Honours / Core Course (CC)

 
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PRINCIPLES OF GENETICS
Unit 1: Mendelian Genetics and its Extension 
Principles of inheritance, Incomplete dominance and co-dominance, Epistasis, Multiple alleles, Isoallele (White eye mutations), Pseudoallele (Lozenge Locus) & Cis-trans test for allelism, Lethal alleles, Pleiotropy, Penetrance & Expressivity

Q. How do Mendel's laws relate to meiosis?
Meiosis is the process of creating gametocytes (sperm and eggs) which have half (haploid) of the DNA of each parent. Mendel’s law of segregation states that the genes of the parents must segregate equally into haploid gametes in such a way that their offspring have an equal likelihood of inheriting either one, which is what, happens during meiosis. Mendel also discovered the law of simple dominance which means that for a pair of genes of a given trait, the dominant gene is always expressed in the offspring. The only way for the recessive gene to be expressed is for the offspring to receive it from both parents. During meiosis, the dominant and recessive traits of the parents segregate equally into haploid gametes and the offspring have equal likelihood of inheriting either one.

Unit 2: Linkage, Crossing Over and Linkage Mapping 
Linkage and Crossing, Complete & Incomplete Linkage, Measuring Recombination frequency and linkage map construction using three factor crosses, Interference and coincidence Sex linkage in Drosophila (White eye locus) & Human (Haemophilia)

Q.The following is a linkage map for three recessive genes located in the same chromosome in a certain species Drosophila measured in percentage recombination frequencies. If the coefficient of coincidence is 0.6, in this case, determine the frequencies of phenotypes expected among 1000 offspring’s of a cross abc/ABC × abc/abc.

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Three linked genes a,b,c occupy the following linkage position on a particular chromosome—

The map distance between a & b -- (SCO %+ DCO %) = 10 m.u
The map distance between b & c – (SCO% + DCO% ) = (20-10) m.u= 10m.u
The expected DCO (%)- 10/100 X  10/100 X 100 = 1%

C.C= Observed DCO % / Expected DCO % = X/1
Or, 0.6 =X/1
So, X- 0.61%
Now, the actual SCO % between a & b = (10-0.6)=9.4% and the actual SCO (%) between b&c = (10-0.6)=0.4%.
Thus, the total CO (%)= (9.4+9.4+0.6)= 19.4%
So NCO % = (100-19.4)= 80.6
According to this problem a cross is made between abc/ABC and abc/abc

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Possible 8 progeny classes along with their expected number, determined from SCO % between a&b and b&c and DCO% are expressed in the tabular form below—

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Unit 3: Mutations 
Types of gene mutations (Classification), Types of chromosomal aberrations (Classification with one suitable example from Drosophila and Human of each), variation in chromosome number; Nondisjunction of X chromosome in Drosophila; Non-disjunction of Human Chromosome 21. Molecular basis of mutations in relation to UV light and chemical mutagens. Mutation detection in Drosophila by attached X method. Biochemical mutation detection in Neurospora.

Q.Why heterozygous individuals show more impact of deletion?
i)Some human disorders are caused by deletions of chromosome segments. 
ii)In many cases, the abnormalities are found in heterozygous individuals, homozygotes for deletions usually die if the deletion is large. 
iii)This distinction tells us that, in humans at least, the number of copies of genes is important for normal development and function. 
iv)Typically, several to many genes are lost in a deletion, so the syndrome that results is caused by the loss of the combined functions of those genes, rather than the loss of just one gene.

Unit 4: Sex Determination 
Mechanisms of sex determination in Drosophila and in man; Dosage compensation in Drosophila & Human.

Q. How do autosomal genes perform under the influence of the SRY gene in sex determination?
(a)Sox9: Activated by SRY, leads to the differentiation of cells that forms the seminiferous tubules that contain male germ cells.
(b)SF1: Involved in the regulation of enzymes affecting steroid metabolism.
(c)WnT4: Leads to development of female gonads; XY males with a duplication of gene for WnT4 shows sex reversal to female.
(d)DMRT1 (Gene for Doublesex and mab-3 related transcription factor 1):
i)This gene is found in a cluster with two other members of the gene family, having in common a zinc finger-like DNA-binding motif (DM domain). 
The DM domain is an ancient, conserved component of the vertebrate sex-determining pathway that is also a key regulator of male development in flies and nematodes, and is found to be the key sex-determining factor in chickens. This gene exhibits a gonad-specific and sexually dimorphic expression pattern, just like the related doublesex gene in fruit flies. Defective testicular development and XY feminization occur when this gene is hemizygous.
ii)The DMRT1 gene is located at the end of the 9th chromosome. This gene is a dose sensitive transcription factor protein that regulates Sertoli cells and germ cell. 
iii)The majority of this gene is located in the testicular cord and Sertoli cells, with a small amount in the germ cells. iv)Two copies of this gene are required for normal sexual development. When a DMRT1 gene is lost the most common disease is chromosome 9p deletion, which causes abnormal testicular formation and feminization. 
v)The DMRT1 gene is critical in the male sex determination and without this gene the default female characteristic takes over and male characteristic is slight or non-existent. 
(e)FGF9 (Gene for Fibroblast Growth Factor family):
i)The protein encoded by this gene is a member of the fibroblast growth factor (FGF) family. FGF family members possess broad mitogenic and cell survival activities, and are involved in a variety of biological processes, including embryonic development, cell growth, morphogenesis, tissue repair, tumor growth and invasion.
ii)FGF9 has also been shown to play a vital role in male development. Once activated by SOX9, it is responsible for forming a feed forward loop with Sox9, increasing the levels of both genes. The absence of Fgf9 causes an individual, even an individual with X and Y chromosomes, to develop into a female, as it is needed to carry out important masculinizing developmental functions such as the multiplication of Sertoli cells and creation of the testis cords.

Unit 5: Extra-chromosomal Inheritance 
Kappa particle in Paramoecium, Shell spiralling in snail 

Q.Give example on influence of the maternal genome in Paramoecium with proper illustrations.
One or the most striking and spectacular cases of cytoplasmic inheritance occur in ParamoecIum aurelia in 1938. Tracy Morton Sonneborn reported that some strains contain kappa particles in the cytoplasm and are known as killer Kappa particles are about 2 µ in diameter and contain DNA and protein.

ii)Individuals not possessing Kappa particles ore sensitive and are killed by a poison paramecin whicn  is secreted by killer individuals. The secretion paramecin is harmless to the killers. The different killer strains have different means of killing their victims.
iii)Most of them do not kill their mates. But there are some strains that instead of Killing from a distance by secretion, kill their mates through close contact. The killer character has a nuclear as well as cytoplasmic basis.
iv)The existence and increase of Kappa particles is determined by the presence of a nuclear dominant gene K. The animals that are homozygous for recessive "k" are sensitive to killing and they cannot themselves become killers. Animals that are homozygous for dominant "K" or heterozygous in normal cytoplasm are potential killers.
v)They are actual killers when their cytoplasm contain kappa particles which in turn produce the lethal poison. vi)In animals of genotype KK Or Kk, kappa particles are transmitted from cell to cell; once they have been lost from a cell, they do not again develop by themselves.
vii)The individuals with genotype kk may also contain Kappa particles of some sort in the cytoplasm, although this state is unstable and eventually the particles disappear
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Q.How spindle orientation is related to maternal genes?
i)Spindle orientation appears to be controlled by maternal genes acting on the developing eggs in the ovary. 
ii)The orientation of the spindle, in turn, influences cell divisions following fertilization and establishes the permanent adult coiling pattern. 
iii)The dextral allele (D) produces an active gene product that causes right-handed coiling. 
iv)If ooplasm from dextral eggs is injected into uncleaved sinistral eggs, they cleave in a dextral pattern. However, in the converse experiment, sinistral ooplasm has no effect when injected into dextral eggs. 
v)Apparently, the sinistral allele is the result of a classic recessive mutation that encodes an inactive gene product.
vi)We can conclude, then, that females that are either DD or Dd produce oocytes that synthesize the D gene product, which is stored in the ooplasm. Even if the oocyte contains only the d allele following meiosis and is fertilized by a d-bearing sperm, the resulting dd snail will be dextrally coiled (right handed).

Unit 6: Genetic Fine Structure 
Complementation test in Bacteriophage (Benzer’s experiment on rII locus)

Q.What do you mean by complementation test? / What is the basic principle of a complementation test?
Complementation test, also called cis-trans test, in genetics, test for determining whether two mutations associated with a specific phenotype represent two different forms of the same gene (alleles) or are variations of two different genes.
To carry out a complementation test, parents that are homozygous for different mutations are crossed, producing
offspring that are heterozygous. If the mutations are allelic (occur at the same locus), then the heterozygous  offspring have only mutant alleles (ab) and exhibit a mutant phenotype:

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If, on the other hand, the mutations occur at different loci, each of the homozygous parents possesses wild-type genes at the other locus (aa b+ b+ and a+ a+ bb); so the heterozygous offspring inherit a mutant and a wild-type allele at each locus. In this case, the mutations complement each other and the heterozygous offspring have the wild-type
phenotype:

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Complementation occurs when an individual possessing two mutant genes has a wild-type phenotype and is an indicator that the mutations are non-allelic genes. When the complementation test is applied to white and apricot mutations, all of the heterozygous offspring have light coloured eyes, demonstrating that white and apricot are produced by mutations that occur at the same locus and are allelic.

Unit 7: Transposable Genetic Elements 
IS element in bacteria, Ac-Ds elements in maize and P elements in Drosophila, LINE, SINE, Alu elements in humans

Q.What is the key difference between composite and non-composite transposons?

The key difference between composite and non-composite transposons is that composite transposons have two flanking insertion sequences while non-composite transposons have inverted repeats instead of flanking insertion sequences.
A transposon is a fragment of DNA which can translocate within the bacterial genome. They are mobile DNA sequences. They move into new locations of the genome. These movements make changes in the sequence of the bacterial genome, causing significant changes in genetic information. They are the transposable genetic elements responsible for establishing new genetic sequences in bacteria. Transposons are also referred to as jumping genes because these jumping sequences can block the transcription of genes and rearrange the genetic material of bacterium. Moreover, they are responsible for the movement of drug resistance, antibiotic resistance genes between plasmids and chromosomes. There are two types of transposons as composite and non-composite transposons.

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