1. Genetics is the study of inheritance
Mendel’s laws of inheritance – Classical genetics
Molecular Genetics – DNA structure and function

2. Mendel’s laws of inheritance
Gregor Johann Mendel ( ) is considered the father of genetics. Proposed the pattern of inheritance
Mendel chose the garden pea to study
the natural laws governing plant hybrids
Mendel carried out two types of crosses
1. Self-fertilization
Pollen and egg are derived from the same plant
Naturally occurs in peas because a modified petal isolates the reproductive structures
2. Cross-fertilization
Pollen and egg are derived from different plants
Required removing and manipulating anthers
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3. Pea flowers

4. Contain the pollen grains, where the male gametes (sperm) are produced
Contain the female gametes (eggs)

5. Provides storage material for the developing embryo

9. Mendel’s Experiments Mendel studied seven traits:
Each trait showed two variants or forms found in the same species
His first experiments crossed only two variants of one trait at a time. This is termed a monohybrid cross
A pea plant contains two discrete hereditary factors, one from each parent
- The two factors may be identical or different
When the two factors of a single trait are different
One is dominant and its effect can be seen
The other is recessive and is not expressed
During gamete formation, the paired factors segregate randomly so that half of the gametes received one factor and half of the gametes received the other
This is Mendel’s Law of Segregation
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10. Mendelian factors are now called genes
Alleles are different versions of the same gene
An individual with two identical alleles is termed homozygous
An individual with two different alleles, is termed heterozygous
Genotype: RR rr
Genotype refers to the specific allelic composition of an individual
Phenotype refers to the outward appearance of an individual
(Red vs white)
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11. Genes to Traits Molecular basis of Alleles
Alleles are alternate forms of the same gene
DNA > Protein -----> Red Pigment
-ATG CCT AGG- - Met-Pro-Arg-
DNA > Protein --X---> No Pigment
-ATG CTT AGG- - Met-Leu-Arg

13. Test Cross : to determine if an individual expressing the dominant trait is a heterozygous or homozygous
Make a cross between the individual expressing the dominant trait with a homozygous recessive individual
Parent TT X tt (test parent)
(Tall) (Dwarf)
Gametes T T t t
F Tt
Tall
Parent Tt X tt (test parent)
Tall (Tall) (Dwarf)
Gametes T t t t
F Tt, t t (1:1)
Tall : Dwarf (1:1)

14. Create an empty Punnett square
Fill in the Punnett square with the possible genotypes of the offspring by combining the alleles of the gametes

15. Dihybrid cross (Two traits)

18. Independent assortment is also revealed by a dihybrid test-cross
TtYy X ttyy
Thus, if the genes assort independently, the expected phenotypic ratio among the offspring is 1:1:1:1
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19. law of Independent assortment (Pattern of Inheritance)
Based on the distribution of the different classes of the F2 progeny in a dihybrid cross, Mendel concluded that the genes assorted independently.
This is the law of Independent assortment
During gamete formation, the segregation of any pair of hereditary determinants is independent of the segregation of other pairs
If one parent is RrYy and the other is rryy, the progeny will be of four types (Test cross)
RrYy, Rryy, rrYy, rryy in the ratio of 1:1:1:1

20. Laws of Inheritance Law of segregation Law of Independent assortment
During gamete formation, the paired factors segregate randomly so that half of the gametes receives one factor and half of the gametes receives the other
If the parent is Rr, the gametes will be R, r in the ratio of 1:1
Law of Independent assortment
During gamete formation, the segregation of any pair of hereditary determinants is independent of the segregation of other pairs
If the parent is RrYy, the gametes could be of four types
RY, Ry, rY, ry in the ratio of 1:1:1:1

21. Pedigree Analysis (Inheritance in Humans)
Pedigree analysis is used to determine the pattern of inheritance of traits in humans
When studying human traits, it is not ethical to control parental crosses
Rather, we must rely on information from family trees or pedigrees
Pedigree analysis is commonly used to determine the inheritance pattern of human genetic diseases
Genes that play a role in disease may exist as
A normal allele
A mutant allele that causes disease symptoms
Disease that follow a simple Mendelian pattern of inheritance can be
Dominant
Recessive
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23. A recessive pattern of inheritance makes two important predictions
1. Two normal heterozygous individuals will have, on average, 25% of their offspring affected
2. Two affected individuals will produce 100% affected offspring
A dominant pattern of inheritance predicts that
An affected individual will have inherited the gene from at least one affected parent
Alternatively, the disease may have been the result of a new mutation that occurred during gamete formation
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26. A family pedigree for albinism

27. An albino family

28. A family pedigree of dwarfism

29. The father (dwarf) must be heterozygous and did not pass the dwarfism gene to his children

30. Dominant versus Recessive Inheritance Pattern?B

31. PROBABILITY AND STATISTICS
The laws of inheritance can be used to predict the outcomes of genetic crosses
For example
Animal and plant breeders are concerned with the types of offspring produced from their crosses
Parents are interested in predicting the traits that their children may have
This is particularly important in the case of families with genetic diseases
Of course, it is not possible to definitely predict what will happen in the future
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32. Probability
The probability of an event is the chance that the event will occur in the future
Probability =
Number of times an event occurs
Total number of events
For example, in a coin flip
Pheads =
1 heads
(1 heads + 1 tails)
= 1/2 = 50%
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33. Sum rule
The probability that one of two or more mutually exclusive events will occur is the sum of their respective probabilities
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34. If two heterozygous (Dede Ctct) mice are crossed
Then the predicted ratio of offspring is
9 with normal ears and normal tails
3 with normal ears and crinkly tails
3 with droopy ears and normal tails
1 with droopy ears and crinkly tail
These four phenotypes are mutually exclusive
A mouse with droopy ears and a normal tail cannot have normal ears and a crinkly tail
Question
What is the probability that an offspring of the above cross will have normal ears and a normal tail or have droopy ears and a crinkly tail?
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35. Applying the sum rule Step 1: Calculate the individual probabilities
P(normal ears and a normal tail) = 9
( )
= 9/16
P(droopy ears and crinkly tail) = 1
( )
= 1/16
Step 2: Add the individual probabilities
9/16 + 1/16 = 10/16
10/16 can be converted to 0.625
Therefore 62.5% of the offspring are predicted to have normal ears and a normal tail or droopy ears and a crinkly tail
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36. Product rule
The probability that two or more independent events will occur is equal to the product of their respective probabilities
Note
Independent events are those in which the occurrence of one does not affect the probability of another
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37. Consider the disease congenital analgesia
Recessive trait in humans
Affected individuals can distinguish between sensations
However, extreme sensations are not perceived as painful
Two alleles
P = Normal allele
p = Congenital analgesia
Question
Two heterozygous individuals plan to start a family
What is the probability that the couple’s first three children will all have congenital analgesia?
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38. Applying the product rule
Step 1: Calculate the individual probabilities
This can be obtained via a Punnett square
P(congenital analgesia) =
1/4
Step 2: Multiply the individual probabilities
1/4 X 1/4 X 1/4 = 1/64
1/64 can be converted to 0.016
Therefore 1.6% of the time, the first three offspring of a heterozygous couple, will all have congenital analgesia
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39. What are the chances of having progeny with the genotype AA bb Cc
Aa Bb CC X Aa bb Cc
What are the chances of having progeny with the genotype AA bb Cc
Chance of AA is ½ x ½ + ¼
Chance of bb is ½ x 1 = ½
Chance of Cc is 1 x ½ = ½
Probability that an offspring will be AA bb Cc is
P = ¼ x ½ x ½ = 1/16

40. What are the chances that T will be an albino?
A couple, denoted R and S, are concerned about the possibility that they will have a child (T) with albinism, an autosomal recessive condition.
S, the prospective father, is an albino, and R, the prospective mother, has 2 albino siblings.
What are the chances that T will be an albino?
The risk depends on two factors: probability that R is heterozygous and the probability that she will transmit this allele to T if she is a carrier
R’s parents must be Aa
Aa x Aa = 1AA : 2 Aa : 1aa = chance that R is Aa is 2/3
Chances that R will pass on a to T is 2/3 x1/2 = 1/3
R is aa
Chances of T being aa is 1/3 x 1 = 1/3