1. Transmission (Classical, Mendelian) Genetics Ch 11
Gregor Mendel
Experiments in Plant Hybridization, 1865
Simple, controlled, data collection, mathematical analysis

2. Pisum sativum, the garden pea
What makes this a good model organism?
easy to grow
hundreds of offspring per cross
short generation time
can self fertilize or cross
Paint pollen (sperm) from one plant onto the female parts of another (emasculated)

3. Mendel’s conclusions Unit factors in pairs - Genes are physical units
2 alleles for each gene
1 allele inherited from each parent

4. Genes and alleles of Pisum sativum
Gene Alleles
Pea color ?
Flower color white, purple
Pod shape constricted, inflated
Pea surface ?
Stem height tall, dwarf

5. 2. Principle of Dominance
One allele is dominant the other is recessive
The dominant allele is expressed in the phenotype

6. Gene for flower color
P allele = purple
p allele = white
GENOTYPES PHENOTYPE
Homozygous dominant =
Heterozygous =
Homozygous recessive =

7. What is the phenotype of all offspring in F1 generation?
3. Random segregation of alleles into gametes
gamete receives ONE allele per gene
random segregation of alleles 50/50
P generation
PP pp
What is the phenotype of all offspring in F1 generation?

8. Note that the P generation is true breeding
P P p
Genotype
Phenotype

9. How did Mendel do it? The Monohybrid cross
YY yy
Which allele is dominant?
What is the genotype of the f1 generation?

10. Cross 2 f1 plants (or let one self-fertilize)
What is the ratio of phenotypes?

11. Results of Mendel’s monohybrid crosses
Parental Strains F2 progeny Ratio
Tall X dwarf tall, 277 dwarf
Round seeds X wrinkled 5474 round, 1850 wrinkled
Yellow seeds X green yellow, 2001 green
Violet flowers X white 705 violet, 224 white
Inflated pods X constricted 882 inflated, 299 constricted
Green pods X yellow 428 green, 152 yellow
Axial flowers X terminal 651 axial, 207 terminal
gene = ?
alleles = ?

12. Test cross (one gene)
A mouse has black fur, what are its 2 possible genotypes?
Test cross mouse to homozygous recessive mouse
If black mouse is BB 
If black mouse is Bb 
Mouse was test crossed and 7 offspring were black 2 were white. What is mouse’s genotype?

13. Autosomal recessive inheritance (bb)
unaffected parents can have affected offspring
May “skip” a generation
Two affected parents cannot have an unaffected child
Not sex related

14. Examples of autosomal recessive traits
Sickle cell disease
Albinism
Cystic fibrosis
O blood type

15. Phenylketonuria (Ch. 4 pg 73)
PKU (1/12,000) Mutation in gene encoding phenylalanine hydroxylase enzyme needed for phe metabolism

16. If plasma phe level is too high, phe is converted
missing phenylalanine hydroxylase enzyme
If plasma phe level is too high, phe is converted
into a phenylpyruvate toxic to brain tissue

17. no tyrosine (little melanin)
Why are these babies normal when born?
Pleiotropic effects
no tyrosine (little melanin)
slow growth
retardation
blue eyes
low adrenaline
No nutrasweet
low phe diet ($5K/yr)

18. “Inborn errors of metabolism” PKU Albinism Alkaptonuria Tyrosinemia
Page 68
1902 Archibald Garrod:
One gene: one enzyme
“Inborn errors of metabolism”
PKU
Albinism
Alkaptonuria
Tyrosinemia
Black urine
arthritis

19. One gene/one enzyme Garrod’s work on alkaptonuria
“Inborn Errors of Metabolism” 1902
Autosomal recessive metabolic disease

20. II, 1 X II, 4 p(aa) p(aa AND a girl)?
Fill in genotypes. If II,1 and II, 4 mate, what is the chance of offspring having PKU?
How do we know
this is autosomal
recessive?
II, 1 X II, 4 p(aa)
p(aa AND a girl)?

21. If III-3 and II-1 mate p (normal child)
p (affected boy)?

22. All people have harmful recessive alleles, small chance
That 2 people with same rare alleles will mate
Consanguinous marriage increases the chance
Bedoin intermarriage

23. Autosomal dominant disorders Aa and AA =affected aa =unaffected
Tend to show up in every generation
2 affected parents can have unaffected child
2 unaffected parents cannot have an affected child

24. Dominant pedigree

25. Achondroplasia -1/20,000 births
Mutation in one allele of FGFR3 gene Chromosome 4
Affects cartilage growth needed for bone lengthening
Most affected individuals are Aa why?
Most cases are spontaneous
(Parents are aa X aa)
Pg. 291

26. P(III, 3 and III, 5 have a child of normal height)
P ( II, 3 and III, 7 have a boy with achondroplasia)

27. Fruit fly nomenclature pg 317 box 12.1
Red eyes is wildtype phenotype, brown is mutant
bw+ = wildtype allele
bw = brown allele
genotype phenotype
red brown

28. Try it:
Wingless is recessive mutant (wg allele)
Genotype of wildtype, heterozygote, mutant?

29. Autosomal
Sex-linked

30. Genetics Home Reference page (National Library of Medicine)
Collagen
Blood clotting factor
Red blood cell enzyme
Dystrophin muscle protein
Color vision gene

31. Sex-linked genes Ch 12 pg 314 – 317, 326 - 328
Human Female = XX
two alleles for each X-linked gene
normal application of recessive and dominance
XHXH
XHXh
XhXh

32. Human Male XY
XHY
XhY

33. X-linked genes Hemophilia (recessive) 1/5000 males
Mutation in gene for clotting factor
Xq28

34. *Criss cross inheritance of X linked traits
Mate III 13 with III 1 Probability of a hemophiliac son?
Mate IV 2 with homozygous normal female p (hemophilia)?
*Criss cross inheritance of X linked traits

35. Dihybrid cross – 2 genes
Mendel’s Law of Independent assortment - each allele for a trait is inherited independently of other alleles
Seeds:
G = yellow allele g = green allele gene?
W = round allele w = wrinkled allele gene?

36. Parents = GGWW X ggww phenotype? gametes?
F1 genotype ?
F1 phenotype ?
F1 Gametes?

39. Note that each gene gives the 3:1 ratio of a monohybrid cross
Yellow/green ratio =
Round/ wrinkled =

40. Forked line method for phenotypes
GgWw X GgWw

41. Test cross G-W- X ggww If all yellow and round:
A pea is round and yellow. What is its genotype?
G-W- X ggww
Note the cross of the “unknown” to a homozygous recessive
If all yellow and round:
If all yellow and some wrinkled:
If all round and some green:
If 1:1:1:1:

42. Probability GgWw X GgWw Product rule-
the probability that two outcomes occur simultaneously is product of their individual probabilities
assumes independent assortment of genes
GgWw X GgWw
What is the probability of a yellow AND wrinkled?
p(G-ww)

43. Trihybrid cross AaBbCc X AaBbCc p(A-B-cc) AabbCcDD X AaBbCcDd
p(triply recessive)

44. Modified Mendelian Ratios
1. INCOMPLETE DOMINANCE
R = red flower (snapdragon)
R’ = white flower
* allele symbols do not connote dominance
* phenotypic ratio = genotypic ratio = ?
P CrCr X CwCw F1 F2

45. Incomplete dominance

46. 2. Codominance Each allele encodes separate gene
product distinct in phenotype of heterozygote
L gene for human blood cell surface protein
LM = M antigen
LN = N antigen

48. A man with the M bloodtype has a child with a woman of the MN bloodtype
Expected ratio of offspring?

49. 3. Multiple alleles (more than 2 alleles for gene in population)
Example: Blood Groups Karl Landsteiner 1900s
Chromosome 9 I gene

50. Blood type genotype ABO blood system = polymorphic I gene
A IAIA or IAi
B IBIB or IBi
AB ?
O ii
What is the mechanism of inheritance of A, B, AB, O?
Autosomal or sex chromosome?

51. ABO dominance and codominance

52. 4. Dominance series – C series/ rabbits
c+ = full color
cch = chinchilla (hypomorphic)
ch = himalayan (hypomorphic)
c = albino (apomorphic allele = nonfunctional)
Himalayan Albino
Chinchilla

53. Genotype phenotype? cch cch cch ch ch c c+ cch c+ = full color
cch = chinchilla (hypomorphic)
ch = himalayan (hypomorphic)
c = albino (apomorphic allele = nonfunctional)

54. 5. Lethal alleles MM’ = manx cat (no tail) M’M’ = lethal
MM = normal spine
MM’ = manx cat (no tail)
M’M’ = lethal
Cross two manx, what is ratio of phenotypes in offspring?
How do breeders obtain manx cats?

55. 6. Epistasis- gene product interactions. Table 13
6. Epistasis- gene product interactions. Table 13.4 page 355 (look at 4 phenotypic classes and fewer than 4)
A product of one gene influences, or masks, the expression of another gene(s)
Modification of dihybrid cross ratio
AaBb X AaBb :3:3:1

56. Epistasis in Cats W = white w = not white B = black b = brown
Mate 2 heterozygous cats
What is the expected ratio?

57. Epistasis in labrador retrievers
B and E color genes (labs)
B black b brown
E color e no color (yellow)
ee is epistatic
Cross two double heterozygotes
Phenotypes of parents?
Phenotypes of offspring? ratio?

58. Polydactyly, dominant
7. Penetrance
% individuals that exhibit phenotype corresponding to genotype
Pp pp
5, , , 6

59. 8. Expressivity (ex. Piebald spotting) – the extent to which a trait is exhibited osteogenesis imperfecta pg. 359

60. Penetrance AND expressivity
NF-1 = Neurofibromatosis1
Autosomal dominant trait N-
50 – 80% penetrance
Expressivity
Pigmented skin to tumors on nerve CT coverings (neurofibromas) on skin, eyes, organs, face
Speech, blood pressure, spine curvature, headaches

62. 9. Quantitative (multifactorial) traits
Vary continuously
Weight, height, IQ

63. Gene expression also affected by:
Sex (baldness)
Temperature (melanin in Siamese cats)
Chemicals (PKU)
Diet (height, cancer)
+ many other factors!