1. Patterns of Inheritance
Chapter 23

2. Gregor Mendel 1822 – 1884 Austrian monk Experimented with garden peas
Provided a basis for understanding heredity

3. Mendel cont’d
Published a paper in 1866 stating that parents pass discrete heritable factors on to their offspring
Factors retain individuality generation after generation
Identified that each trait is inherited by a pair of factors, one from each parent
One form of a factor may be dominant over an alternative form
Reasoned that each egg and sperm must contain only 1 copy of a factor for each trait

4. Mendel cont’d. Mendel’s law of segregation
Each individual has two factors (genes) for each trait
The factors segregate (separate) during the formation of gametes
Each gamete contains only one factor from each pair of factors
Fertilization gives each new individual 2 factors for each trait

5. Modern Genetics – Genes
Sections of chromosomes which give instructions for one characteristic or protein.
Located at the same point or locus, on each member of a homologous pair
All together make up the organism’s genome.
Controls the physical characteristics of a species

6. Modern Genetics - Alleles
Alternative forms of the same gene on each chromosome
One allele comes from each parent
Dominant allele
Masks other traits present
Only 1 dominant allele needs to be present for a certain trait to be expressed
Represented by a capital letter
Recessive Allele
2 copies of the recessive allele need to be present for trait to be expressed
Represented by a lower case letter

7. Modern Genetics – Alleles cont’d
Gene locus
Fig. 23.2

8. Modern Genetics – Alleles cont’d
Genotype
Genetic composition of a specific trait
Homozygous dominant – 2 dominant alleles
Heterozygous – 1 dominant allele and 1 recessive allele
Homozygous recessive 2 recessive allele
Phenotype
Physical expression of a specific trait
Homozygous dominant or heterozygous  Dominant Trait
Homozygous recessive  recessive trait

9. Single Gene Inheritance
Simplest situation
One gene carries all the information responsible for one trait
Widow’s Peak
Alternative forms of alleles for hairline shape
Widow’s peak is dominant to straight hair line
W=allele for widow’s peak
w= allele for straight hairline

10. Widow’s peak
Fig. 23.3

11. Genotype related to phenotype
Table 23.1

12. Single Gene Inheritance cont’d.
Monohybrid cross
Looks at inheritance of one trait only
A punnett squares used to find all possible combinations of alleles.

13. Single Gene Inheritance cont’d.
Example 1:
If a homozygous woman with a widow’s peak reproduces with a man with a straight hairline, what kind of hairline will their children have?

14. Single Gene Inheritance cont’d.
Example:
If two heterozygous parents reproduce what kind of hairline will their children have? W w WW Ww ww

15. Single Gene Inheritance cont’d.
Genetic Ratios
Express ratio’s of possible outcomes
Genotypic ratio
Homozygous Dominant: Heterozygous: Homozygous Recessive
Phenotypic ratio
Dominant trait : recessive trait
Often expressed as probability
The probability resets and is the same for each pregnancy!
Having one child with a trait has no effect on future children.

16. Single Gene Inheritance cont’d.
Look back to example 1
What are the genotypic and phenotypic ratio’s?
WW x ww
Genotypic Ratio
0 : 4 : 0
Phenotypic Ratio
1:0
100% Widow’s peak
Ww x Ww
Genotypic Ratio
1: 2: 1
Phenotypic Ratio
3 : 1
Probability is ¾ Widow’s Peak
75%
Probability ¼ Straight
25%

17. Single Gene Inheritance cont’d
Determining Genotype
No way to distinguish between a homozygous dominant individual and a heterozygous individual just by looking
They are phenotypically the same
Test cross may help us determine
Used in breeders of plants and animals
Cross unknown with a recessive individual
We know one parent genotype, this will help us determine the other genotype
If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous

18. Single Gene Inheritance cont’d
Example : In rats, large ears is dominant to small ears. A rat breeder has a female rat with large ears, which she breeds with a male rat with small ears. In the first litter, all rats are born with large ears.
What is the genotype of the female rat?
In a second litter from the same parents, 4 baby rats have large ears, one has small ears.

19. Single Gene Inheritance cont’d
Practice problems
Both a man and a woman are heterozygous for freckles. Freckles are dominant over no freckles. What is the chance that their child will have freckles?
Both you and your sibling have attached ear lobes, but your parents have unattached lobes. Unattached earlobes (E) are dominant over attached (e). What are the genotypes of your parents?
A father has dimples, the mother of his children does not, and all 5 of their children have dimples. Dimples (D) are dominant over no dimples (d). Give the probable genotypes of all persons concerned.

20. Homework (WHAT??? It’s BIO!)
Bikini Bottom Beach Genetics

21. Independent Assortment
Mendel reasoned from the results of his pea plant crosses that each pair of factors assorts independently into gametes
Each trait is passed down individually. The allele you receive for any one gene is not related to any other alleles you receive.
We can now explain this through independent alignment and crossing over in meiosis. (Fig 23.6)

22. Independent Assortment cont’d
Fig. 23.6

23. Independent Assortment cont’d
Called the Law of Independent Assortment
Each pairs of factors assorts independently (without regard to how the others separate)
All possible combinations of factors can occur in the gametes

24. Independent Assortment cont’d
Practice problems
For each of the following genotypes, give all possible gametes WW
WWSs Tt
Ttgg
AaBb
For each of the following, state whether the genotype or a gamete is represented D Ll Pw
LlGg

25. Dihybrid Cross
Punnet squares, considering two-trait crosses at one time.
Example:
The traits for hairline and finger length are both single gene traits. As before, widows peak is dominant over straight hairline. Having short fingers is considered dominant over long fingers.
Two parents who are both heterozygous for both traits have children.
Determine the genotypic and phenotypic ratios for their children.

26. Dihybrid cross cont’d Possible Gametes? Genotypic Ratio?
Phenotypic Ratio?

27. Dihybrid Crosses cont’d
Determining Ratio’s
Product rule of probability
The chance of 2 or more independent events occurring together is the product of their chance of occurring separately
In our example:
Probability of widow’s peak = ¾
Probability of short fingers= ¾
What is the probability of widow’s peak AND short fingers?
¾ x ¾ = 9/16

28. Dihybrid Crosses cont’d
Recall from our single trait crosses
Probability of widow’s peak = ¾ Probability of short fingers= ¾
Probability of straight hairline= ¼ Probability of long fingers= ¼
Using the product rule
Probability of widow’s peak and short fingers = X =
Probability of widow’s peak and long fingers = X = 3/16
Probability of straight hairline and short fingers = ¼ X ¾ = 3/16
Probability of straight hairline and long fingers = ¼ X ¼ = 1/16
These values are standard for all heterozygous crosses! You don’t need to memorize them, but should be able to figure them out in your head!

29. Dihybrid Crosses cont’d
Using the product rule
Probability of widow’s peak and short fingers
Probability of widow’s peak and long fingers
Probability of straight hairline and short fingers
Probability of straight hairline and long fingers

30. Dihybrid Crosses cont’d
Two-trait test cross
Cross an individual with the dominant phenotype for each trait with an individual with the recessive phenotype of both traits
W?S? x wwss WS W? ?S ??
wwss
WwSs
Ws?s
?wSs
?w?s

31. Dihybrid Cross cont’d
Attached earlobes are recessive, What genotype do children have if one parent is homozygous for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline?
If an individual from this cross reproduces with another of the same genotype, what are the chances that they will have a child with a straight hairline and attached earlobes?
A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant traits) and a woman who does not. What are the genotypes of all persons concerned?

32. Polygenic Inheritance
Not fully understood by geneticists.
Generally:
One trait controlled by 2 or more genes at different loci
The higher the number of dominant alleles you possess, the stronger the expression of the trait.
Result is a continuous range of phenotypes
Distribution resembles a bell curve
The more gene pairs involved, the more continuous the pattern of variation
Ex: human height, skin pigmentation, eye colour

33. Polygenic inheritance cont’d
Fig. 23.9

34. Polygenic Inheritance cont’d
Skin color
Controlled by many gene pairs and many alleles
Let’s assuming a simple model of two alleles at 2 loci
A and B
If two heterozygous parents have children, children can range from very light to very dark
Genotype
Phenotype
AABB
Very Dark Skin
AABb or AaBB
Dark Skin
AaBb, AAbb, or aaBB
Medium brown skin
Aabb, or aaBb
Light Skin
aabb
Very light skin

35. Polygenic Inheritance cont’d
Eye colour is controlled by 3 genes we have identified
We suspect there are more
Not a clear dominant and recessive
Brown allele

36. Environmental Influences on Inheritance
Environment can influence gene expression and therefore phenotype
Ex: sunlight exposure on skin; coat color in Himalayan rabbits
Human twin studies
Polygenic traits are most influenced
“nature vs. nurture”
Identical twins separated at birth are studied
If they share a trait in common even though raised in different environments, it is likely genetic

37. Coat color in Himalayan rabbits
Fig

38. Incomplete Dominance Incomplete dominance
Heterozygous individual has a phenotype intermediate to the two homozygous individuals
Ex: Curly-haired Caucasian woman and a straight- haired Caucasian man produce wavy-haired children
When 2 wavy-haired people have children, the phenotypic ratio is 1 curly: 2 wavy: 1 straight

39. Incomplete dominance
Fig

40. Codominance Multiple allele inheritance
The gene exists in several allelic forms, but each person still has only 2 of the possible alleles
Occurs when both alleles are equally expressed
Ex: type AB blood has both A antigens and B antigens on red blood cells

41. Codominance cont’d. ABO blood types IA = A antigens on RBCs
IB = B antigens on RBCs
i = has neither A nor B antigens on RBCs
Both IA and IB are dominant over I, IA and IB are codominant
Phenotype
Genotype A
IAIA or IAi B
IBIB or Ibi AB
IAIB O ii

42. Codominance cont’d. Paternity testing- ABO blood groups often used
Can disprove paternity but not prove it
Rh factor- another antigen on RBCs
Rh positive people have the antigen
Rh negative people lack it
There are multiple alleles for Rh negative, but all are recessive to Rh positive

43. Inheritance of blood type
Fig

44. Practice Problems
A polygenic trait is controlled by three pairs of alleles. What are the two extreme genotypes for this trait?
What is the genotype of the lightest child that could result from a mating between two medium-brown individuals?
A child with type O blood is born to a mother with type A blood. What is the genotype of the child? The mother? ‘what are the possible genotypes of the father?
From the following blood types determine which baby belongs to which parents:
Baby 1 type O Mrs. Doe type A Mrs. Jones type A
Baby 2 type B Mr. Doe type A Mr. Jones type AB

45. Sex-linked inheritance
Sex chromosomes
22 pairs of autosomes, 1 pair of sex chromosomes
X and y
In females, the sex chromosomes are XX
In males, the sex chromosomes are XY
Note that in males the sex chromosomes are not homologous
Traits controlled by genes in the sex chromosomes are called sex-linked traits
X chromosome has many genes, the Y chromosome does not

46. Sex-linked inheritance cont’d.
X-linked traits
Red-green colorblindness is X- linked
The X chromosome has genes for normal color vision
XB = normal vision
Xb – colorblindness
Genotypes
Phenotypes
XBXB
female with normal vision
XBXb
carrier female, normal vision
XbXb
colorblind female
XBY
male with normal vision
XbY
colorblind male

47. Cross involving an X-linked allele
Fig

48. Practice Problems
Both the mother and the father of a colorblind male appear to be normal. From whom did the son inherit the allele for colorblindness? What are the genotypes of the mother, father, and the son?
A woman is colorblind. What are the chances that her son will be colorblind? If she is married to a man with normal vision, what are the chances that her daughters will be colorblind? Will be carriers?
Both the husband and the wife have normal vision. The wife gives birth to a colorblind daughter. Is it more likely the father had normal vision or was colorblind? What does this lead you to deduce about the girl’s parentage?
What is the genotype of a colorblind male with long fingers is s=long fingers? If all his children have normal vision and short fingers, what is the likely genotype of the mother?

49. Inheritance of linked genes
The sequence of individual genes on a chromosome is fixed because each allele has a specific locus
All genes on a single chromosome form a linkage group
When linkage is complete, a dihybrid produces only 2 types of gametes
Any time traits are inherited together, a linkage group is suspected
If very few recombined phenotypes appear in offspring, linkage is also suspected

50. Inheritance of linked genes
Crossing over between 2 alleles of interest, can result in 4 types of gametes
Occurrence of crossing over can indicate the sequence of genes on a chromosome
More frequent between distant genes
Fig

51. Practice Problems
When AaBb individuals reproduce, the phenotypic ratio is about 3:1. What ratio was expected? What may have caused the observed ratio?
The genes for ABO blood type and for fingernails are on the same homologous pair of chromosomes. In an actual family, 45% of offspring have type B blood and no fingernails, and 45% have type O blood and fingernails; 5% have type B blood and fingernails, and 5% have type O blood and no fingernails. What process accounts for the recombinant phenotypes?