1. Chapter 14: Mendel and the Gene Idea
14.1 Mendel used the scientific approach to identify two laws of inheritance
14.2   The laws of probability govern Mendelian inheritance
14.3 Extensions to Mendelian Genetics: Inheritance patterns are often more complex than predicted by simple Mendelian genetics
14.4   Many human traits follow Mendelian patterns of inheritance

2. Before we begin: Define these terms:
Gene: functional unit of heredity - a segment of DNA located in a specific site on a chromosome that gives the instrcutions to make one (or more) enzyme or other protein.
Dive into the chromosome here! !
Phenotype: Observable characteristics of an organism, for example, hair type, eye color, height.
Genotype: The genetic identity of an individual, including alleles that do not show as outward characteristics.
Locus: The position on a chromosome where a gene, or some other sequence, is located.
Allele: One of two or more alternative forms of a gene; for example, allele "A" may produce a tall plant, while allele "a" gives a short plant.
Dominant allele: An allele that is fully expressed in the phenotype, regardless of the characteristics of its counterpart allele.
Recessive allele: An allele which is not expressed in the phenotype unless two copies are present in the individual, ie, in homozygous recessive condition.
Homozygous: the state of having two identical alleles of a particular gene (eg AA, aa).
Heterozygous: the state of having two different alleles of a particular gene (eg Aa).
Mendel’s Law of Segregation: Homologous chromosomes separate during the formation of gametes, and are distributed to different gametes, so that every gamete receives only one member of the pair.
Mendel’s Law of Independent Assortment: Homologous chromosomes are random distributed at the metaphase plate such that a mixture of maternal and paternal homologous assort to each gamete.

3. Figure 14. 3 and 14.5 Mendel’s Peas
1. Parental Generation (P): Purple flowers x white flowers: Mendel bred (cross-pollinated) two types of pea plants that were true-breeding for flower color.
2. F1 Generation: When Mendel grew seed from this First Filial generation (F1) and looked at their flower color, he saw 100% purple flowers. Result: In the (F1) generation, the white trait was masked.
3. F2 Generation Mendel went one step farther, allowing the F1 to “self”. Result: The white trait re-appeared in the F2 generation in a ratio of ~3 purple plants to 1 white.
Next Slide: Mendel’s Genius: He realized that these results were explainable if three things were true.

4. Figure 14.4: Mendel’s Genius: He realized that these results were explainable if three things were true. He hypothesized that:
1. Every trait (like flower color) is controlled by two
"heritable factors” (genes), one from each parent
-Each form of the gene is called an allele, and
produces a slightly different protein
2. If the two alleles differ, one is dominant
(observed in appearance or physiology) and
one is recessive (only observed when
an individual has two copies of the recessive allele).
Dominant traits mask the appearance of recessive traits.
3. Alleles are randomly segregated into eggs and sperm
when the homologues separate during Meiosis I, allowing
all possible combinations of factors to occur in the gametes.
Mendel's Law of Segregation - When gametes (eggs or sperm) are formed, the two factors (alleles) separate, and only one factor (allele) is present in each gamete. [ie: the homologues separate during meiosis]
Alleles controlling
Flower color:
Purple
White
Pea shape:
Wrinkled
Round
Homologous Chromosomes

5. Table 14.1 The Results of Mendel’s F1 Crosses for Seven Characters in Pea Plants
Result: For a monohybrid cross (single trait), Mendelian Genetics predicts that a 3:1 ratio (Dominant : recessive) will be seen in the F2 generation

6. Figure 14.6 Phenotype versus genotype3 1 2
Phenotype (Observable characteristics)
Purple
White
Genotype (Gene alleles) PP
(homozygous) Pp
(heterozygous) pp
Ratio 3:1
Ratio 1:2:1

7. Doing a genetic cross (Monohybrid Cross = 1 gene):
(Most of you have done this in lab already)
Geneticists use letters to represent alleles.
*Dominant trait = Capital letter ; recessive trait = a lowercase letter.
*The same letter is used to indicate both alleles.
*Remember, the definition of “Dominant” means…”SEEN IN THE PHENOTYPE”
Examples
= Flower color: P= purple, p= white
= Seed color: Y= yellow, y = green
= Seed shape: W = wrinkled, w = round
In Humans...
= Widow's peak: W = widow's peak, w = continuous hairline (which are you?)
= Freckles: F = Freckles, f = no freckles (which are you?)
= Earlobes: E = unattached, e = attached (which are you?)
= Cystic fibrosis C = no CF, c = cystic fibrosis
E-Z steps for doing genetics problems:
1. Indicate the genotype of the parents using letters
2. Determine what the possible gametes are
3. Determine the genotype and phenotype of the children after reproduction. To consider every type of offspring possible, use a Punnett Square in which all possible types of sperm are lined up vertically and all types of eggs are lined up horizontally:
4. Fill in the squares by "multiplying" the alleles from mom and dad:

8. Human Genetics: Recessively Inherited Disorders
1. Autosomal recessive human disorders: Show up only in the homozygous recessive person (aa). With two carrier parents, there is a 1/4 chance of an affected child. There is 2/4 chance of a carrier child (50%)
A.Cystic fibrosis: Mutation in CFTR (Chloride channel)
B.Tay-Sachs: Mutation in Hexaminidase A gene. (Jewish - Ashkenazi)
C. Sickle-cell disease: Mutation in
hemoglobin gene. The most common
inherited disease of African-Americans
(1:400 affected).
With these parents, what is the chance of
- giving birth to an affected child?
- giving birth to a child who is also a carrier?
Heterozygote advantage; For some
disease alleles, being a heterozygote
(‘carrier’) offers protection against another disease.
Examples:
1. Sickle cell disease: Ss carriers are resistant
to malaria
2. Cystic fibrosis: Cc carriers are resistant to cholera
unaffected
carrier mother

9. Frequency of Hbs allele
Heterozygote advantage
One in 10 Africans are carriers (heterozygotes) for the sickle cell trait, even though the recessive condition, SCD, can be fatal. Why is the incidence so high?
Sadly, over 1 million people / year
(mostly children, pregnant women)
die from Malaria (2,700 per day)….
People who are
HbBHbB = normal hemoglobin, but may succumb to malaria if they live in a region with endemic malaria.
HbSHbS = have sickle cell disease SCD and are anemic, in pain, and may suffer strokes or death.
HbBHbS = heterozygotes suffer mild symptoms of SCD, but if bitten by a mosquito carrying the malaria parasite, the parasite life cycle is interrupted due to the difficult-to-digest, irregular HbS
In parts of the world where malaria
is common, the incidence of SCD is common because there is an advantage to being a carrier.
P. falciparum malaria
Frequency of Hbs allele
1–10%
10–20%

10. Human Genetics: Dominantly Inherited Disorders
Autosomal dominant human disorders: child will show the phenotype if he / she receives just 1 allele from either parent (OR if a spontaneous mutation appears in the zygote).
A. Achondroplasia (dwarfism): Cause: Mutation in FGFR3 (fibroblast growth factor receptor 3 gene) severely limits bone growth. AA = Dominant lethal - fatal. Aa = dwarfism. aa = no dwarfism % of all people in the world are (aa). (Fig 14.15)
B. Polydactyly (extra fingers or toes): PP or Pp = extra digits, aa = 5 digits. 98% of all people in the world are (pp).
C. Huntington's chorea Cause: Mutation of Huntingtin gene damages nerves and brain. Dominant lethal (HH = fatal). Heterozygotes (Hh) live to be ~40 or so, then their nervous system starts to degenerate. (Woody Guthrie).
D. Marfan's syndrome - Cause: Defect in Fibrillin gene involved
in connective tissue; affects skeletal system, eyes, and CV
system. (Pleiotropic). *Abraham Lincoln, Flo Hyman
With these parents, what is the chance of
- giving birth to an affected child?
- giving birth to a child who is also a carrier?
E. Progeria - Cause: Mutation of Lamin-A Gene. Defective Lamin-A makes the nucleus unstable, leading to aging, CV disease, stroke (Note:
since children with Progeria don’t live to reproduce, this is
an example of a spontaneous (rather than inherited), Dominant disorder.

11. Fig 14.7: Is there a way to distinguish between a purple
heterozygote (Pp) and a purple homozygous dominant (PP)?
Yes = A testcross:
Mate the unknown genotype
(Purple, but PP or Pp?)
with a recessive genotype:
white (pp) in this case.
If offspring are 100% purple:
the unknown genotype was
homozygous dominant (PP),
to begin with; and
100% of the offspring will be
heterozygotes.
If offspring are 50% purple
And 50% white:
a heterozygote (Pp),
To begin with.
So now you know!

12. AA or Aa = Purple; aa = white BB or Bb = Tall; bb = short
Fig 14.8: Studying Two Traits at once: (Dihybrid Cross)
AA or Aa = Purple; aa = white BB or Bb = Tall; bb = short
Result: Classic 9:3:3:1 ratio: 9 Purple & Tall, 3 Purple & short, 3 white & Tall, 1 white & short
One really important thing that Mendel noticed from this type of cross was that the two traits (like flower color, height) are inherited independently - not together as a unit. This has become known as Mendel's Law of Independent Assortment - Genes for various traits assort into gametes independently (due to homologues lining up randomly at the metaphase plate).

13. (See Fig 14.9) A simpler way to do a Dihybrid Cross or beyond...
(or, how to avoid making anything larger than a simple Punnett square)
Mendelian genetics reflect the laws of probability
For a dihybrid cross - the chance that 2 independent events will occur together is the product of their chances of occurring separately.
The chance of yellow (YY or Yy) seeds = 3/4 (the dominant trait)
The chance of round (RR or Rr) seeds  = 3/4 (the dominant trait)
The chance of green (yy) seeds = 1/4 (the recessive trait)
The chance of wrinkled (rr) seeds = 1/4 (the recessive trait)
So... (Remember multiplication with fractions..its easy! Just multiply across the top and the bottom! )
The chance of yellow and round = 3/4 x 3/4 = 9/16
The chance of yellow and wrinkled = 3/4 x 1/4 = 3/16
The chance of green and round = 1/4 x 3/4 = 3/16
The chance of green and wrinkled = 1/4 x 1/4 = 1/16
9:3:3:1 - Sound familiar? And no Punnett square needed (whew). Therefore, you can avoid doing a Punnett square if you can reduce the problem to a series of probability statements.
Next, you guessed it, a tri-hybrid cross

14. Figure 14.10 Incomplete dominance in snapdragons
Incomplete Dominance: 2 incompletely dominant alleles (no ‘recessives’) CR and CW
The F1 heterozygote (CRCW) is
intermediate in phenotype (pink!)
compared to the two parentals.
Why does this happen?
Just like with Purple:white, this comes about when one of the alleles lacks the
ability to produce a pigment, but in this case, the loss IS detectable in the phenotype.
Note that traits do not ‘blend’: all 4 colors result in the F2.
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

15. Table 14.2 Determination of ABO Blood Group by Multiple Alleles (Co-dominance)
ABO Blood Types 3 alleles;
(Each dominant allele codes for the linkage of a glycophorin sugar molecule to the surface of RBCs.)
2 are dominant (IA and IB ),
1 is recessive (i)
This results in 4 possible phenotypes Which do YOU have?

16. Figure 14.12 Polygenic inheritance: skin color
2 or more genes act additively to produce
a phenotype
Ex: human height,
skin color 
AaBbCc
aabbcc
Aabbcc
AaBbcc
AABbCc
AABBCc
AABBCC
20⁄64
15⁄64
6⁄64
1⁄64
Fraction of progeny
Skin pigmentation

17. Pleiotropy: (Greek: pleion, more)
How 1 gene can affect multiple aspects of the phenotype. Ex: Sickle Cell Disease, Cystic Fibrosis,
Marfan’s Disease
Many organ systems affected, since each mutation changes a protein involved in many bodily functions

18. 9 Black : (BBEE, BbEE, BbEe)
Fig Epistasis: Presence of alleles at one locus affect the expression of alleles at a second locus. (Epi = above)
Other examples: Albinism in humans [In class] Can use probability to solve these; Variations on a 9:3:3:1 result
(9:3:4; 9:7, 15:1) Make sure you can work an epistasis genetics problem! Bonus: What genotype was the yellow lab BB, Bb, or bb?
The nose knows!
4 Yellow :
(BBee, Bbee, bbee)
3 Chocolate
(bbEE, bbEe)
9 Black : (BBEE, BbEE, BbEe)

19. Fig 15-10: Sex-linked disorders in humans: X Chromosome:
Females: Inherit 2 X chromosomes, one from each parent - 2 alleles for all genes (Fig 15.9)
Males: Inherit one X chromosome (from their mother), so only 1 allele for every gene, Technically, males are hemi-zygous - neither homozygous nor heterozygous at the X!
If that one allele is mutated, the male will be affected (show the trait).
Daughters are almost always unaffected (carrier females) but have a 50% chance of giving birth to an affected son and a 50% chance of giving birth to a carrier daughter.
Affected females are more rare because they must inherit two recessive alleles.
List the 5 possible genotypes and phenotypes:
Color blindness: sex-linked recessive
Defect in opsin gene;
XA = normal vision, Xa = color blind
Duchenne muscular dystrophy: Affects 1:3500 US. Males. Children lack Dystrophin (X chromosome).
Progressive muscle weakness and loss of coordination
leads to death in early adulthood.
Hemophilia: Affected children
lack the blood clotting factor VIII. Queen Victoria
was a carrier; thus her sons had a 50% chance of
being affected and her daughters had a 50%
chance of being carriers.

20. (randomly inactivated chromosome)
Fig 15.11: Dosage compensation in females: X-Chromosome Inactivation (XCI)
Females normally inactivate one of their 2 Xs
in early embryonic development (XCI). The inactivated X-chromosome condenses into a Barr Body
This inactivation is a random event, thus females
(this means all K101 females) are Mosaics
50% of our cells have one active X, and
50% have the other active X.
Female calico kitty:
mosaic!
Dosage Compensation = inactivation
of 1 X chromosome per cell that
equalizes the expression of X-linked genes in males and females 1
Nucleus 2 2
Barr body
(randomly inactivated chromosome)
Followed by mitosis 3
50% of cells % of cells

21. Just a mention about Multifactorial Diseases: (and see Fig 14.13)
Cancer, Autism and Schizophrenia are multi-factorial - they are not 'caused' by one dominant or recessive gene, but may be the result of several inherited genetic susceptibilities PLUS environmental factors.
Example: Cancer: Can be an inherited susceptibility - loss of tumor suppressors (BRCA1, BRCA2, p53), oncogene activation, OR a spontaneous mutation PLUS environmental factors .
Example: Autism: at least 4 genes linked to this neurodevelopmental condition, particularly on Chromosomes 2, 3, 7, 15, and X. Some have to do with language development, some with brain development, some with producing brain neurotranmitters.
Example: Schizophrenia: several genes linked to the development of this disorder, including Chromosome 22 plus a combination of family genetics and environmental factors.

22. Four haploid cells produced
Two homologous
chromosomes
undergo synapsis
in meiosis
Fig 15.5: “Linked Genes”
Mendel showed that genes are inherited independently, but in 1910, Thomas Hunt Morgan (Nobel Prize 1933) and Alfred Sturtevant showed that Independent assortment does not apply if 2 genes are very close together “Linked” on a chromosome. Crossing Over during meiosis results in recombination between genes. The closer two genes are on a chromosome, the more likely that they are linked and will travel together. The farther apart they are, the more likely they will be separated.
Can use this information to create maps of gene order on chromosomes (See Fig for more info)
Crossing-over
between a pair of
homologous
chromatids
Meiosis I
Meiosis II
Parental
Recombinant
Recombinant
Parental
Four haploid cells produced

23. BV bv Bv bV bv BbVv bbvv Bbvv bbVv 575 575 575 575 965 944 206 185
Making a physical map of a chromosome based on recombination frequency (crossing over): Morgan and Sturtevant constructed many “two-point” testcrosses (see Fig 14-7) to get a estimate of how closely linked genes were on the Drosophila chromosomes.
Recombination frequency: Frec (or % crossing over))
391 recombinants X 100= 17%
2300 offspring
Thus, we can say that these genes are 17 “Centimorgans” or “map units” apart (cM or m.u.)
Try it: Problem 1
DdEe x ddee
65 Ddee, 25 ddEe, 256 DdEe, 266 ddee
Problem 2:
AaEe x aaee
424 Aaee, 430 aaEe, 70 AaEe, 76 aaee
Make a map of these 3 genes,
assuming E is in the middle
Grey,
normal wings
BbVv BV bv Bv bV
Grey,
normal
Black,
vestigial
Grey,
vestigial
Black,
normal bv
Black,
vestigial wings
bbvv
BbVv
bbvv
Bbvv
bbVv
Expected results:
independent assortment
575
575
575
575
Parentals Recombinants
965
944
206
185
Actual results

24. BV bv Bv bV bv BbVv bbvv Bbvv bbVv 575 575 575 575 965 944 206 185
Making a physical map of a chromosome based on recombination frequency (crossing over): Morgan and Sturtevant constructed many “two-point” testcrosses (see Fig 14-7) to get a estimate of how closely linked genes were on the Drosophila chromosomes.
Recombination frequency: Frec (or % crossing over))
391 recombinants X 100= 17%
2300 offspring
Thus, we can say that these genes are 17 “Centimorgans” or “map units” apart (cM or m.u.)
Try it: Problem 1
DdEe x ddee
65 Ddee, 25 ddEe, 256 DdEe, 266 ddee
1. Circle the parentals
2. Add the recombinants = = 90
3. Add all the offspring = 612 total
4. 90 / 612 X 100 = D-E are 14.7 m.u. apart
Problem 2:
AaEe x aaee
424 Aaee, 430 aaEe, 70 AaEe, 76 aaee
2. Add the recombinants = = 854
3. Add all the offspring = total
/ X 100 = 85.3
E-A are 85.3 m.u. apart
Make a map of these 3 genes, assuming E is in the middle
D E A
Grey,
normal wings
BbVv BV bv Bv bV
Grey,
normal
Black,
vestigial
Grey,
vestigial
Black,
normal bv
Black,
vestigial wings
bbvv
BbVv
bbvv
Bbvv
bbVv
Expected results:
independent assortment
575
575
575
575
Parentals Recombinants
965
944
206
185
Actual results

25. Hardy-Weinberg ‘Equlilbrium’ The Hardy-Weinberg equation : Basic assumptions 1. Mutation is not occurring 2. Natural selection is not occurring 3. The population is infinitely large 4. All members of the population breed 5. All mating is totally random 6. Everyone produces the same number of offspring 7. There is no migration in or out of the population (This is never fully true for any population, but If you know the allele frequency, Can determine the genotype frequencies Allele Frequency: Dom + rec = 1 (p + q = 1) Genotype Frequency: p2 + 2pq + q2 = 1

26. Hardy Weinberg: Basic Relations 1. Two alleles at a gene - A and a 2
Hardy Weinberg: Basic Relations 1. Two alleles at a gene - A and a 2. Frequency of the A allele = p 3. Frequency of the a allele = q 4. Allele frequencies: p + q = 1 5. Genotype frequencies: p2 (AA) + 2pq (Aa) + q2 (aa) = 1 Problem 1. Allele B is present in a large, random mating population at a frequency of 54 % 1. What is the frequency of allele b? ______________ 2. What proportion of the population is expected to be homozygous for B (BB)? __________ ______ homozygous for b (bb)? __________ heterozygous? (Bb) ______________ Problem 2. In a large, random-mating population, the frequency of homozygous recessive (aa) individuals for sickle cell disease is 90 per 1000 individuals. What is the frequency of the recessive disease aa? ______________ 2. the recessive allele a? ______________ 3. the dominant allele A? ______________ 4. the heterozygous individual Aa? ______________ 5. the homozygous dominant individual AA? ______________ p q p2 pq q2

27. Hardy Weinberg: Basic Relations 1. Two alleles at a gene - A and a 2
Hardy Weinberg: Basic Relations 1. Two alleles at a gene - A and a 2. Frequency of the A allele = p 3. Frequency of the a allele = q 4. Allele frequencies: p + q = 1 5. Genotype frequencies: p2 (AA) + 2pq (Aa) + q2 (aa) = 1 Problem 1. Allele B is present in a large random mating population at a frequency of 54% p 1. What is the frequency of allele b? q _0.46________ 2. What proportion of the population is expected to be -homozygous for B (BB)? p2___0.291_____ -homozygous for b (bb)? q2___0.216_____ -heterozygous? (Bb) 2pq_ 2(.54)(.46) = 0.495__ Problem 2. In a large, random-mating population, the frequency of homozygous recessive (aa) individuals for sickle cell disease is 90 per 1000 individuals. 1. What is the frequency of the recessive disease aa? q290/1000 = What is the frequency of the recessive allele a? q sq.rt (0.09) = What is the frequency of the dominant allele A? p = What is the frequency of the heterozygous individual Aa? 2pq2(.3)(.7) = What is the frequency of the homozygous dominant individual AA? p2 = 0.49 p q p2 pq q2

28. The DNA sequence of the Human X chromosome
(Not on exam): EXCITING and SURPRIZING new results in Nature, March 17, 2005 about the Sequence of the X-chromosome and X-chromosome inactivation.
Nature 434, (17 March 2005); doi: /nature03479
X-inactivation profile reveals extensive variability in X-linked gene expression in females
LAURA CARREL AND HUNTINGTON F. WILLARD
In human females, up to 15% of all genes on the ‘inactive’ X escape XCI, and are expressed from both the active and the inactive X. An additional 10% of X-linked genes show variable patterns of inactivation and are expressed to different extents in different women.
Nature 434, (17 March 2005); doi: /nature03440
The DNA sequence of the Human X chromosome
The International Human Genome Sequence Consortium
We have determined 99.3% of the euchromatic sequence of the human
X-chromosome. We found 1,098 genes in this sequence, of which 99 encode proteins expressed in the testis.
Nature 434, (17 March 2005); doi: /434279a
Genome biology: She moves in mysterious ways
CHRIS GUNTER
The human X chromosome is a study in contradictions. The detailed sequence of the X, and a survey of inactivated genes in females, help to illuminate this unique 'evolutionary space'…

29. Objectives:
Define the terms phenotype, genotype, locus, allele, dominant allele, recessive allele, homozygous, and heterozygous.
Describe Mendel’s Laws (principles) of segregation and independent assortment.
Solve genetics problems involving monohybrid, dihybrid, (trihybrid) and test crosses.
Use probability (3/4 x 3/4…) to predict the outcomes of genetic crosses.
Be able to discuss the difference between autosomal dominant and autosomal recessive disorders, being able to give specific examples and work genetics problems.
Solve genetics problems involving incomplete dominance, codominance / multiple alleles, and epistasis;
Pleiotropy: discuss how a single gene can affect many features of the organism.
Discuss the the inheritance of X-linked genes in mammals.
Discuss what it means to be hemizygous, and what is means to be a mosaic. What is a Barr Body, and what is dosage compensation?
Define linkage, and relate it to specific events in meiosis.
Show how data from a test cross involving alleles of two loci (‘two point test cross’) can be used to distinguish between independent assortment and linkage.
Know how to calculate allele frequency using the Hardy-Weinberg equation.
Explain how prenatal testing, carrier screening, and newborn screening can be used to detect genetic conditions before and after birth.