC bbCC BBCC 9⁄16 3⁄16 4⁄16  Sperm Eggs EX: Coat color in mice B = Black b = brown C = color deposited in coat c = color NOT deposited cc-mouse looks white even though it has color genes

Polygenic Inheritance
Many human characters Vary in the population along a continuum and are called quantitative characters Quantitative variation usually indicates polygenic inheritance An additive effect of two or more genes on a single phenotype

POLYGENIC traits are recognizable by their expression as a gradation of small differences (a continuous variation). The results form a bell shaped curve.

Nature and Nurture: The Environmental Impact on Phenotype
Another departure from simple Mendelian genetics arises When the phenotype for a character depends on environment as well as on genotype The norm of reaction Is the phenotypic range of a particular genotype that is influenced by the environment

Environment influences Phenotype “Nature vs Nurture”
Siamese cats and Himalayan rabbits have dark colored fur on their extremities Allele that controls pigment production is only able to function at the lower temperatures of those extremities.

Integrating a Mendelian View of Heredity and Variation
Multifactorial characters Are those that are influenced by both genetic and environmental factors An organism’s phenotype Includes its physical appearance, internal anatomy, physiology, and behavior Reflects its overall genotype and unique environmental history Even in more complex inheritance patterns Mendel’s fundamental laws of segregation and independent assortment still apply

Sex Linked Genes Genes carried on the X chromosome are called X-linked traits. Red-green colorblindness, hemophilia, an Duchenne muscular dystropy are examples of X-linked traits.

Hairy pinnae SRY gene initiates male sex determination Y-LINKED GENES:
Genes carried on the Y chromosome Y-linked genes only show up in MALES SRY gene initiates male sex determination Hairy pinnae

NON-HOMOLOGOUS partners
X and y chromosomes NON-HOMOLOGOUS partners

Pedigrees are diagrams that show how genes are passed on in families over several generations
Pedigrees can be used to predict future offspring in families with genetic disorders

Drawing a pedigree chart

recessive to the normal working allele.
A mutation in an allele that causes a protein to be NON-FUNCTIONAL would appear recessive to the normal working allele. Examples of autosomal recessive GENETIC DISORDERS: Phenylketonuria (PKU) Tay-Sachs Disease Cystic Fibrosis

Phenylketonuria (PKU)
CAUSE: Mutation in gene for an enzyme the breaks down an amino acid called phenylalanine Build up causes mental retardation

Phenylketonuria (PKU)
ALL babies are tested for PKU before they leave the hospital. Treatment: Need a diet low in phenylalanine to extend life and prevent mental retardation

CYSTIC FIBROSIS thick mucous CAUSE:
Loss of 3 DNA bases in a gene for the ion channel protein that transports Cl- ions Salt balance is upset Causes a build up of thick mucous in lungs and digestive organs, Leads to: Respiratory and digestive complications, increased susceptibility to infections thick mucous Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006

Carrier Heterozygous individual
That carries one recessive allele for a genetic disorder Doesn’t show the disorder themselves, but can pass it on to offspring

TAY-SACHS DISEASE Autosomal Recessive
CAUSE: Mutation in gene for an enzyme the breaks down a kind of lipid in the developing brain As these lipids build up in brain infant suffers seizures, blindness, loss of motor & mental function > > > leads to early death. Found more frequently in people with Jewish, Mediterranean, or Middle Eastern ancestry

Disorders caused by autosomal codominant alleles: SICKLE CELL DISEASE
CAUSE: A changed to T in gene for Hemoglobin (protein in red blood cells that carries oxygen in blood)

SICKLE CELL DISEASE SYMPTOMS:
Red blood cells become sickle shaped under low oxygen condition in persons with two sickle cell alleles (ss) Ss=Sickle cell trait Normally healthy, but can suffer some sickle cell episodes

Cells stick in capillaries Loss of blood cells (anemia)
SICKLE CELL DISEASE Circulatory problems Cells stick in capillaries Loss of blood cells (anemia) Organ damage (brain, heart, spleen) Can lead to DEATH

HUNTINGTON’S DISEASE is AUTOSOMAL DOMINANT
CAUSE: Extra CAG repeats at end of gene on chromosome 4 The more repeats the more severe the symptoms.

HUNTINGTON’S DISEASE Begins in middle age
Huntington’s brain Begins in middle age Causes progressive loss of muscle control and mental function 1 in 10,000 people in U.S. have Huntington’s disease Normal brain

A person with Huntington’s disease has a _____ chance of passing the disorder on to their offspring. 50% Problem: Symptoms of disorder usually don’t show until ____________ . . . so you don’t know you have it until ________ you have had children. MIDDLE AGE AFTER

ACHONDROPLASIA (One kind of Dwarfism)
CAUSE: Autosomal Dominant gene 1 in 25,000 births DD = lethal Dd = dwarf phenotype dd= = normal height 200,000 “little people” worldwide One of oldest known disorders – seen in Egyptian art

ACHONDROPLASIA (One kind of Dwarfism)
Normal size head and torso; short arms and legs Problem with way cartilage changes to bone as bones grow

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