1. DNA and Inheritance

2. Biology 30 Genes & Heredity
Only one copy of each gene is on a chromosome.
Your traits are a result of the interactions of the genes from both parents.

3. Pioneer of Genetics: Gregor Mendel
Gregor Mendel, an Austrian monk, conducted some simple but significant experiments in his monastery garden. ( ). Why did Mendel work with the garden pea?
Garden peas have a number of different characteristics that can be expressed in ONLY one of two ways.
(ex. yellow vs. green pea)

4. Mendel’s Characters
The 7 characteristics that Mendel studied

5. Mendel crossed the F1 hybrids together to produce an F2 generation.
The recessive trait “reappeared”. F1 F2

6. He found a 3:1 ratio of dominant to recessive
1/4 homozygous dominant
1/2 heterozygous
1/4 homozygous recessive
3 round to
1 wrinkled

7. Rr r R R RR Rr Rr r rr Rr 1/4 homozygous dominant
1/4 homozygous recessive r rr Rr

8. Dominant genes –
Determine the expression of the genetic trait in the offspring.
(Capitalize this letter).
Recessive genes –
Are “masked” by dominant genes.
(Lower case letter).

9. Mendel’s Laws of Heredity
First Law (Law of Unit Characters)
Inheritance is governed by genes that exist in the individual and are passed on to offspring. These factors (genes) occur in pairs, one gene comes from the female and one gene comes from the male. The alternate forms of the same genes are known as alleles.

10. Second Law (Law of Dominance)
One allele of a
gene, masks the
effect of another.
This process is
known as the
principle of
dominance.
The dominant
expression is
seen and the
recessive gene is not seen (remains hidden).

11. 8 4 2 Third Law (Law of Segregation & Recombination)
A pair of genes (the 2 alleles or 2 letters) segregate/separate during the formation of sex cells (meiosis). As a result, each parent can only contribute one member (allele) of a pair of genes to their offspring. 8 4 2

12. Genetic Terms Define the following terms: Genotype – Phenotype –
Homozygous (“purebred”) –
Heterozygous –
Alleles –

13. Monohybrid Cross The combining of single contrasting traits:
eg. Crossing a tall plant (TT) with a short plant (tt)
a special chart called a Punnett square, helps geneticists organize the results of a cross between the sex cells of two individuals
from a Punnett square, we can predict the genotypes & phenotypes of the offspring:

14. Draw a Punnett square for a cross between a heterozygous round-seed plant and a wrinkled-seed plant.Rr rr r Rr rr r
In the F1 generation ½ are round ½ are wrinkled

15. Rr r R R RR Rr Rr r rr Rr 1/4 homozygous dominant
1/4 homozygous recessive r rr Rr

16. Draw a Punnett square for a cross between two heterozygous round-seed plants.RR Rr R Rr rr r
In F1 generation ¾ are round ¼ wrinkled

17. Draw a Punnett square for the cross between a heterozygous purple flower with a white flower.Ww ww w Ww ww w
In the F1 generation ½ are purple ½ white

18. Phenotypic Ratio:
50% of flowers are purple
50% of flowers are white
Genotypic Ratio:
50% of flowers are Ww
50% of flowers are ww

19. Test Cross
A test cross is often performed to determine the genotype of a dominant phenotype. When would we need to know this?
There would be some offspring produced that have white wool and some with black.
eg. A sheep farmer wants to ensure that all of his flock will have white hair (black wool is brittle & hard to dye). He chooses a white ram to mate with the flock. What if the ram is heterozygous?

20. To ensure that the ram is not, a test cross can be performed to ensure that the ram is homozygous for the white phenotype
A test cross is always performed between the unknown genotype and a homozygous recessive (known) genotype.

21. Possible outcomes:
If 50% of the offspring are black and the other half white, the unknown genotype must have been heterozygous (Ww).
If 100% of the offspring are white, then the unknown genotype must be homozygous dominant (WW).

22. Mendel showed that by performing a. test cross with the homozygous
Mendel showed that by performing a test cross with the homozygous recessive, the genotype could be determined.
YY x yy Yy

23. Yy x yy yy Yy Y y
This test cross produces 1/2 dominant and 1/2 recessive phenotypes

24. Gene Interactions
There are three main types of gene interactions:
1) Multiple Alleles
There are many traits that are controlled by more than two alleles for example, eye colour in Drosophila is controlled by four possible alleles.
The following phenotypes and dominance hierarchy is possible:
wild type (red) > apricot > honey > white

25. phenotype genotype wild type E1 Apricot E2 honey E3 white E4
When dealing with multiple alleles, it is no longer necessary to use upper & lower case letters  both letters & upper case numbers are used.
phenotype genotype
wild type E1
Apricot E2
honey E3
white E4
Predict the genotypes and phenotypes of the F1 generation from the mating of wild type (E1E4) with apricot (E2E3).

26. Predict the genotypes and phenotypes of the F1 generation from the mating of wild type (E1E4) with apricot (E2E3). E1 E4
E1E2
E2E4 E2
E1E3
E3E4 E3
In the F1 generation ½ are wild type; ¼ apricot; ¼ honey

27. Do Now
phenotype genotype wild type E1 Apricot E2 honey E3 white E4 With the above information, make two heterozygous crosses what phenotypes do you get?

28. 2) Incomplete Dominance
In some heterozygotes, both alleles of a pair are expressed in the phenotype. These alleles are said to be equally dominant. This lack of dominance is known as incomplete dominance.
eg.
P1 Black  White
F Grey
P2 Grey × Grey F2

30. Ex. Snapdragons
An interaction between the alleles in the heterozygote shows an intermediate phenotype. WW Ww ww

31. 3) Codominance
A form of incomplete dominance where two alleles are expressed in such a way that the effect of each is noticed separately in the phenotype.
Both parental phenotypes can be distinguished in the heterozygote offspring.

32. The expression of one allele does not mask the expression of another.
Ex. A red bull crossed with a white cow = roan calf (calf has intermingled white & red hair)

33. Roan cattle and horses have both coloured and white hair.
Blue Roan
Strawberry Roan

34. The ABO blood group system is another example.
Inheritance of blood groups is determined by the gene “I” which has three different alleles, only two of which can occur at the locus at once. The alleles are responsible for producing antigens on the surface of the red blood cells, which determines the blood group.
Alleles A and B are co-dominant so that when they are both present, both A and B antigens are produced. Both A and B are dominant to O.

35. Blood Types (A, B, O, AB)
Allele IA – formation of blood factor A (antigen A)
Allele IB – formation of blood factor B (antigen B)
Allele I – no factors result
Genotype Blood Type
IAIA or IAi A
IBIB or IBi B
ii O
IAIB AB

36. IA an IB are codominant and i is recessive.
Type A
Type B
Type AB
Type O
IA an IB are codominant and i is recessive.
Rh antigens are straight dominant vs. recessive.

37. Blood types of North America
Blood types of North America with RH.

38. Example:
A mother with blood type A has a child with Blood type O. The father is blood type B. Indicate the genotypes of the parents.
Father can be IBIB or IBi and mother can be IAIA or IAi. To have a child that is type O (ii) both mother and father must contribute the i allele, therefore the father must be IBi and the mother is IAi

39. Sex Chromosomes
So far, what do you know about sex chromosomes?
In addition to their role in determining sex, the sex chromosomes, especially X chromosomes, have genes for many characters unrelated to sex.
We call these sex-linked alleles.

40. Female cells can differ from male cells in two ways:
1. Female cells show dark spots of chromatin (called Barr Bodies) during interphase, male cells do not.
2. Female cells contain 2 X chromosomes and males contain only one X chromosome.

41. The Y chromosome carries few genes.
There are very few genes on the Y chromosome that are common on the X chromosome, and because of that, little crossing over may occur between an X and a Y.

42. eg.) Calico cats
Male cats tend to be black (XBY) or orange (X0Y). Female cats can be black (XBXB), orange (X0X0) or calico (XBX0) – a mixture between black and orange.
Very few male cats can be calico, why?
Those who do, carry a hidden X chromosome, and are likely sterile.

43. A male embryo does not differ from a female fetus until the 6th/7th week of pregnancy.
At this point, the “testes determining factor” (TDF) gene on the Y chromosome is activated.
The TDF gene initiates the production of a protein that stimulates the testes to begin secreting male hormones.

44. Examples of sex linked traits.
a. Hemophilia - lack or deformity of blood clotting factor VII or IX.

45. b. Red Green colorblindness

46. c. Pattern baldness - sex influenced not sex-linked.
i. Humans carry two alleles for baldness.
ii. In females the allele for baldness is recessive but in males, due to testosterone, it is dominant.

47. HB - baldness allele
HN - normal hair
Male Female
HNHN Normal Normal
HBHN Bald Normal
HBHB Bald Bald
So heterozygote females avoid baldness.

48. We can also perform monohybrid crosses between sex chromosomes.
For example:
Brown eye color (B) is dominant to blue (b).
Eye color is carried on the X chromosome.
Homozygous dominant female XBXB (brown)
Heterozygous female XBXb (brown)
Homozygous recessive female XbXb (blue)
Dominant male XBY (brown)
Recessive male XbY (blue)

49. XB Xb XBXb XbXb Xb XBY XbY Y
Draw a Punnett square for a cross between a heterozygous female with a recessive male. Calculate the phenotypic & genotypic ratios. XB Xb
XBXb
XbXb Xb
XBY
XbY Y
In the F1 generation:

50. Phenotypic ratios:
1 brown eyed girl: 1 brown eyed boy: 1 blue eyed girl: 1 blue eyed boy
Genotypic ratios:
1XBXb: 1XbXb: 1XBY: 1XbY

51. Xb Xb XBXb XBXb XB XbY XbY Y Example #2
Is it possible to get a blue eyed female from crossing a blue eyed female with a brown eyed male? Explain. Xb Xb
XBXb
XBXb XB
XbY
XbY Y
No it is not possible, all females would be browned eyed

54. Pedigrees A chart or register showing a line of ancestors
Circles represent females, squares represent males, solid circles & squares represent those who have the trait being studied.
Horizontal lines between circles and squares represent mating between male and female.
A vertical line joins parents and children.
Pedigrees are often used to study sex-linked traits such as color-blindness and hemophilia

56. PEDIGREE SUMMARY
This will help you determine a trait in a pedigree as with autosomal or X-linked:
Autosomal Dominant
must be in each generation
affected individuals transmit to minimum ½ of their offspring
males and females are equally affected
cannot have carriers, they will all be affected!

57. Autosomal Recessive
may skip a generation
affected offspring generally have normal (but heterozygous) parents
male and female are equally affected

58. X-linked Dominant
very likely to be observed in each generation
females pass on to half of either sex
no transmission from father to son (only daughters)
X-linked Recessive
affect males more than females
no transmission from father to son
daughters of males are carriers
females pass onto ½ sons
affected females have affected fathers and carrier mothers

60. Pedigree analysis

61. Probability
Probability is the likelihood of an event happening.
Probability can be expressed by the following formula:
Probability = # of chances for an event
# of possible combinations

62. What are the chances of getting heads?
Therefore, when a coin is tossed, there are two possibilities – heads or tails.
What are the chances of getting heads?
What is the probability of two coins being tossed and getting heads?
Coin 1 –
Coin 2 –
1 head/2 head/tail = 50 % chance
½ = 50 % chance
½ = 50 % chance

63. The Rule of Independent Events –
Product of the probabilities of two separate events:
The Product Rule –
Chance has no memory. This means that previous events will not affect future events. Ex. If you tossed two heads in a row, the probability of tossing heads again will still be ½.
½ × ½ = ¼ therefore a 25% chance
The probability of two or more independent events occurring together is the product of the individual probabilities if each individual event occurs separately.

64. Dihybrid Crosses
Mendel also studied two separate traits with a single cross by using the same procedure he had used for studying single traits.

65. Mendel crossed a purebred yellow round pea with a purebred green wrinkled pea.
Pure breeding round = RR
Pure breeding wrinkled = rr
Pure breeding yellow = YY
Pure breeding green = yy
Genotype for the yellow, round parent is RRYY
Genotype for the green, wrinkled parent is rryy

66. P1 YYRR x yyrr
Purebred yellow round × Purebred green wrinkled
The entire F1 generation has the genotype: YyRr and is phenotypically yellow round!
Male gametes
Female gametes F1

67. Now let’s cross the F1 generations with one another and see what we get…

68. 9/16 Yellow Round
3/16 Yellow wrinkled
3/16 green Round
1/16 green wrinkled
The phenotypic ratio 9:3:3:1 is the ratio you will find in all heterozygous dihybrid crosses!

69. Example:
In summer squash, white fruit color is dominant “W” and yellow fruit color is recessive “w”. Another allele produces disc shaped fruit “S” while its recessive allele “s” yields sphere-shaped fruit. If a pure breeding white disc variety is crossed with a homozygous yellow sphere variety, the F1 are all white disc hybrids. If the F1 generation is allowed to mate, what would be the expected phenotypic ratio in the F2 generation?
Try this!

70. P1 WWSS x wwss
F1 All offspring will be WwSs
P2 WwSs x WwSs F2

71. Rhesus Factor & Birth P1 : Female Rh- × Male Rh+
Baby is Rh+ because father is. Mother’s blood produces antibodies upon birth, (since blood mixes at birth). First baby is okay.
Second pregnancy- mom’s antibodies can now move across the placenta and cause baby’s RBC’s to clump (agglutinate) if second baby is also Rh+. This decreases oxygen delivery in the baby – “blue baby.”

72. What can be done?
Mom can be given an injection of a drug that inhibits antibody production immediately after delivery.
What happens if this is undetected?
Baby could be given a blood transfusion while in the womb. Fairly uncommon.

73. There will be a 50% chance. Blood Types & Rhesus Factor Question
R – dominant allele (Rh+)
r – recessive allele (Rh-)
Example: A woman homozygous for blood type A and heterozygous for the rhesus allele, Rh+, has a child with a man with type O blood who is Rh-. What is the probability that their child will have blood type A, Rh+?
There will be a 50% chance.

74. Do Now
Biology creatures have many strange traits. Blinking eyes “B” is dominant and burning eyes are recessive “b”. Another allele produces pleasant smelling pheromones “P” while its recessive allele “p” yields rotting fruit pheromones. If a pure breed blinking pleasant smelling biology creature is crossed with a heterozygous blinking pleasant smelling biology creature, what would be the expected phenotypic ratio?

75. Techniques used in order to produce the genotype or phenotype that you want:
Selective Breeding: the crossing of desired traits from plants or animals to produce offspring with both characteristics

76. Inbreeding: the process by which breeding stock is drawn from a limited number of individuals possessing desirable phenotypes.
Hybridization: Blending of desirable but different traits.

77. Gene Interaction
1) Many traits studied by Mendel were controlled by one gene.
2) Some traits are regulated by more than one gene; many of your characteristics are determined by several pairs of independent genes –
polygenic.
eg.) skin color, eye color and height, feather colour in parakeets
3) One trait controlled by more than one allele (multiple alleles).
Eg. Blood types, Drosophilia eye colour

78. 4) Genes that interfere with the expression of other genes are called epistatic.
Example:
allele B produces a black coat color in dog
b produces a brown coat color
A second gene, W prevents the formation of pigment
w does not prevent color

79. genotype wwBb would be black
genotype WwBb would appear white
W allele masks the effect of the B color gene
If wwBb is crossed with a WwBb, state the phenotypes produced.
8/16 = white
6/16 = black
2/16 = brown

80. 5) Complementary Interaction occurs when two different genotypes interact to produce a phenotype that neither is capable of producing by itself.
Example:
Allele R produces a rose comb in chickens
Allele P (on a different chromosome) produces a pea comb.
R and P alleles both present = walnut comb
The absence of rose and pea alleles results in an individual with a single comb

81. Chromosomal Theory The chromosomal theory is as follows:
Chromosomes carry genes, the units of hereditary
Paired chromosomes segregate during meiosis. Each sex cell or gamete has half the number of chromosomes found in a somatic cell
iii) Chromosomes sort independently during meiosis. Each gamete receives one of the pairs and that one chromosome has no influence on the movement of a member of another pair
iv) Each chromosome contains many different genes

82. Chromosome Mapping and Gene Linkage
A single chromosome contains many genes linked together and so does the other chromosome in the homologous pair.
The sequence of genes on each chromosome pair should match each other exactly.
Gene linkage reduces the chance for genetic recombination and variety among the offspring.
Parts of a chromosome holding many genes, may separate and switch places with the matching part of the other chromosome = crossing over.

84. The closer genes are to each other, the less likely they will separate during crossing over = linked genes.
Scientists use crossover frequencies on genes to determine their positions on chromosomes
eg.) if the crossover frequency of a gene is 5%, then the two genes are 5 map units apart.

85. Crossover frequency is determined by the following formula:
crossover % = number of recombinations x 100
total number of offspring
Gene markers are usually recessive genes that are easily observed in offspring and can be used to identify other genes found on the same chromosome.

86. By using crossover frequencies, we can determine gene maps.
Gene maps show the relative positions of genes on a chromosome (loci).
Gene maps are constructed by:
- ordering fragments of DNA
- studying chromosomal alterations
- performing crosses to see how frequently crossing over occurs between fragments.

87. Problem 1: 3 genes A, B, C
AB – 12% CB – 7% AC – 5%
A B C
12 map units
7 map units
5 map units
Problem 2: AB - 3% BC - 28% AC - 31%
31 map units A C B
28 map units
3 map units

88. Problem 3: Genes X Y Z X - 10 15 Y 10 - 5 Z 15 5 - 15 map units Z X Y

89. Crossover Frequency of Some Genes on Chromosome #6
Genes Cross-over Frequency
Diabetes(1) and Ovarian cancer (2) 21%
Diabetes (1) and RH blood group(3) 12%
Ragweed allergy (4) and RH blood group(3) 10.5
RH blood group(3) and ovarian cancer (2) 9%
Ragweed allergy (4) and ovarian cancer (2) 19.5
Hint: Start here

90. Transposons (see handout)
Gene Therapy: when defective genes are replaced with normal genes in order to cure genetic diseases
Human Genome Project: to determine the complete sequence of the 3 billion DNA subunits (bases), identify all human genes, and make them accessible for further biological study.