Genetics and Molecular Biology
Botany Major, Semester V
BT 502C
Unit I Genetics-I
1 Mendelian Genetics:
Principles of segregation and independent assortment,concept of Dominance, Incomplete dominance, Codominance, Multiple allele, Penetrance, Expressivity, Pleotropism, Phenocopy effect and Atavism.
2. Determining allelism of mutants
Complementation test, Definition of Cistron,Muton & Recon, Concept of pseudoallele, Fine structure of gene -- structure of phase rII locus.
3. Gene interactions with modified dihybrid ratios
12:3:1
9:7
9:4:3
9:6:1
13:3
15:1
Unit II Genetics -II
1. Cytoplasmic Inheritance:- Features, Plastid inheritance - leaf colour in Mirabilis jalapa, Mitochondrial inheritance -:Poky &Petite mutation, Maternal effect - shell coiling in snail.
2. Linkage :- Definition of complete linkage & incomplete linkage, coupling phase, repulsion phase, linked group,
Crossing over:- definition and cytological basis of crossing over ( Creigton & McClintock Experiment).
3. Recombination - Basic concept, Recombination frequency, Two point & Three point Test cross, Gene mapping from three point test cross data, coefficient of correlation , interference.
Unit-III Genetics-III
1. Chromosomal abberation:- Numerical changes:-( aneuploidy and euploidy), Polyploidy types.
Structural changes: Definition and types of deletion, Duplication, Inversion, Translocation
Meiotic behaviour of inversion and translocation heterozygote,
Position Effect.
2 Sex determination : Mechanism of sex determination in Human and Drosophila.
3. Sex linkage: Sex linked inheritance, Dosage compensation & Lyon's hypothesis, Sex limited and Sex influenced traits, problems on Sex linkage( Haemophilia & Colour blindness).
Unit IV Genetics IV
1 Population genetics :- Concept of Gene pool, Allele frequency and genotype frequency, Hardy- Weinberg law, Conditions for HW equilibrium, Numerical problems based on HW equation.
2. Factors affecting changes in gene frequency: Migration, Mutation,Selection and Genetic Drift:- definition and effects on gene frequency.
3.Quantitive inheritance - Characters of quantities traits, Heritability- Narrow sense and Broad sense Heritability ,
Polygenic inheritance: Regulation of kernel colour in Wheat
Cistron is a segment of DNA coding for a polypeptide .
(It is a segment of DNA containing the genetic information for the sequence of amino acids of a polypeptide chain i.e. the blueprint of a protein is actually stored in DNA as a cistron).
(2)Define intron and exon .
In Eukaryotes, the structural gene is monocistronic and coding sequences are interrupted by intervaining sequences.
Coding sequences are expressed as exons and intervaining sequences are expressed as introns in pre -mRNA/ hn RNA during transcription process .Introns are excised by spliceosome at processing time and only exons are appeared and joined to make a mature/processed mRNA.
The monocistronic structural genes are also known as split gene.
(3) Define transcription .
The process of copying/transfer of genetic information from template strand of DNA into mRNA maintaining "the principle of complimentarity" is known as transcription .
(4) Why both strands of DNA are not copied during transcription ?
There are two reasons :
(i) If two strands of a structural genes act as template then the two transcribed mRNA will be of different sequences , as a result of that transcribed proteins will differ from each other . Hence , a gene will be coding two different genes ,which contradicts the definition of a gene.
(ii) The transcribed mRNA molecules will be complementary to each other , hence, they will form ds RNA and translation pracess will prevented .
Unit II : Genetics II
Chapter: : Maternal Effects and Cytoplasmic Inheritance
*.1.Maternal inheritance
The phenomenon of dextral and sinistral shell coiling in Lymnaea peregra was first discovered,analysed and published by A.E. Boycott and colleagues in 1923 and 1930. It is one of the best example of the maternal inheritance.
In Lymnaea peregra , the female gamete is physically larger than the male gamete and supplies the cytoplasm for the developing embryo. This cytoplasm contains factors originally transcribed from the female nuclear genes.
Maternal effect refers to a phenomenon where an offspring's phenotype is determined entirely by the nuclear genotype of the mother ,regardless of the offspring's own genotype or paternal influence.These traits are controlled by factors ( such as mRNAs or proteins) that are encoded by the nuclear genes in the mother and deposited into the cytoplasm of the egg.
Reciprocal cross of the above cross:
F2 generation (dd) : Although these individuals possess the homozygous recessive genotype (dd) , they exhibit a dextral ( right - handed) phenotype . This occurs because their mother ( the F1 generation with a Dd genotype) deposited maternal genes products-- specifically the functional dextral protein or m RNA into the egg cytoplasm during oogenesis. Consequently,the mother's phenotype ( Dd producing dextral offspring) dictates the physical trait of the F2 progeny regardless of their individual nuclear genotype.
F3 generation (dd) : These dd offsprings develop a sinistral (Left -handed) phenotype. Their mothers belong to the F2 generation ( dd genotype) and, despite displaying a dextral phenotype themselves due to their maternal inheritance,their underlying genotype is homozygous recessive (dd). Therefore, these F2 mothers produce eggs lacking the dextral factors, resulting in the expression of the maternal genotype( dd) as a sinistral phenotype in the F3 generation.
Key Principles: The phenotype of an individual for this trait is entirely determined by the genotype of the mother , delayed by one generation.
*.2.Organellar inheritance/Plastid inheritance in Mirabilis jalapa
Plastid inheritance was first discovered by C. Correns in Mirabilis jalapa in 1908.There are three types of branches in this plant:-
(i)Complete green branches and leaves having chloroplasts,
(ii)White branches and leaves without chloroplasts and contains leucoplasts,and
(iii)Variegated branches with both kinds of plastids.
The following reciprocal crosses proved that phenotype of offsprings depend on the type of eggs , not on the nature of pollens.
Cross I : A cross between eggs from variegated plant/ branch ( female) and pollens from green plant/ branch ( male) :
Results:
(i) If the egg cytoplasm contains chloroplasts only then all F1 appear as green,
( ii) If the egg cytoplasm contains both of chloroplasts and leucoplasts then F1 all offspring appear as variegated., and
( iii) When the eggs have leucoplasts only then all offsprings appear as white. These white seedlings die early due to the lack of chloroplast and consequently lack of the photosynthesis process.
Cross II : It is reciprocal to the previous one. Eggs from green plant/ branches and pollens from variegated plant/ branches:
Results:
Only one kind of F1 offsprings appear which are green.
Conclusion: Since the results of the reciprocal crosses are phenotypically completely different from each other , it is proved that it is a case of maternal line and leaf variation is inherited non-Mendelianly through the cytoplasmic DNA or plastidDNA.
The following experiment also proves that plastid inheritance in Mirabilis jalapa is depend on the nature of female branchs :
In such cases the phenotype of F1 offssprings depend upon phenotype of branch on which flowers are pollinated.
*.3.Mitochondrial Inheritance
'Petite in Yeast
The petite phenotype is primarily caused by mutations in the mitochondrial genome , loss of mitochondria, deletions in mtDNA , or host cell genome mutation( neutral petite).
Genetic basis of petite: Vegetative petite which completely lack rho factor are called neutral petite. A vegetative petite having a defective rho factors is called suppressive petite.This rho is a cytoplasmic factor. The neutral petites are not transmitted and lack mtDNA, while suppressive petites are transmitted to a fraction of vegetative diploid progeny and contain mtDNA , that is often grossly altered in base composition with respect to wild mtDNA.
Experiments /Crosses:
A cross between a neutral petite yeast ( with defective mtDNA) and a wild type yeast:
Inference: Upon sporulation and tetrad analysis it is found that all four spores of an ascus are wild-type and there is no segregation of petite trait into 2:2 ratio, completely violating Mendel's first law .
Another cross between Suppressive petite and wild type:
Result: Formation of suppressive petite yeast in F1 generation.
Inference: In this cross resulted F1 generation are petite, because the mutated or deleted mtDNA actively overrides or ' swamps out' the wild type mitochondrial genome.Since, all four spores are petite or in irregular ratios in a single ascus, the mutation is extra nuclear .
Physiological basis of petite:
Inner chamber of mitochondria is the site of TCA cycle because of the presence of respiratory enzymes. In petite yeast there is lack of Cyt a, Cyt a3, Cyt b and a number of other changes in mitochondrial respiratory enzymes, or mitochondria with incomplete developement , hence petites have a defective aerobic respiratory mechanism. As a result of that petites fail to grow on carbon source like sucrose and only produce smaller colonies when grown on sugars like glucose.
2. Linkage
Definition of Linkage: Linkage is the tendency of genes or DNA segments located close to one another on the same chromosome to be inherited together during meiosis.
Classification of Linkages: Linkages are broadly classified into two primary categories based on the presence or absence of recombinant ( non- parental) combinations in subsequent generations :-
(i) Complete Linkage: Complete linkage occurs when two or more linked genes are inherited together with zero recombination, appearing exclusively in their parental combinations across successive generations.
* Molecular basis: The genes exhibiting complete linkage are situated extremely close to each other on the same chromosome.
* Crossing over: Because of there physical proximity, chiasma formation and subsequent crossing over donot occur between their loci during meiosis.
* Progeny output: Generates 100% parental phenotypes and strictly zero non- parental ( recombinant) progeny.
Classical example : Linkage observed in male Drosophila.
Income Linkage: Incomplete linkage is characterized by the production of both parental combinations and a measurable portion of non- parental (recombinant) combinations.
*Molecular basis: The genes are located at a distance from one another along the chromosome.
* Crossing over: During Prophase I of meiosis, , the physical breakage and exchange of chromosomal segments (crossing over) occur between non-sister chromatids of homologous chromosomes.
*Progeny output -Results in a mixture of both parental and recombinant phenotypes, with the frequency of recombination directly proportional to the physical distance between the genes.
Coupling phase & Repulsion phase of linkage
Coupling phase:
Definition: When two dominant alleles of linked genes reside on the same homologous chromosome, it is known as the coupling phase of linkage.
Genetic effect: The coupling arrangement of alleles tends to keep parental combinations together which resulted in a higher frequency of non-recombinant /dominant parental phenotype in offsprings in higher than expectations.
Repulsion phase:
Definition: When a dominant allele of one gene and a recessive allele of a second generation resides on the same homologous chromosome,it is known as the repulsion phase of linkage.
Genetic effect:: This arrangement tends to produce higher frequencies of recombinant/ non-parental phenotypes in the offspring compared to independent assortment.
Bateson and Punnet's experiment ,1906:
Bateson and Punnet worked with sweet pea which are presented below:
Moreover,the frequency of parental progeny turned 43.75%+43.75%= 87.50%. Simultaneously, frequency of recombination appears 6.25%+6.25%=13.50% . The frequency of recombination less than 50% implies that the genes are linked on the same chromosome. Bateson and Punnet concluded that as gene P and L were from same parent so show the tendency to remain together in progenies and they termed the arrangement as cis- configuration or coupling phase.
CROSSING OVER
Cytological basis of crossing over :
In 1931,C.Stern ( using Drosophila) and H.S. Creighton and B.McClintock (using maize) independently provided empirical proof for the cytological basis of crossing over.By utlizing cytological markers----specially , structrully altered chromosomes resulting from chromosomal aberrations---- they were able to visually distinguish homologous partners and correlate physical exchange with genetic recombination.
Three Point Test Cross & Gene Mapping
A Three point testcross is a fundamental genetic mapping technique used to determine the linear order and calculate the map distances/ recombination frequencies among three linked genes on a single chromosome.
Problems:
1. In a three-point testcross (CshWx/cShwx x cshwx/cshwx following data was obtained:
Calculate the recombination values and prepare a linkage map showing relative distances and linear order between the genes.
Also calculate Coefficient of Coincidence and interference.
Solution
Step I : Determination of Gene Order
(1) Identification of Parental classes(Non-recombinsnts):
The highest frequencies shown in the given data is:
CshWx = 2777
cShwx= 2708
-------------------------------
Total = 5485
So, these are the parental classes.
(2) Identification of the Double Cross Over (DOC) classes:
CShWx = 4
cshwx = 3
--------------------------------
Total = 7
These classes show lowest frequencies,hence , these are the DCO classes.
(3) Determination of Gene Order:
By comparing alleles of DCo with the parental classes configuration it has been observed that alleles of gene Sh / sh flipped relative to the parents indicates the gene Sh/sh is placed in between middle C/c and Wx/wx genes.
Here, C (c) and WX (wx) remain with their original parental partners, while Sh (sh) has switched position .
Therefore, Sh is in the middle and the gene sequences will be :
C_______sh______WX
c_______Sh_______wx
Step II : Calculation of Recombination Frequencies/ Map Distance:
(1)Recombination Frequencies between C--Sh:
116+123+4+3
= -----------------------------x100
7000
= 3.51%
(2) Recombination Frequencies between Sh--Wx
643+626+4+3
=-------------------------------x100
7000
= 18.23%
(3) Recombination Frequencies between C--Wx
116+123+643+626+2(4+3)
=--------------------------------------------------x100
7000
= 21.74%
Since, 1map unit(mu) or cM (Centimorgan) equals to 1% recombination,then
Distance between C & Sh = 3.51cM ,
Distance between Sh & Wx = 18.23 cM ,and
Distance between C & Wx = 21.74 cM
Gene Map:
<-- 3.51 cM ---> <- 18.23 cM-->
C_________________Sh_______________Wx
<------------------------21.74cM----------------->
Fig: A linkage map prepared from results obtained from given data in the problem.
Coefficient of coincidence
Observed DCO
CoC=---------------------------------------------x100
Expected DCO
4+3
= -----------------------------------------
0.0351 x 0.1823 x 7000
7
= -----------------------------------
44.79
= 0.1562
Interference
Interference= 1-- 0.1562
= 0.8437
or , 84.37% indicating that about 84.37% of expected DCOs were prevented due to positive interference.
2. In a three-point testcross (ABC/abc x abc/abc), following data are obtained ( only phenotypes are given)
ABC .....230
abc.......240
aBc........96
AbC......104
ABc......138
abC......142
aBC........12
Abc........08
_______________________________
Total. 970
(a) Calculate the recombination values and prepare a linkage map showing relative distances and linear order between the genes.
(b) Calculate Coefficient of Coincidence and interference.
Solution:
Determination of Gene order:
Highest frequency:
ABC.......230
abc........240
----------------------------------------
Total= 470
Lowest frequency:
aBC......12
Abc......08
-----------------------------------------
Total = 20
Therefore, Parental classes( non- recombinant) = ABC & abc
Double Cross Overs = aBC & Abc
Comparison of parental classes and DCOs:
Calculation of Recombination Frequencies /Map Distance :(1) Recombination Frequencies between B--A
96+104+12+8
= --------------------------------------x100
970
22.68%
(2) Recombination Frequencies between A--C
138+142+12+8
=----------------------------------------x 100
970
= 30.92%
(3) Recombination Frequencies between B--C
96+104+138+142+2(12+8)
=------------------------------------------------x100
970
= 53.60%
Since, 1 mu or 1 cM = 1% Recombination Frequencies, therefore,
Distance between B--A = 22.68 cM
Distance between A--C = 30.92 cM
Distance between B--C= 53. 60 cM
Gene Map:
<---22.68 cM--> <----30.92--cM--->
B A C
-----------------------------------------------------
<--------------53.60 cM---------------------->
Or,
<---30.92 cM------> <-----22.68cM---->
C A B
-----------------------------------------------------
<----------------------53.60 cM --------------->
Coefficient of Coincidence:
Observed DCO
CoC =------------------------------------------
Expected DCO
12+8
= -------------------------------------------
0.2268 x 0.3092 X 970
20
= ------------------------------------
68.02
= ~0.2940
Interference :
1 - CoC = 1-- 0.2940
= 0.7060
Or, 70.60% , indicating that about 70.60% of expected DCOs were prevented due to positive interference.
Thank you Sir
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