Mendelian Genetics : 10 (28) 1 Mendelian Genetics Dr Meera Narasimha, ICFAI University, Hyderabad Outline : 10 (28) 2 Outline Mendel’s Laws of Heredity
Probability versus Possibility
Steps in Solving Heredity Problems
Double Factor Crosses The first geneticist: Gregor Mendel : 10 (28) 3 The first geneticist: Gregor Mendel Mendel was a monk who was the first to describe the basic patterns of inheritance.
Studied inheritance in garden pea plants
Studied several different phenotypes
Identified the concepts of dominance and recessiveness
Didn’t know about genes or chromosomes
Identified patterns by mathematical analysis of the data Dominant and recessive traits in pea plants : 10 (28) 4 Dominant and recessive traits in pea plants Slide 5: 10 (28) 5 Self Pollination: Pollen is transferred from stamen to stigma on the same plant.
Cross Pollination: Pollen is transferred from the stamen of one plant to the stigma of a different plant
Pure Line: A plant homozygous for a number of characteristics. Slide 6: 10 (28) 6 Mendel’s experiment : 10 (28) 7 Mendel’s experiment Parental (P) generation
A pure-breeding purple-flowered plant mated with a pure-breeding white-flowered plant
CC x cc
First filial generation (F1)
All offspring had purple flowers (Cc).
They were allowed to self-pollinate.
Cc x Cc Mendel’s experiment : 10 (28) 8 Mendel’s experiment Second filial generation (F2)
¾ of the offspring were purple.
of the offspring were white.
3:1 ratio, purple: white
Mendel saw this pattern with any of the traits he studied. White White ¼ Mendel’s Laws of Inheritance : 10 (28) 9 Mendel’s Laws of Inheritance Organisms have two pieces of genetic information for each trait.
We know these as alleles.
The Law of Dominance: Two different alleles for a given trait and Allele that express overshadows the other allele is dominant.
Some alleles mask other alleles.
Gametes fertilize randomly. Mendel’s Laws of Inheritance : 10 (28) 10 Mendel’s Laws of Inheritance The Law of Segregation: During gamete formation, alleles control a trait separate from each other and retain individuality.
Alleles separate into gametes during meiosis. Mendel’s Laws of Inheritance : 10 (28) 11 Mendel’s Laws of Inheritance Law of Independent Assortment: Members of one gene pair separate from each other independently of other gene pairs. Probability vs. possibility : 10 (28) 12 Probability vs. possibility Probability is the mathematical chance that an event will happen.
Expressed as a percent, or a fraction
Probability = the # of events that can produce a given outcome/the total # of possible outcomes.
The probability of two or more events occurring simultaneously is the product of their individual probabilities.
Possibility states that an event can happen; probability states how likely the event is to happen. Monohybrid Crosses : 10 (28) 13 Monohybrid Crosses Single Factor Cross: mono= one ; hybrid= combination.
6 examples, using the Punnett square for monohybrid crosses, one should be able to do any problem.
Table: 10.1 Solving genetics problems: Single factor crosses : 10 (28) 14 Solving genetics problems: Single factor crosses The pod color of some pea plants is inherited so that green pods are dominant to yellow pods.
A pea-plant that is heterozygous for green pods is crossed to a pea plant that produces yellow pods.
What proportion of the offspring will have green pods? Yellow pods Steps involved in solving Heredity Problems : 10 (28) 15 Steps involved in solving Heredity Problems In humans, the allele for Tourette syndrome (TS) is inherited as an autosomal dominant allele.
If both parents are heterozygous, what is the probability that they can have a child without Tourette syndrome ? With Tourette syndrome ? Steps involved in solving Heredity Problems: Single Factor Crosses : 10 (28) 16 Steps involved in solving Heredity Problems: Single Factor Crosses Five basic steps are involved.
Step 1: Assign a symbol for each allele.
A capital letter is used for a dominant allele and a small letter for a recessive allele.
Tourette =T and t = no tourette.
Allele Genotype Phenotype
T= Tourette TT Tourette syndrome
t= Normal Tt Tourette syndrome
tt Normal Step 2: Determine the Genotype of Each Parent and Indicate a Mating : 10 (28) 17 Step 2: Determine the Genotype of Each Parent and Indicate a Mating Both parents are heterozygous, the male and female genotypes are Tt.
The X between the two genotypes used to indicate a mating.
Tt X Tt Step 3: Determine all possible gametes from each parent : 10 (28) 18 Step 3: Determine all possible gametes from each parent Heterozygous Parent in diploid stage is Tt and produce two types of gametes. 50% with T and 50% with t.
Both male and female has same genotype they produce gametes two types with each type 50% ( T and t gametes ) Create a Punnett Square : 10 (28) 19 Create a Punnett Square Punnett square is a box figure that allows to determine the probability of genotypes and phenotypes of the progeny of a particular cross.
Put the gametes from male parent are listed on the left side of the square.
Female gametes are listed on the top.
Simulate random fertilization by crossing the possible gametes.
This will determine offspring phenotypes. Punnett square : 10 (28) 20 Punnett square Male genotype Female genotype
Possible female gametes T & t
Possible male T t
gametes T & t T
t Step 4: Determine All the Gene Combinations that Can Result when these Gametes unite : 10 (28) 21 Step 4: Determine All the Gene Combinations that Can Result when these Gametes unite To determine the possible combinations of alleles that could occur as a result of this mating, fill each empty square with the alleles that can be donated from each parent .
All the allelic combinations that can result when these gametes unite. T t
T TT Tt
t Tt tt Step 5: Determine offspring phenotypes and calculate probability : 10 (28) 22 Step 5: Determine offspring phenotypes and calculate probability Use the gene key to determine the phenotype of the offspring you predicted.
Revisit the question to calculate the answer to the question.
What proportion of offspring will produce without TS? With TS?
The answer is 25% ( Without TS) and 75% (With TS). Problem: Dominant/Recessive PKU : 10 (28) 23 Problem: Dominant/Recessive PKU Normal: Conversion of Phenylalanine into Tyrosine.
PKU: Unable to convert Phenylalanine into Tyrosine and accumulation of phenylalanine prevents normal development of nervous system and may be mentally retarded. Problem: Dominant/Recessive PKU : 10 (28) 24 Problem: Dominant/Recessive PKU The normal condition is to convert phenylalanine to tyrosine is dominant over the condition for PKU.
If one parent is heterozygous and the other parent is homozygous for PKU, what is the probability that they will have
A child that is normal?
A child with PKU? Problem on PKU contd., : 10 (28) 25 Problem on PKU contd., Step 1: Symbol N for normal and n for PKU.
Allele Genotype Phenotype
N= Normal NN Normal metabolism of Phenylalanine
n= PKU Nn Normal metabolism of Phenylalanine
nn PKU disorder
Step 2: Nn X nn
Step 3: n
n Problem on PKU contd., : 10 (28) 26 Problem on PKU contd., Step 4:
Step 5: 50% progeny will be normal and 50% will have PKU. Solution pathway : 10 (28) 27 Solution pathway Normal PKU Nn nn N n
n n n n N n Nn nn nn Nn Nn
50% nn 50% Sample problem: Codominance : 10 (28) 28 Sample problem: Codominance If a pink snapdragon is crossed with a white snapdragon, what phenotypes can result?
What is the probability of each phenotype?
Step1: FW = White flowers FR = Red Flowers
FWFW White Flower
FWFR Pink Flowers
FRFR Red Flowers White Flowers Solution pathway: Codominance : 10 (28) 29 Solution pathway: Codominance Problem: X-Linked : 10 (28) 30 Problem: X-Linked In humans, Normal color vision is dominant and color deficiency is recessive.
Both alleles are on X chromosome.
Male has one copy of X chromosome and female Two copies of X chromosome.
Female to be color bind requires both the alleles for color blind but in male single allele of color blind results in color blindness. Solution pathway: X-linked inheritance : 10 (28) 31 Solution pathway: X-linked inheritance A male who has normal color vision mates with a female who is heterozygous for normal color vision.
What type of children can they have in terms of these traits?
What is the probability for each type? Double factor crosses : 10 (28) 32 Double factor crosses Dihybrid crosses track the inheritance of two traits.
Mendel used dihybrid crosses to identify the law of independent assortment.
States that alleles of one character separate independently of alleles of another character
Only true when the genes for the two characters are on different chromosomes Solving double factor crosses : 10 (28) 33 Solving double factor crosses When solving a double factor cross, you must obey the law of segregation and the law of independent assortment.
Each gamete must receive only one copy of each gene.
All combinations of alleles for A and B must be considered.
Consider an individual whose genotype is AaBb.
Gametes could receive AB, Ab, aB or ab. A sample double factor cross : 10 (28) 34 A sample double factor cross In humans the allele for free earlobes is dominant over the allele for attached earlobes.
The allele for dark hair dominates the allele for light hair.
If both parents are heterozygous for earlobe shape and hair color, what types of offspring can they produce, and what is the probability for each type? Solving the double factor cross : 10 (28) 35 Solving the double factor cross In a dihybrid cross we deal with two traits at a time.
For example, in heterozygous individuals with earlobes and color of hair, the gametes can combine in 16 different ways. The probability for a given phenotype will be 9:3:3:1. Solving the double factor cross : 10 (28) 36 Solving the double factor cross Solving the double factor cross : 10 (28) 37 Solving the double factor cross