1. Theoretical Genetics Theoretical Genetics 1 1 1

7. Definitions This image shows a pair of homologous chromosomes.
Name and annotate the labeled features.
Theoretical Genetics 7

8. Definitions This image shows a pair of homologous chromosomes.
Name and annotate the labeled features.
Genotype
The combination of alleles
of a gene carried by an organism
Homozygous dominant
Having two copies of the same dominant allele
Phenotype
The expression of alleles
of a gene carried by an organism
Homozygous recessive
Having two copies of the same recessive allele. Recessive alleles are only expressed when homozygous.
Centromere
Joins chromatids in cell division
Codominant
Pairs of alleles which are both expressed when present.
Alleles
Different versions of a gene
Dominant alleles = capital letter
Recessive alleles = lower-case letter
Heterozygous
Having two different alleles.
The dominant allele is expressed.
Carrier
Heterozygous carrier of a
recessive disease-causing allele
Gene loci
Specific positions of genes on a chromosome
Theoretical Genetics 8

9. Making Babies
1. Count the chromosomes in your envelope - there should be 46 in total.
2. Shuffle the chromosomes, so that they are well
mixed up. Which aspects of meiosis and sexual
reproduction give genetic variation?
3. Now arrange them in a karyotype (don't turn them
over - leave them as they were).
4. What is the gender of your baby?
Explain how gender is inherited in humans.
Crossing-over in prophase I
Random orientation in metaphase I and II
Random fertilisation
Activity from:
Theoretical Genetics 9

10. Making Babies Crossing-over in prophase I
Random orientation in metaphase I and II
Random fertilisation
List all the traits in a table. Use the key above to determine the genotypes and phenotypes of your offspring. Draw a picture of your beautiful child’s face!
Identify traits which are polygenic, involve gene interactions and some which are linked.
Activity from:
Theoretical Genetics 10

11. Explain this
Mendel crossed some yellow peas with some yellow peas. Most offspring were yellow but some were green!
Mendel from:
Theoretical Genetics 11

12. Segregation
“alleles of each gene separate into different gametes when the individual produces gametes”
The yellow parent peas must be heterozygous. The yellow phenotype is expressed.
Through meiosis and fertilisation, some offspring peas are homozygous recessive – they express a green colour.
Mendel did not know about DNA, chromosomes or meiosis.
Through his experiments he did work out that ‘heritable factors’ (genes) were passed on and that these could have different versions (alleles).
Mendel from:
Theoretical Genetics 12

13. Segregation F0 F1 Y y Y y Y or y Y or y
“alleles of each gene separate into different gametes when the individual produces gametes” F0
Genotype:
Y y
Y y
Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes.
Y or y
Y or y
Gametes:
Punnet Grid:
gametes F1
Genotypes:
Phenotypes:
Phenotype ratio:
Mendel from:
Theoretical Genetics 13

14. Monohybrid Cross F0 F1 Y y Y y Y or y Y or y Crossing a single trait.
Genotype:
Y y
Y y
Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes.
Fertilisation results in diploid zygotes.
A punnet grid can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F1).
Y or y
Y or y
Gametes:
Punnet Grid:
gametes F1
Genotypes:
Phenotypes:
Phenotype ratio:
Mendel from:
Theoretical Genetics 14

15. Monohybrid Cross F0 F1 Y y YY Yy yy Y y Y y Y or y Y or y
Crossing a single trait. F0
Genotype:
Y y
Y y
Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes.
Fertilisation results in diploid zygotes.
A punnet grid can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F1).
Y or y
Y or y
Gametes:
Punnet Grid:
gametes Y y YY Yy yy F1
Genotypes:
Phenotypes:
Phenotype ratio:
Mendel from:
Theoretical Genetics 15

16. Monohybrid Cross F0 F1 Y y YY Yy yy Y y Y y Y or y Y or y 3 : 1
Crossing a single trait. F0
Genotype:
Y y
Y y
Alleles segregate during meiosis (anaphase I) and end up in different haploid gametes.
Fertilisation results in diploid zygotes.
A punnet grid can be used to deduce the potential outcomes of the cross and to calculate the expected ratio of phenotypes in the next generation (F1).
Ratios are written in the simplest mathematical form.
Y or y
Y or y
Gametes:
Punnet Grid:
gametes Y y YY Yy yy F1
Genotypes: YY Yy Yy yy
Phenotypes:
Phenotype ratio:
3 : 1
Mendel from:
Theoretical Genetics 16

17. Monohybrid Cross
What is the expected ratio of phenotypes in this monohybrid cross? F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Homozygous recessive
Homozygous recessive
Punnet Grid:
gametes F1
Genotypes:
Phenotypes:
Phenotype ratio:
Theoretical Genetics 17

18. Monohybrid Cross F0 F1 y yy y y y y All green
What is the expected ratio of phenotypes in this monohybrid cross? F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
y y
y y
Homozygous recessive
Homozygous recessive
Punnet Grid:
gametes y yy F1
Genotypes: yy yy yy yy
Phenotypes:
Phenotype ratio:
All green
Theoretical Genetics 18

19. Monohybrid Cross
What is the expected ratio of phenotypes in this monohybrid cross? F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Homozygous recessive
Heterozygous
Punnet Grid:
gametes F1
Genotypes:
Phenotypes:
Phenotype ratio:
Theoretical Genetics 19

20. Monohybrid Cross F0 F1 Y y Yy yy y y Y y 1 : 1
What is the expected ratio of phenotypes in this monohybrid cross? F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
y y
Y y
Homozygous recessive
Heterozygous
Punnet Grid:
gametes Y y Yy yy F1
Genotypes: Yy Yy yy yy
Phenotypes:
Phenotype ratio:
1 : 1
Theoretical Genetics 20

21. Monohybrid Cross
What is the expected ratio of phenotypes in this monohybrid cross? F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Homozygous dominant
Heterozygous
Punnet Grid:
gametes F1
Genotypes:
Phenotypes:
Phenotype ratio:
Theoretical Genetics 21

22. Monohybrid Cross F0 F1 Y y YY Yy Y Y Y y All yellow
What is the expected ratio of phenotypes in this monohybrid cross? F0
Key to alleles:
Y = yellow
y = green
Phenotype:
Genotype:
Y Y
Y y
Homozygous dominant
Heterozygous
Punnet Grid:
gametes Y y YY Yy F1
Genotypes: YY YY Yy Yy
Phenotypes:
Phenotype ratio:
All yellow
Theoretical Genetics 22

23. Test Cross
Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. F0
Key to alleles:
R = Red flower
r = white
Phenotype:
Genotype:
R ?
r r
unknown
Homozygous recessive
Possible outcomes: F1
Phenotypes:
Unknown parent = RR
Unknown parent = Rr
gametes
gametes
Theoretical Genetics 23

24. Test Cross F0 F1 r r R Rr R Rr rr R ? r r All red Some white, some red
Used to determine the genotype of an unknown individual. The unknown is crossed with a known homozygous recessive. F0
Key to alleles:
R = Red flower
r = white
Phenotype:
Genotype:
R ?
r r
unknown
Homozygous recessive
Possible outcomes: F1
All red
Some white, some red
Phenotypes:
Unknown parent = RR
Unknown parent = Rr
gametes r R Rr
gametes r R Rr rr
Theoretical Genetics 24

25. Dihybrid Crosses Dihybrid Crosses 25 25 25

26. Dihybrid Crosses & Gene Linkage
Mendel’s Law of Independent Assortment
“The presence of an allele of one of the genes in a gamete has no influence over which allele of another gene is present.”
This only holds true for unlinked genes (genes on different chromosomes).
Dihybrid Crosses & Gene Linkage 26

27. Dihybrid Crosses & Gene Linkage
Mendel’s Law of Independent Assortment
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough
“The presence of an allele of one of the genes in a gamete has no influence over which allele of another gene is present.”
meiosis
This only holds true for unlinked genes (genes on different chromosomes). SY sY Sy sy
Dihybrid Crosses & Gene Linkage 27

28. Dihybrid Crosses
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci?
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Heterozygous at both loci
Heterozygous at both loci
Genotype:
Punnet Grid:
gametes F1
Dihybrid Crosses & Gene Linkage 28

29. F1 Dihybrid Crosses F0 SsYy SsYy
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci?
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous at both loci
Genotype:
SsYy
SsYy
Punnet Grid:
gametes F1
Dihybrid Crosses & Gene Linkage 29

30. F1 Dihybrid Crosses F0 SsYy SsYy SY Sy sY sy SSYY SSYy SsYY SsYy SSyy
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci?
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous at both loci
Genotype:
SsYy
SsYy
Punnet Grid:
gametes SY Sy sY sy
SSYY
SSYy
SsYY
SsYy
SSyy
Ssyy
ssYY
ssYy
ssyy F1
Dihybrid Crosses & Gene Linkage 30

31. F1 Dihybrid Crosses F0 SsYy SsYy SY Sy sY sy SSYY SSYy SsYY SsYy SSyy
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
What is the predicted phenotype ratio for a cross between two pea plants which are heterozygous at both loci?
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous at both loci
Genotype:
SsYy
SsYy
Punnet Grid:
gametes SY Sy sY sy
SSYY
SSYy
SsYY
SsYy
SSyy
Ssyy
ssYY
ssYy
ssyy F1
Phenotypes: 9 Smooth, yellow : 3 Smooth, green : 3 Rough, yellow : 1 Rough, green
Dihybrid Crosses & Gene Linkage 31

32. Dihybrid Crosses
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Heterozygous at both loci
Heterozygous for S, homozygous dominant for Y
Genotype:
Punnet Grid: F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 32

33. F1 Dihybrid Crosses F0 SsYy SsYY
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous for S, homozygous dominant for Y
Genotype:
SsYy
SsYY
Punnet Grid: F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 33

34. F1 Dihybrid Crosses F0 SsYy SsYY SY sY Sy sy
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous for S, homozygous dominant for Y
Genotype:
SsYy
SsYY
Punnet Grid:
gametes SY sY Sy sy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 34

35. F1 Dihybrid Crosses F0 SsYy SsYY SY sY SSYY SsYY Sy SSYy SsYy ssYY sy
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous for S, homozygous dominant for Y
Genotype:
SsYy
SsYY
Punnet Grid:
gametes SY sY
SSYY
SsYY Sy
SSYy
SsYy
ssYY sy
ssYy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 35

36. F1 Dihybrid Crosses F0 SsYy SsYY SY sY SSYY SsYY Sy SSYy SsYy ssYY sy
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
Calculate the predicted phenotype ratio for:
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, yellow
Smooth, yellow
Heterozygous at both loci
Heterozygous for S, homozygous dominant for Y
Genotype:
SsYy
SsYY
Punnet Grid:
gametes SY sY
SSYY
SsYY Sy
SSYy
SsYy
ssYY sy
ssYy
6 Smooth, yellow : 2 Rough, yellow F1
Phenotypes: 3 Smooth, yellow : 1 Rough, yellow
Present the ratio in the simplest mathematical form.
Dihybrid Crosses & Gene Linkage 36

37. Dihybrid Crosses SsYy SsYy SsYy SsYY SsYy Ssyy SSyy ssYY SSYY ssyy
Common expected ratios of dihybrid crosses.
SsYy
SsYy
SsYy
SsYY
Heterozygous at both loci
Heterozygous at both loci
Heterozygous at both loci
Heterozygous at one locus, homozygous dominant at the other SY Sy sY sy
SSYY
SSYy
SsYY
SsYy
SSyy
Ssyy
ssYY
ssYy
ssyy SY sY
SSYY
SsYY Sy
SSYy
SsYy
ssYY sy
ssYy
3 : 1
9 : 3 : 3 : 1
SsYy
Ssyy
SSyy
ssYY
= All SsYy
Heterozygous at both loci
Heterozygous/
Homozygous recessive
SSYY
ssyy
= all SyYy Sy sy SY
SSYy
SsYy
SSyy
Ssyy sY
ssYy
ssyy
Ssyy
ssYy
= 1 : 1 : 1 : 1
4 : 3 : 1
Dihybrid Crosses & Gene Linkage 37

38. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Rough, yellow
Genotype:
Punnet Grid: F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 38

39. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Rough, yellow
ssYy
Genotype:
Punnet Grid:
gametes sY sy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 39

40. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Rough, yellow
ssYy or ssYY
Genotype:
Punnet Grid:
gametes sY sy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 40

41. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Rough, yellow
ssyy
ssYy or ssYY
Genotype:
Punnet Grid:
gametes sY sy
All sy F1
Phenotypes:
Remember: A test cross is the unknown with a known homozygous recessive.
Dihybrid Crosses & Gene Linkage 41

42. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Rough, yellow
ssyy
ssYy or ssYY
Genotype:
Punnet Grid:
gametes sY sy
All sy
ssYy
ssyy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 42

43. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A rough yellow pea is test crossed to determine its genotype.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Rough, yellow
ssyy
ssYy or ssYY
Genotype:
Punnet Grid:
gametes sY sy
All sy
ssYy
ssyy F1
Phenotypes:
Some green peas will be present in the offspring if the unknown parent genotype is ssYy.
No green peas will be present in the offspring if the unknown parent genotype is ssYY.
Dihybrid Crosses & Gene Linkage 43

44. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, green
Genotype:
Punnet Grid: F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 44

45. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, green
ssyy
Genotype:
Punnet Grid:
gametes
All sy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 45

46. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, green
ssyy
SSyy
Genotype:
Punnet Grid:
gametes Sy
All sy
Ssyy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 46

47. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, green
ssyy
SSyy or Ssyy
Genotype:
Punnet Grid:
gametes Sy sy
All sy
Ssyy
ssyy F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 47

48. Dihybrid Crosses & Gene Linkage
Consider two traits, each carried on separate chromsomes (the genes are unlinked).
In this example of Lathyrus odoratus (sweet pea),
we consider two traits: pea colour and pea surface.
A smooth green pea is test crossed. Deduce the genotype.
Smooth green = nine offspring. Rough green = one offspring.
Key to alleles:
Y = yellow
y = green
S = smooth
s = rough F0
Phenotype:
Smooth, green
ssyy
SSyy or Ssyy
Genotype:
Punnet Grid:
gametes Sy sy
All sy
Ssyy
ssyy F1
Phenotypes:
No rough peas will be present in the offspring if the unknown parent genotype is SSyy.
The presence of rough green peas in the offspring means that the unknown genotype must be Ssyy.
The expected ratio in this cross is 3 smooth green : 1 rough green. This is not the same as the outcome. Remember that each reproduction event is chance and the sample size is very small. With a much larger sample size, the outcome would be closer to the expected ratio, simply due to probability.
Dihybrid Crosses & Gene Linkage 48

49. Dihybrid Crosses & Gene Linkage
Sooty the Guinea Pig
Key to alleles*:
C = colour c = albino
A = agouti a = black
R = round ears r = pointy ears
L = long whiskers l = short whiskers
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty news story from the BBC:
* C and A genes are real. The rest are made up for this story.
Dihybrid Crosses & Gene Linkage 49

50. Dihybrid Crosses & Gene Linkage
Sooty the Guinea Pig
Key to alleles:
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty has soft fur and sharp nails.
In one of his matings with a rough-furred, smooth-nailed female, the following guinea piglets are produced:
6 x rough fur, sharp nails; 3 x soft fur sharp nails.
Deduce Sooty’s genotype. F0
Phenotype:
Rough fur, smooth nails
Soft fur, sharp nails
Genotype:
Punnet Grid: F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 50

51. Dihybrid Crosses & Gene Linkage
Sooty the Guinea Pig
Key to alleles:
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty has soft fur and sharp nails.
In one of his matings with a rough-furred, smooth-nailed female, the following guinea piglets are produced:
6 x rough fur, sharp nails; 3 x soft fur sharp nails.
Deduce Sooty’s genotype. F0
Phenotype:
Rough fur, smooth nails
Soft fur, sharp nails
Genotype: ssnn
Punnet Grid:
Possible
Gametes
All sn F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 51

52. Dihybrid Crosses & Gene Linkage
Sooty the Guinea Pig
Key to alleles:
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty has soft fur and sharp nails.
In one of his matings with a rough-furred, smooth-nailed female, the following guinea piglets are produced:
6 x rough fur, sharp nails; 3 x soft fur sharp nails.
Deduce Sooty’s genotype. F0
Phenotype:
Rough fur, smooth nails
Soft fur, sharp nails
Genotype: ssnn SSNN or SsNN or SsNn
Punnet Grid:
Possible
Gametes SN Sn sN sn
All sn F1
Phenotypes:
Dihybrid Crosses & Gene Linkage 52

53. Dihybrid Crosses & Gene Linkage
Sooty the Guinea Pig
Key to alleles:
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty has soft fur and sharp nails.
In one of his matings with a rough-furred, smooth-nailed female, the following guinea piglets are produced:
6 x rough fur, sharp nails; 3 x soft fur sharp nails.
Deduce Sooty’s genotype. F0
Phenotype:
Rough fur, smooth nails
Soft fur, sharp nails
Genotype: ssnn SSNN or SsNN or SsNn
Punnet Grid:
Possible
Gametes SN Sn sN sn
All sn
SsNn
Ssnn
ssNn
ssnn F1
Phenotypes:
Soft fur
Sharp nails
Soft fur
Smooth nails
Rough fur
Sharp nails
Rough fur
Smooth nails
Dihybrid Crosses & Gene Linkage 53

54. F1 Sooty the Guinea Pig F0 SN Sn sN sn All sn SsNn Ssnn ssNn ssnn
Key to alleles:
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty has soft fur and sharp nails.
In one of his matings with a rough-furred, smooth-nailed female, the following guinea piglets are produced:
6 x rough fur, sharp nails; 3 x soft fur sharp nails.
Deduce Sooty’s genotype. F0
Phenotype:
Rough fur, smooth nails
Soft fur, sharp nails
Genotype: ssnn SSNN or SsNN or SsNn
Punnet Grid:
Possible
Gametes SN Sn sN sn
All sn
SsNn
Ssnn
ssNn
ssnn F1
Phenotypes:
Soft fur
Sharp nails
Soft fur
Smooth nails
Rough fur
Sharp nails
Rough fur
Smooth nails
Only these two phenotypes have been produced.
Sooty has only produced SN and sN gametes.
Dihybrid Crosses & Gene Linkage 54

55. It is most likely that his genotype is SsNN.
Sooty the Guinea Pig
Key to alleles:
S = soft fur s = rough fur
N = sharp nails n = smooth nails
Sooty has soft fur and sharp nails.
In one of his matings with a rough-furred, smooth-nailed female, the following guinea piglets are produced:
6 x rough fur, sharp nails; 3 x soft fur sharp nails.
Deduce Sooty’s genotype. F0
Phenotype:
Rough fur, smooth nails
Soft fur, sharp nails
Genotype: ssnn SSNN or SsNN or SsNn
Punnet Grid:
Possible
Gametes SN Sn sN sn
All sn
SsNn
Ssnn
ssNn
ssnn F1
Phenotypes:
Soft fur
Sharp nails
Soft fur
Smooth nails
Rough fur
Sharp nails
Rough fur
Smooth nails
Only these two phenotypes have been produced.
Sooty has only produced SN and sN gametes.
It is most likely that his genotype is SsNN.
10.2 Dihybrid Crosses & Gene Linkage 55

56. F1 Sooty the Guinea Pig F0 Key to alleles: Deduce Sooty’s genotype.
R = round ears r = pointy ears
L = long whiskers l = short whiskers
Deduce Sooty’s genotype.
Offspring = five with pointy ears and long whiskers F0
Phenotype:
Pointy ears, short whiskers
Pointy ears, long whiskers
Genotype:
Punnet Grid: F1
Phenotypes:
10.2 Dihybrid Crosses & Gene Linkage 56

57. F1 Sooty the Guinea Pig F0 rL rl All rl Key to alleles:
R = round ears r = pointy ears
L = long whiskers l = short whiskers
Deduce Sooty’s genotype.
Offspring = five with pointy ears and long whiskers F0
Phenotype:
Pointy ears, short whiskers
Pointy ears, long whiskers
Genotype: rrll rrLL or rrLl
Punnet Grid:
Possible
Gametes rL rl
All rl F1
Phenotypes:
10.2 Dihybrid Crosses & Gene Linkage 57

58. F1 Sooty the Guinea Pig F0 rL rl All rl rrLl rrll Key to alleles:
R = round ears r = pointy ears
L = long whiskers l = short whiskers
Deduce Sooty’s genotype.
Offspring = five with pointy ears and long whiskers F0
Phenotype:
Pointy ears, short whiskers
Pointy ears, long whiskers
Genotype: rrll rrLL or rrLl
Punnet Grid:
Possible
Gametes rL rl
All rl
rrLl
rrll F1
Phenotypes:
Pointy ears
Long whiskers
Pointy ears
Short whiskers
10.2 Dihybrid Crosses & Gene Linkage 58

59. F1 Sooty the Guinea Pig F0 rL rl All rl rrLl rrll Key to alleles:
R = round ears r = pointy ears
L = long whiskers l = short whiskers
Deduce Sooty’s genotype.
Offspring = five with pointy ears and long whiskers F0
Phenotype:
Pointy ears, short whiskers
Pointy ears, long whiskers
Genotype: rrll rrLL or rrLl
Punnet Grid:
Possible
Gametes rL rl
All rl
rrLl
rrll F1
Phenotypes:
Pointy ears
Long whiskers
Pointy ears
Short whiskers
Only this phenotype has been produced.
Sooty has only produced rL gametes.
10.2 Dihybrid Crosses & Gene Linkage 59

60. It is most likely that his genotype is rrLL.
Sooty the Guinea Pig
Key to alleles:
R = round ears r = pointy ears
L = long whiskers l = short whiskers
Deduce Sooty’s genotype.
Offspring = five with pointy ears and long whiskers F0
Phenotype:
Pointy ears, short whiskers
Pointy ears, long whiskers
Genotype: rrll rrLL or rrLl
Punnet Grid:
Possible
Gametes rL rl
All rl
rrLl
rrll F1
Phenotypes:
Pointy ears
Long whiskers
Pointy ears
Short whiskers
Only this phenotype has been produced.
Sooty has only produced rL gametes.
It is most likely that his genotype is rrLL.
10.2 Dihybrid Crosses & Gene Linkage 60

61. Codominance
Some genes have more than two alleles.
Where alleles are codominant, they are both expressed.
Human ABO blood typing is an example of multiple alleles and codominance.
The gene is for cell-surface antigens (immunoglobulin receptors).
These are either absent (type O) or present.
If they are present, they are either type A, B or both.
Where the genotype is heterozygous for IA and IB, both
are expressed. This is codominance.
Key to alleles:
i = no antigens present
IA = type A anitgens present
IB = type B antigens present
4.3 Theoretical Genetics 61

62. More about blood typing
A Nobel breakthrough in medicine.
Antibodies (immunoglobulins) are specific to antigens.
The immune system recognises 'foreign' antigens and produces antibodies in response - so if you are given the wrong blood type your body might react fatally as the antibodies cause the blood to clot.
Blood type O is known as the universal donor, as it has not antigens against which the recipient immune system can react. Type AB is the universal recipient, as it has no antibodies which will react to AB antigens.
Blood typing game from Nobel.org:
Images and more information from:
4.3 Theoretical Genetics 62

63. Incomplete Dominance
Heterozygotes have an appearance somewhat in between the phenotypes of the dominant and the recessive. (Mix)
Example: snapdragons (flower)
red (RR) x white (rr)
RR = red flower
rr = white flower R R r r

64. Incomplete Dominance R R r Rr Rr All Rr = pink (heterozygous pink) r

65. Incomplete Dominance

66. 10.3 Polygenic Inheritance (AHL)
A single characteristic controlled by multiple genes.
Polygenic inheritance gives rise to continuous variation in the phenotype.
Use these two examples in the exam.
Human Skin Colour
Wheat kernel colour
Other examples:
Susceptibility* to heart disease, certain types of cancer, mental illnesses.
The Autism Spectrum.
Autism is a pervasive developmental disorder that presents on a scale (known as the Childhood Autism Rating Scale). It is not as clearly polygenic as the above examples - it is suspected that gene interactions and environmental factors play a large role.
*susceptibility is not deterministic, but it is beneficial to know if you are at elevated genetic risk of these illnesses.
10.3 Polygenic Inheritance (AHL) 66

67. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Polygenic inheritance gives rise to
continuous variation in the phenotype.
Globally we observe continuous variation in skin colours. Skin colour is the result of pigments, such as melanin, being produced - the darker the skin, the greater the protection against the harmful effects of the Sun.
Skin colour is though to be controlled by up to four
separate genes, each with their own alleles. This is too large for us to deal with simply, so we'll look at two genes with two alleles each.
Image from:
Watch this TED Talk and think about the following questions:
What is melanin and what purpose does it serve?
What skin tone were early humans most likely to have? Why does this change with latitude as humans migrated towards the poles?
What are the relative advantages and disadvantages of light and dark skin, depending on climate?
TOK/ Aim 8: Why have people historically discriminated based on skin colour? How could the Natural Sciences educate people to think twice about their prejudices?
Nina Jablosnki breaks the illusion of skin colour, via TED.
10.3 Polygenic Inheritance (AHL) 67

68. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Example: 2 genes (A and B), 2 alleles each
Assume: genes are not linked (separate chromosomes)
In polygenics, alleles can be:
Contributing (they add to the phenotype)
Non-contributing (they do not add to the phenotype)
How many genotypes are possible?
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin
Remember that alleles segregate during meiosis.
Alleles of unlinked chromosomes orient randomly.
There is also random fertilisation of gametes.
So many combinations! or or or
gametes
10.3 Polygenic Inheritance (AHL) 68

69. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Example: 2 genes (A and B), 2 alleles each
Assume: genes are not linked (separate chromosomes)
In polygenics, alleles can be:
Contributing (they add to the phenotype)
Non-contributing (they do not add to the phenotype)
How many genotypes are possible?
Nine:
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin
Notice that the possible combinations of genotypes gives rise to continuous variation in the phenotype.
This population follows a normal distribution.
10.3 Polygenic Inheritance (AHL) 69

70. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Is it possible for twins to be:
Different colours?
YES.
As long as they are non-identical twins. Two eggs will have been fertilised by individual sperm cells.
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin
Each gamete carries a different combination of alleles, so it is possible that the twins have noticeably differently-coloured skin.
Couple has differently-coloured twins – for the second time! From Associated Press
10.3 Polygenic Inheritance (AHL) 70

71. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Is it possible for twins to be:
b Lighter or darker than both parents?
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin F0
Phenotype:
AABb
AaBb
Genotype:
Punnet Grid:
gametes F1
Genotypes:
Phenotypes:
10.3 Polygenic Inheritance (AHL) 71

72. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Is it possible for twins to be:
b Lighter or darker than both parents?
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin F0
Phenotype:
AABb
AaBb
Genotype:
Punnet Grid:
gametes AB Ab aB ab F1
Genotypes:
Phenotypes:
10.3 Polygenic Inheritance (AHL) 72

73. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Is it possible for twins to be:
b Lighter or darker than both parents?
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin F0
Phenotype:
AABb
AaBb
Genotype:
gametes AB Ab
AABB
AABb
AAbb aB
AaBB
AaBb ab
Aabb
Punnet Grid: F1
Genotypes:
Phenotypes: 73
10.3 Polygenic Inheritance (AHL)

74. 10.3 Polygenic Inheritance (AHL)
Polygenic Inheritance of Skin Colour
Is it possible for twins to be:
b Lighter or darker than both parents?
YES.
Key to alleles:
A = add melanin
a = don’t add melanin
B = add melanin
b = don’t add melanin F0
Phenotype:
AABb
AaBb
Genotype:
gametes AB Ab
AABB
AABb
AAbb aB
AaBB
AaBb ab
Aabb
Punnet Grid:
darker
lighter F1
Genotypes:
Phenotypes: 74
10.3 Polygenic Inheritance (AHL)