GENETICS

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GENETICS Reading Assignment 11.1, 11.2 and
11.3 In this lecture we will introduce the ways
in which the genetic information is passed
between generations and how it can be expressed
in terms of models. The variations produced by
sexual reproduction serves as a basis for
evolutionary selection, preserving the most
desirable properties in a particular
environmental context. Genetic material is mixed
in sexual reproduction. This may result
off-springs that are different from
parents. Ultimate source of variation is mutation
of genes. Mutation takes place after many
generations. As such there are two types of
reproductions Asexual cell reproduction (
Mitosis)- Asexual reproduction of a
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cell results from copying and equal distribution
of genetic material of a single cell. In such
case, each resultant daughter cell possesses the
same genes as the parent cell. However, there
could be mutation after several generations. If
we consider multi-cellular environment, the
daughter cell may take functions different from
the parent cell by selectively turning genes
off. This process creates various tissues of a
multi-cellular organism. (for examples cells of
skin, liver, kidney, nerve-tissues are
multi-cellular organisms and are reproduced by
asexual reproduction and take different
functions). Some cells rarely divide muscle
cells, fat cells and cells of central nervous
system. Asexual reproduction can generate
daughter cells that may differs from each other(
example above).
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Sexual Reproduction ( Meiosis and
fertilization)-sexual reproduction involves the
creation of an offspring that contains genetic
contribution from two parents. A type of cell
division called meiosis halves the chromosome
number of a germinal cell to produce sperm or
egg. A sperm and egg combine in fertilization to
restore the double chromosome number. The new
offspring has genetic information from two
sources for every characteristic. The ways in
which these two sets of information combine to
produce a single property is complex and is
studied through classical genetics. Human
chromosome consists of 23 homologous pairs or two
sets of 23 each. We get one set each from our
parents. In all there are 46 chromosomes in a
single human cell. This number is found in most
of the cells in human body.
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Exceptions are red blood cells- they lack nuclei,
and cells of reproduction systems - known as
germinal cells. Any cell in our body, which are
not germinal are called somatic cells. This
category includes virtually entire bulk of our
body. In Somatic cell the chromosome number is 46
( 23 pairs). These cells include skin, blood,
nervous system, muscle cells, etc. A sperm and
egg combines in fertilization process to restore
the double chromosome number. The new offspring
now has genetic characteristics from both the
parents. This is contributed by all 23 elements
from each cell.
adult
sperm
sperm
fertilization
mitosis
zygote
egg
adult
egg
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• A special kind of cell division called meiosis
  creates gametes having half the number of
  chromosomes found in a somatic cell.
• The chromosomes are not partitioned at random,
  however, every gamete winds up with with exactly
  one random representation of each homologous
  pair, giving it one basic set of 23 chromosomes.
• Each such cell is called a haploid
• A cell that has two basic set of chromosomes is
  said to be diploid. Somatic cells are diploid
  and germinal cells are haploid.
• Meiosis in humans converts a diploid cell ( ch 46
  ) to haploid ( ch 23).
• Classical Genetics
• Describes the ways the genetic material of two
  parents combines to produce a single observable
  property.

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Let us consider a simple example, a red flowered
plant and a white flowered plant usually produce
an offspring with a single color in its flower.
What that color will be is not predictable unless
a geneticist has already studied flower-color in
that plant - because there are about a dozen ways
that parental genes can combine.
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Now let us cross a true-breeding red flower with
a true- breeding white flower. The genetic
information of red-flower color is symbolized by
R and each plant has a pair of chromosome (
homologous pair). One from each parent. Similarly
for the white ( true-breeding ) plant, the
genetic information is represented by w. Meiosis
produces gamete containing one and only one
chromosome, representative of each homologous
pair. A gamete from each parent combine at
fertilization to re-establish diploid condition (
R with w ). The offspring has flower color
genetic information Rw and it turns out that the
pea- plant produces only red flower,
indistinguishable from the original true-breeding
red flower.
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• Evidently, red has masked-out white. Red is
  dominant to white or white is recessive to red.
• At this stage let us define some new terms
• Variant form of information for one property (
  from chromosome)- symbolized by R and w are
  called alleles - color alleles.
• The allelic composition is the organisms
  genotype RR or WW are called homozygous and
  genotype Rw as heterozygous.
• What the organism actually looks like - red or
  white - phenotype
• Thus, the initial cross ( parental) between
  homozygous red and homozygous white, results F1
  generation - which is heterozygous red flowered
  plant.

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• To obtain F2 generation, self-cross the F1
  generation , which is crossing it with one just
  like itself ( Rw cross with Rw).
• You would obtain 1 RR, 1 ww and 2 Rw, with a
  ratio of red to white color is 31.
• ( see figure on the next transparency.)
• An experiment just described, involving a single
  property, like the color of the flower, is called
  mono-hybrid cross.
• You can also make a di-hybrid cross involving two
  properties flower color and stem length, which
  are unlinked ( independent).
• R red flower, w white flower
• L Long Stem, s short stem
• (a) Parental generation
• RRLL X wwss

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Parental gametes
RL
ws
Rw Ls
(b) Self cross Rw Ls X Rw Ls F-1 Gametes will
be RL RL wL wL Rs Rs ws ws F2
Generation after self cross will be RL
wL Rs ws RL RRLL RwLL RRLs RwLs wL RwLL ww
LL RwLs wwLs Rs RRLs RwLs RRss Rwss ws RwLs
wwLs Rwss wwss
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• How many Red and White flower ratio? 12 4
  31
• The details are Red flower - Long stem - 9
• Red flower - short stem - 3
• White flower - long stem - 3
• white flower - short stem - 1
• The overall ratio is 9 3 3 1 , long stem
  short stem also 3 1
• The ratios are known as phenotype ratios.
• Note that including the second property in our
  modeling process has not affected the flower
  color ratio.
• We had expected this, as we know that these two
  properties are un-related or are independent.
• The ratios 31 and 9 3 3 1 are also known as
  Mendelian ratios.

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• The ultimate cause of the genetic variation is
  mutation of genes. However, there is a small
  chance that a given locus ( property) will mutate
  between two generations.
• The possible combination of chromosomes in
  gametes for two pairs are 22 4.
• For human beings there are 23 pairs which gives
  2 23 possible combinations chromosomes in their
  gametes. This explain the variations in human
  beings in a family - even in various generations.
  These variations are not due to mutation of
  genes.
• It is complex to study human genetics. Let us
  consider very simple case.
• Consider a single locus for which there are two
  alleles, say A and a.
• Hence there will be three distinct genotypes AA,
  aa and Aa. AA and aa are homo-zygotes and Aa
  hetero-zygotes.

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If one parent is AA and the other is Aa, then the
possible zygote genotypes resulting from a mating
can be represented by event tree as follows
start

contribution by first parent
A
A
1/2
1/2
1/2
1/2
A
contribution by second parent
A
1/2
a
a
1/2
AA
Aa
AA
Aa
Note that AA, Aa , aa and aA ( same as Aa) have
equal probabilities. Pr ( AA ) 1/2 , Pr ( Aa
) 1/2
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Random Allelic Combination Let nAA denote the
number of AA genotypes in a population and
likewise naa denote the number aa genotypes.
Also nAa and naA denote Aa and aA types of
genotypes respectively. Note that Aa type
genotype and aA type are the same. So we simply
use Aa type genotypes for both the
variations. The size of the entire population is
given by N nAA naa 2 nAa Now let na
and nA denote the number of a and A alleles,
carried by the population. Then na nA 2 N (
since there are 2 alleles in each genotype) Let
pAA, paa, and pAa denote their corresponding
fractions of population. Then pAA paa pAa
1
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Similarly, Let pa and pA be fraction of
populations of alleles of types a and A
respectively, then, pa pA 1 pA nA/2N
(2 nAA 2 nAa)/ 2N pAA pAa pA pAA
pAa Similarly, ,pa paa pAa Now,
imagine all the alleles of the population are
pooled and two are selected at random to form one
pair. Assume that the pool is large enough, so
that the removal one allele would not affect the
probability of subsequent selection. The
probability of selecting AA pair is given
by pA.pA., similary for the aa pair is pa. pa
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Male gametes
Female
A(pA)
pA.pA
pA.pa
a(pa)
pA.pa
pa.pa
This table shows Mendelian inheritance
probability. Random mating of allelic fraction in
terms of probabilities is given by( it is also
known as Mendelian inheritance probabilities
) AA x AA pAA . pAA AA x Aa 2pAA
. paA AA x aa 2pAA . paa Aa x Aa
2pAa . pAa Aa x aa 2pAa .
paa aa x aa 2paa . paa
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Suppose initially 70 0f population is AA, (pA
0.7), and 30 0f population is aa, (pa
0.3). After one generation the probabilities of
various combinations will be prob of AA 0.70 x
0.70 0.49 prob Aa 2. (0.7)(0.3) 0.42 prob
of aa (0.3)(0.3) 0.09 total prob is
1.00 now suppose that male has chromosome pair
XY ( non-homologous pair) and female has XX
(homologous pair). Let us consider color
blindness as one of the properties. Color
blindness chromosome in male ( X- chromosome)(pX
0.08) This property in female ( with XX
chromosome ) will be pX. pX
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• 0.08 x 0.08 0.0064 0.64 ( actual
  figure is lower, which can be explained by the
  theory).
• Now let us consider sex linked genes allelic
  fraction that are different between males and
  females ( let us say that we are considering this
  property in X- chromosome).
• Let, F fraction of alleles A in females
• then f ( 1-F) will be fraction of a alleles
  in females.
• Similarly, Let M fraction of alleles A in
  males
• then m ( 1-M) will be fraction of a alleles
  in males.
• Assume that there are equal number of males and
  females.
• Then, population fraction of alleles A will be
  pA (MF)/2.
• Similarly, pa (m f)/2. Check that pA
  (1-pa).
• This fraction remains constant from generation
  to generation.

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• However, the number of M and F may change from
  generation to generation.
• To follow these fractions, through several
  generations, we have to keep track of F and M ( f
  and m can be calculated from F and M).
• Let Fn and Mn refer to the generation n, with n
  0, corresponding to generation 0 the initial
  existing fraction of F and M.
• Mn1 Fn ( male have chromosome XY). They
  get X from female.
• Fn1 1/2(Fn Mn ). Females have chromosome
  XX, one each from male and female.
• Frequency in females will be average of two sexes
  in proceeding generation. Since each sex
  contributes one X chromosome.

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Yn1 T. Yn
Let us consider particular alleles in male (
X-chromosome allele is different coming from a
different geographical location). XY XX male f
emale ( females do not have that allele in the
current generation) after first generations, they
will get X(mother)Y(father) in males and X (
mother) X ( father) in females. This particular
allele will be transferred in all females. Male
gets his X-chromosome from mother. Males in first
generation would not have this allele at all.
Apply the equation Xn1 Xn T You can use above
equation for future generations.
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According to Mendels theory - characters in
individuals are determined by discrete genetic
units that are inherited down through
generations. Mendels First Law The two members
of the gene pairs segregate (separate) from each
other into gametes, such that one-half of the
gametes carry one member of the pair and the
other one-half of the gametes carry the other
member of the gene pair. For example Aa gene-pair
will have gametes A and a respectively. The
union of one gamete from each parent to form the
first cell (called as zygote) of a new progeny
individual is random.
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Example Parent AA x aa sex cells A
and a (gametes) F1 Aa ( zygote) Aa x
Aa F2 generation 1/2 A 1/2 a 1/2 A 1/4
AA 1/4 Aa 1/2 a 1/4 Aa 1/4 aa The ratio of
the new genetic pairs will be AA Aa aa
121 Mendels Second Law During the gamete
formation, the segregation of one gene pair is
independent of other gene pairs.
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Let us consider a pea plant with two properties(
di-hybrid ratios). Round and wrinkled seeds (
R,r) Yellow and green seeds ( Y, y) Assume that
the parents are pure classes 1. Round and yellow
( RR YY) 2. Wrinkled and green ( rr yy
) Wrinkled and green properties are recessive to
round and yellow respectively or round and yellow
colors are dominant to wrinkled and
green. RR red YY yellow Rr red rr wrinkled y
y green Yy yellow
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Pure classes are RR YY x rr yy. Let us
cross them. The gametes will be RY and ry in
equal numbers the first generation will be
RY ry RY RRYY Rr Yy ry Rr Yy rr
yy The F1 generation will consist of RRYY, Rr Yy
and rr yy genes in the ration of 121, and 3/4
will be round yellow and 1/4 wrinkled green. Now
let us consider only the class Rr Yy and self
cross it F1 Rr Yy x Rr Yy The gametes will
be RY, Ry, rY and ry p(RY ) 1/2 x1/2 1/4
p ( Ry) p(rY) p(ry)
Rr
Yy
Yy
RY Ry rY ry
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The F2 generation can be represented by a grid (
known as punnett). Table Symbolic representation
of genetic and phenotype constitution of Fe
generation.
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The ratio of round yellow, round green, wrinkled
yellow and wrinkled green will be 9 3 3
1. This concept can be extended to tri-hybrid
ratios ( such as Aa Bb Cc x Aa Bb Cc ) We ask
what portion of progeny from the above cross will
be AA bb Cc Answer p( AA bb Cc) p(AA) x
p(bb) x p(Cc) 1/4 x 1/4 x 1/2 1/32
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A branch diagram is very useful for solving some
problems. For example the 9331 phenotypic
ratio can be obtained from branch diagram. F1
Rr Yy are selfed Composition of F2 By product
rule 3/4 of these 9/16 round yellow 3/4
round will be yellow(Y-) (R- Y-) ( R-) 1/4 of
these will be 3/16 round green green (
yy) ( R- yy) 3/4 of these will be 3/16
wrinkled yellow 1/4 wrinkled yellow ( Y-) (rr
Y- ) (rr) 1/4 of these will be 1/16 wrinkled
green green( yy) ( rr yy)