ICT Role in 21st Century Education & its Challenges.pptx
Genetic Competition Exam
1. Dr. Tahir Abbas Baloch
Ph.D. Zoology (Punjab University, Lahore)
tahirabbasbloch@gmail.com 03339810179
Dr. Tahir Abbas Baloch You Tube
https://www.youtube.com/channel/UC-
9mb8kyILhPKKl9PwAdvbw?view_as=subscriber
Previous Lecture Genetics: 10, 11,12
Genetics, Mendelian Inheritance
Dr. Tahir Abbas Baloch
2. Dr. Tahir Abbas Baloch You Tube
https://www.youtube.com/channel/UC-
9mb8kyILhPKKl9PwAdvbw?view_as=subscriber
Dr. Tahir Abbas Baloch
MDCAT, NUMS, PPSC
13
3. A A
B b
HOMOZYGOUS
HETEROZYGOUS
HETEROZYGOUS (AABb)
A A
bB
A A
bB
A A
bB
Un – replicated
Replicated
ALLELES
NOT- ALLELES
NOT- ALLELES
Bacteria, have a single ring of DNA, have one allele per gene per organism
Chiasma/Synaptonemal Complex
Synapse/Synapsis
Centromere
Primary
Constriction
Sec Constriction
Junk Gene –non coding/Heterochromatin/Euchromatin
TELOMERS TELOMERS
Dr. Tahir Abbas Baloch
coupling/cisphase
trans phase
REPULSION PHASE
Crossing Over
Recombinants
*Coupling
Reaction
Sister Chromatids
Daughter Strands
Replicated Forms
transposonorjumpinggene
4. GENES A, B, D, E, F, H, I, R, T, W, X, Y, Z, Rh, SRY, W+…… Dominant Genes
a, b, d, e, f, h, i, r, t, w, X, Y, Z, Rh, SRY, ……….…… Recessive Genes
AA, BB, DD, EE, FF, HH, II, RR, TT, W+ W+, XX, YY, ZZ, …… Allele
Aa, Bb, Dd, Ee, Ff, Hh, Ii, Rr, Tt, W+ w, Xx, Yy, Zz, …………. Allele
aa, bb, dd, ee, ff, hh, ii, rr, tt, w w, xx, yy, zz, ………………... AlleleALLELES SINGLE ALLES
AA1A2a, BB1b, CC1C2C3, D1D2D3D4 --------------
AaBbCc ---------------------------------------------------- POLYGENES/ Polygeny/Polygenic
AA, Aa, aa,
A1A1, AA2,
A1A2, AA1
A1a, A2a
Gene Pool of Populations
A, A1, A2, a
4
Individual = 9
Genes= 2n= 18
AA, Aa, aa,
A2A1, AA1,
A1A1, AA1
A1a, aA1
Gene Flow
A2
AA
A2A1
Bottle Neck Effects
Genetic Drift
AA, AA2, A2A2, A1A1,
AA2, A1A2, AA1, A1A,
A2A, AA, AA2, A2A2,
A1A1, AA2, A1A2, AA1,
A1A, A2A, A1A2, AA1
A, A1, A2
3
Population Density
PopulationPressure
A
Gene
a
A1
A2
Alles
Mutation
Genetic Diversity
Gene Pool DeclineDr. Tahir Abbas Baloch
DOUBLE ALLES
MULTIPLE ALLES
5. GENE:
unit of heredity transferred from parent to offspring.
sequence of nucleotides in DNA /RNA determines order of monomers in polypeptide
basic physical and functional unit of heredity.
unit of hereditary information that occupies a fixed position on a chromosome.
bacteria gene is more complex than eukaryotes.
plasmids, contain (usually) non-essential genes.
viral genes are mostly present in a single molecule of nucleic acid.
Females have more genes & more complex genome than male due to XX
LOCUS is a position of genes on a chromosome.
GENE MARKER
(Probe)
DNA MARKER
Molecular Marker
Morphological
Marker
Transposable element (transposon, or jumping gene) DNA sequence that can
change its position within a genome.
Gene silencing is regulation of gene in a cell to prevent their expression.
LOCUST Dr. Tahir Abbas Baloch
Wilhelm Johannsen coined the word “ GENE "
6. Wilhelm Johannsen describe the Mendelian units of heredity (gene), expressed as phenotype and
genotype.
Mendel is Considered Father of Genetics
SelfingDusting
Dr. Tahir Abbas Baloch
7. Mendel First Law
Mendel Law of Inheritance
Law of Dominance
Monohybrid Law
Law of Purity of Gametes
Phenotypic Ratio 3: 1
Genotypic Ratio 1: 2: 1
Genotypic Ratio of Homozygous Dominant 1/4
Genotypic Ratio of Heterozygous 2/4 = ½= 0.5
Genotypic Ratio of Homozygous Recessive 1/4
Dr. Tahir Abbas Baloch
9. Dr. Tahir Abbas Baloch
Mendel First Law
Round = ¾ Yellow = ¾
Wrinkle = ¼ Green = ¼
Product Rule
Round x Yellow = ¾ x ¾ = 9/16
Round x Green = ¾ x ¼ = 3/16
Wrinkle x Yellow = ¾ x ¼ = 3/16
Wrinkle x Green = ¼ x ¼ = 1/16Tall = ¾ Short = ¼
Round x Tall = ¾ x ¾ = 9/16
Round x Short = ¾ x ¼ = 3/16
Wrinkle x Tall = ¾ x ¼ = 3/16
Wrinkle x Short = ¼ x ¼ = 1/16
The probability of two or
more independent events
are equal to the product of
their individual probabilities.
10. 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1 genotypes = 9 Total = 169 : 3 : 3 : 1 phenotypes = 4
F 2
F 1
P 1
LAW OF INDEPENDENT ASSORTMENT
Male gametes 4
Femalegametes4
4 x 4 = 16
Meiosis
Gametogenesis ½ 0.5
DihybridSelf Cross F1 x F1
11. AaBb
Dihybrid
aabb
Recessive
AB, Ab, aB, ab
Gametes
ab
Gametes
AaBb, Aabb, aaBb,aabb
1 : 1 : 1 : 1
P 1
F 1
Diallelic Characters
Dihybrid Characters
Test Cross
Complete Dominance
Law of Independent
Assortment
Single Crossing Over
Recombination
Dr. Tahir Abbas Baloch
14. Ratio, Proportion, Probability
Monohybrid First Law Phenotype in F 2 is 3:1, Genotypic 1:2:1
Dihybrid/Second Law Phenotype 9:3:3 :1, Genotype 1:2:2:4:1:2:1:2:1
Incomplete Dominance Phenotype & Genotype 1:2:1
Trihybrid/Polygeny Phenotype 1:6:15:20:15:6:1
Monohybrid First Law, Phenotype in F 1 is 1:1:1:1
Dihybrid Second Law, Phenotype in F 1 is 1:1:1:1
Trihybrid Second Law, Phenotype in F 1 is 1:1:1:1
Test Cross for Homozygous is always 100%
Test Cross for Heterozygous is # 100% but not more than 50 %
15. T t Aa Bb Rr Yy Complete Dominance
Dominant 100 % in F1, Recessive in F 1 0 %
R1 R2 R1R2 In Complete Dominance
Dominant 50 % in F1, Recessive in F 1 50 %
A B AB
Dominant 100 % in F1, Recessive in F 1 100 %
Co-Dominance
A a
Traits - I
Traits - II
Pleiotropy
Chromosome 9 Chromosome 19
EPISTASIS Non-Homologous Dominance
Polygeny/Contineously Varying Traits
AaBbCc /Trihybrids
F 1
F 2
1:6:15:20:15:6:1
Phenotypes = 7
Multiple Allele
Blood Group
Gene ABO
Dr. Tahir Abbas Baloch
16. Dr. Tahir Abbas Baloch
Sutton and Boveri, states chromosomes are the vehicles of genetic heredity; beave as
segregation, independent assortment, and linkage.
Chromosomal Theory
Gene theory - genes are the basic units in which characteristics are passed from one
generation to the next.
Neutral Gene theory, the behavior of mutant genes in populations is determined by
random genetic drift/bottle neck effect.
Thomas Hunt Morgan won the Nobel Prize in 1933 prove that rule by Drosophila.
17. LINK
Gene to Gene Linkage Gene to Chromosome Link
Genetic Linkage
Linkage Group
Homologous Pairs
Genes of Chromatid
Autosomal Linkage Sex Linked Genes
Male = Female Male # Female
X-Linked Genes Y-Linked Genes
Only in MaleX-Linked Dominant Genes
X-Linked Recessive GenesMale ≥ Female
Female ≥ Male
Pseudoautosomal Gene
on mammalian X and Y
chromosomes e.g.
Mammary Gland Genes
Hemophilia
Red-green
Blindness
White Eye
Color in
Drosophila
Baldness
Hypophosphatemic ricket
Blood Group Gene
18. X X
W+ W+
Female Red Eye
X X
W+ w
Hetrozygous, Carrier Female Red Eye
X Y
W+
Male Red Eye
X X
w w
Female White Eye
X Y
w
Male white Eye
X X
X , X
Eggs
X Y
X , Y
Sperms
HOMOGAMETIC HETROGAMETIC
WILD TYPE (Red Eye)
AUTOSOME
A = 2n-2
Sex Chromosome
ALLOSOMES
19. X X
W+ W+
Female Red Eye
X X
W+ w
Female Red Eye
X X
w w
Female White Eye
X Y
W+
Male Red Eye
X Y
w
Male White Eye
Female/Male Red Eye Female/Male Red Eye
2 Female/1 Male Red Eye
2 Female Red Eye
2 Male White Eye
1 Male White Eye
Female/Male Red Eye
Female / Male White Eye
Female / Male White Eye
Parental Cross Test Cross
Test Cross
Back Cross
Reciprocal Cross Confirmatory Cross
Self Cross = F 1 x F 1
= F 2 x F 2
X X
W+ W+
Female Red Eye
X X
W+ w
Hetrozygous, Carrier Female Red Eye
X Y
W+
Male Red Eye
X X
w w
Female White Eye
X Y
w
Male white Eye
Dr. Tahir Abbas Baloch
20. X X
W+ w
Hetrozygous, Carrier Female Red Eye
X Y
w
Male white Eye
X X
W+ W+
Female Red Eye
X Y
w
Male white Eye
Parental Cross P 1
X , X
W+ W+
X , Y
w
X X
W+ w
Hetrozygous, Carrier Female Red Eye
X Y
W+
Male Red Eye
SelfCrossF1xF1
X , X
W+ w
X , Y
W+
X X
W+ W+
Female Red Eye
X Y
W+
Male Red Eye
X X
W+ w
Hetrozygous, Carrier Female Red Eye F 2
X X
W+ w
Hetrozygous, Carrier Female Red Eye
X Y
w
Male White Eye
TEST CROSS
X , X
W+ w
X , Y
w
Gametes by Meiosis
X X
W+ w
Hetrozygous, Carrier Female Red Eye
X Y
W+
Male Red Eye
X X
w w
Female White Eye
X Y
w
Male White Eye
X Y
w
Male White Eye
RECIPROCAL
CROSSX X
w w
Female
White Eye
X Y
W+
Male Red Eye
X , X
w w
X , Y
W+
P1 = F(R) x M (W) → 2 F (R,C), 2 M (R)
Parental Cross 2:2, 1: 1, 100 %
F1xF1 = F(R,C) x M (R) → 1 F (R,C), 1 M (R)
Self Cross 1 F (R) , 1 M (W)
3:1 (R,W)/75%, 25%, 100% (F), 50% (M)
Test Cross = F(R,C) x M (W) → 1 F (R,C), 1 M (R)
1 F (W) , 1 M (W)
2:2 (R,W) or 1: 1 50% (F), 50% (M)
Reciprocal Cross = F(W) x M (R) → 2 F (R,C), 2 M (W)
1:1 (R,W) or 50%, 100% (F), 100% (M white)
One gene one enzyme
One gene polypeptide
Sex Determination
Chromosomal Theory
Gene Theory
Sutton Theory
T.H Morgan Work
X-Linked Recessive Trait Page Prepared By Dr. Tahir Abbas Baloch
F 1
21. X X
W+ W+
Female Red Eye
X Y
w
Male White Eye
Gametes
X , X
W+ W+
X , Y
w
X X
W+ w
Female Red Eye
Carrier
X Y
W+
Male Red Eye
2 F (R) : 2 M (R)
1 : 1
overall 100% Red eyes
ParentalCross
Reciprocal Cross
Phenotypically inverse to P1
Male Red Eye
overall 50% Red eyes
Female 100% Red
Male 100 % White Eye
Test Cross
Test Cross
22. X X
H h
Heterozygous, Carrier Female
X Y
h
Hemophiliac Male
X X
H H
Normal Female
X Y
h
Hemophiliac Male
Parental Cross P 1
X , X
H H
X , Y
h
X X
H h
Heterozygous, Carrier Female
X Y
H
Normal Male
X , X
H h
X , Y
H
X X
H H
Normal Female
X Y
H
Normal Male
X X
H h
Heterozygous, Carrier Female F 2
X X
H h
Heterozygous, Carrier Female
X Y
h
Male Hemophiliac
TEST CROSS
X , X
H h
X , Y
h
Gametes by Meiosis
X X
H h
Heterozygous, Carrier Female
X Y
H
Normal Male
X X
h h
F Hemophiliac
X Y
h
M Hemophilic
X Y
h
Male Hemophiliac
RECIPROCAL
CROSSX X
h h
Female
Hemophiliac
X Y
H
Normal Male
X , X
h h
X , Y
H
P1 = F(N) x M (H) → 2 F (N,C), 2 M (H)
Parental Cross 2:2, 1: 1, 100 %
F1xF1 = F(N,C) x M (N) → 1 F (N,C), 1 M (N)
Self Cross 1 F (N) , 1 M (H)
3:1 (N,H)/75%, 25%, 100% (F), 50% (M)
Test Cross = F(N,C) x M (H) → 1 F (N,C), 1 M (N)
1 F (H) , 1 M (H)
2:2 (N,H) or 1: 1 50% (F), 50% (M)
Reciprocal Cross = F(H) x M (N) → 2 F (N,C), 2 M (H)
1:1 (N,H) or 50%, 100% (F), 100% (M H)
One gene one enzyme
One gene polypeptide
Sex Determination
Chromosomal Theory
Gene Theory
Sutton Theory
T.H Morgan Work
X-Linked Recessive Trait Page Prepared By Dr. Tahir Abbas Baloch
F 1
23. Gene recombination is naturally occurring during meiosis in eukaryotes, involves the pairing of homologous
chromosomes and information transfer between the chromosomes.
A
B
a
b
AB ab Ab aB
AB ab
Non-Parental α Crossing Over α Recombination α Genetic Map α Gene Distance
A B
20
A B
20
C
15
35
A B
15
A B
15
C
?
55 ? = 40
A
B
a
b
A
B
a
b
A
B
Synapsis Chiasma
A
B B
a
AB aB
a
bb
A
Ab ab
NP (recombinant)
P P
a
b
a
b
ab
P
AB aB Ab ab
abAa
Bb
aa
Bb
Aa
Bb
aa
bb
PP NP NP
Recombinant Frequency = Σ Non-Parental x 100 = Genetic Map
Total
4x1=4
P = 40+40=80 NP = 10+10=20
24. A
B
AB
O
IA IA
IA i
IB IB
IB i
IA IB
ii
+
DD
Dd _
dd
+
DD
Dd _
dd
+
DD
Dd _
dd
+
DD
Dd _
dd
PHENOTYPES
GENOTYPE (ABO)GENOTYPE (ABO-Rh)
12
6
4
Dr. Tahir Abbas Baloch
26. I I
A A
I i
A
I I
B B
I i
B
I I
A B
i i
H H
H h
h h
A
A
O
B
B
O
AB
AB
O
O
O
O
9
19
EPISTASIS
BOMBAY
PHENOTYPE
ABO Group
27. I I
A A
I i
A
I I
B B
I i
B
I I
A B
i i9
I I
A A DD Dd dd DD Dd dd DD Dd dd DD Dd dd DD Dd dd DD Dd dd
I i
A
I I
B B
I i
B
I I
A B
i i
1
DD,Dd,dd
DD,Dd,dd
DD,Dd,dd
DD,Dd,dd
DD,Dd,ddDD,Dd,dd
Dr. Tahir Abbas Baloch
29. Dr. Tahir Abbas Baloch
DD dd
D d
Dd
+
-
+
+
+
+
anti-Rho(D) test / positive
Dd dd
D, d d
DD, Dd
+
30. AA : XO M
AA : XY M
AAA : XY Metamale, Suppermale
AA : XX F
AA : XXY F
AA : XXX F
Sex Determination in Drosophila
X : A Genic Ratio System
Humans: XXY = M
Humans: XO = F
31. Ascaris contains 24 bivalent chromosomes,
n = 24 male 19 bivalents and 5 univalent.
Male 35 , Autosome 13 pairs AA , X = 8, Y
Gametes 13A + 8X, 13A + Y
Female 42, Autosome 13 pairs AA, X = 16
Gametes 13A + 8X XO in Drosophila is sterile male
XO in Man is Turner Syndrome is sterile Female
XO in Bug is Fertile Male
Drophila male genotype XY, XO
AA : XO M (sterile male)
AA : XY M
AAA : XY Metamale, Sterile, Suppermale
AA : XX F
AA : XXY F (fertile female)
AA : XXX F (Metafemale)
Metafemale A metafemale is a low viability Drosophila fruit
fly with a female phenotype in which the ratio of X
chromosomes to sets of autosomes exceeds .
Metamale (supermale) is a low viability Drosophila fruit fly
with a male phenotype in which the ratio of X chromosomes
to sets of autosomes (A) is less than 0.5.
COMPOUND SEX CHROMOSOMES
DimorphicBivalent is one pair of chromosomes in a tetrad,
with crossing over.
Univalent chromosome that is unpaired in
prophase 1 and not show crossing.
These are POLYPLOIDS with odd numbers
of chromosome sets.
32. Dr. Tahir Abbas Baloch
What is gene ?
What is gene action?
What is gene action in Neurospora?
What is gene action in Man?
What is one gene one enzyme ?
What is one gene one protein ?
What is one gene polypeptide?
Prove that proper position of amino acids is essential for proper activity of proteins ?
Write the genetic base/reason of sickle cell anemia?
Why Neurospora was selected for genetic experiments?
33. Dr. Tahir Abbas Baloch
E1
B C D ProductA
E2 E3 E4
1 2 3
PRECURSOR
NEGATIVE/FEED BACK INHIBITOR
Gene Gene Gene
Chromosome
On
Off
AutomaticSwitch
Water
Flow
A B CAll Ok
Fault Ok
Fault Fault
Ok
Fault Fault Fault
34. Dr.TahirAbbasBaloch
NEUROSPORA
Ascomycetes
Ascosporangium/Ascus
8 - Ascospores
X-Rays
Non-Mutant/Normal Grow in Minimal Medium
Vitamins, Sugar, Salt, Water
LETHAL DEAD SPORE NEVER GROW
Mutant Culture Not Grow in Minimal
Medium
SUPPLEMENTS ?
Carbohydrates Nucleic
Acids
Lipid Amino Acids- Proteins Arginine
Histidine
Proline
Methionine
Leucine
Tryptophan
Valine
Lysine
Glycine
Arginine
Ornithine Citruline Arginine
41. Dr. Tahir Abbas Baloch
N 14
Cesium
Chloride
MESELSON & STAHL EXPERIMENT
N 14
N 15
N 15
N 15 N 15 N 15
N 14 N 14 N 14
t~ 0 sec T= 20 Mint T= 40 Mint
REPLICATION
OF
DNA
44. Dr. Tahir Abbas Baloch
Non Template or Coding Strand or Sense Strand
Anti-Coding Strand or Anti - Sense Strand or Template Strand
A T C G A A T T C C C G G
T A G C T T A A G G G C C
A T C G
T A G C
A U C G A A U U C C C G G
A U C G
Sense code, Non sense Code, Codon
DNA
DNA
m RNA
Anti sense Code – t RNA
Decode
A U C
U A G
t RNA
(m RNA)
(t RNA)
5
3
3’ 5’
Towards 5' end of the sense strand is upstream
Towards 5' end of the RNA is upstream.
5’
3’
5’ 3’
TRANSCRIPTION
45. Dr. Tahir Abbas Baloch
Bacteria, sequence at -35 has TTGACA sequence
Promoter
PROMOTER SEQUENCE PROKARYOTES EUKARYOTES
TATAAT (TATA box) -10 -25
TTGACA (CAAT box) -35 -75, -70
Transcription of DNA begins for a particular gene (i.e., positions
upstream are negative numbers counting back from -1
Telomeres is TTAGGG
Palindromic Sequence is ACCTAGGT
Towards 5' end of the sense strand/Coding is upstream
Towards 5' end of the RNA is upstream.
Towards 3‘ end of the template strand is upstream
46. Dr. Tahir Abbas Baloch
DNA
Promotor
TTGACA-- TAC ACC GGT TAA ATC
AUG UGG CCA AUU UAG
m-RNA
15 Nucleotides
5 Genetic codes
4 Amino acids
Tripeptide Genetic codes = Amino acids
(without stop code)
Genetic codes = Amino acids +1
(with stop code)
21 Nucleotide
TRANSCRIPTION
47. Dr. Tahir Abbas Baloch
mRNA
5’
Leader Sequence
Leader RNA
AUG
Start Code
UAG
60 S
40 S
80 S
APE
1 Amino acyl t RNA ase Amino Acid
Charge t RNA
Methionine1
Anticode
Initiating Factor
5’
AUG UAG
APE
2 Amino acyl t RNA ase
2 Amino Acid
Charge t RNA
1
Initiating Factor
UAC
UAC
Decoding
2 Charge t RNA
2 Amino Acid
TRANSLATION
INITIATION
48. Dr. Tahir Abbas Baloch
5’
AUG UAG
APE
3 Amino acyl t RNA ase
3 Amino Acid
Charge t RNA
1
Initiating Factor
UAC
Decoding
3 Charge t RNA
3 Amino Acid2 Charge t RNA
Peptide Bond
2
5’
AUG UAG
APE
n Amino acyl t RNA ase
n Amino Acid
Charge t RNA
1
Initiating Factor n Charge t RNA
n Amino Acid
Peptide Bond
2 Stop Code
Elongation Factor
1
3
2
3
ELONGATION
49. Dr. Tahir Abbas Baloch
5’
AUG UAG
APE
n Amino acyl t RNA ase
1
Initiating Factor
Peptide Bond
2 Stop Code
Elongation Factor
2
3
3
4
APE
1
Free Polypeptide
n
n
n
Releasing Factor
TERMINATION
UAG
UAA
UGA
Stop Code
Non Sense
3
Total Genetic Codes 4
n
Total Genetic Codes 64
Total Sense Codes 61
Start Code AUG
50. Dr. Tahir Abbas Baloch
Adenine
C5H5N5
Purine
C5H4N4
Guanine
C5H5N5O
CytosinePyrimidine Thymine Uracil C5H4N4O3
Uric Acid
51. 1. The ___________ is considered as father of genetics.
a) Darwin c) Lamarck d) Malthusb) Mendel
1. The ___________ is considered as father of nucleic acid.
a) Darwin b) Mendel c) Lamarck
1. The physical and structural unit of inheritance is _________
a) DNA b) Chromosome c) Gene d) Nucleotide
The __________ is the basic unit of chromosome and chromatin.
a) Nucleotide b) Nucleosome c) Gene d) Genome
The genetic code is AUG, then its decoding anticode is ________
a) TAC b) ATC c) UAC d) UAG
Which of the following is a correct definition of genetics?
Transmission of traits from parent to offspring
Study of variation between members of a species
Study of DNA Genes and traits determined by genes
If coding strand is AATTCCGCTA then mRNA will be __
a) AAUUCCGCTA b) TTUUGGCGAT c) AAUUCCGCUA
d) TTAAGGCGAT
1. The _____________ store information for protein synthesis.
DNA Cytoplasm r RNA t RNA
d) Friedrich Miescher
Dr. Tahir Abbas Baloch
62. Fingers / Toes = 5
Bone in Left Palm = 5
Bone in Right Palm = 5
Bone in Left Sole = 5
Bone in Right Sole = 5
Lumbar Vertebrae = 5
Kingdoms of Bio = 5
Prime End of DNA = 5
Head of m RNA = 5
Common Senses = 5
3rd Prime Number = 5
Starfish Appendages = 5
Pentemerous Petals = 5
Pentadactyl = 5
Lobes of Lungs = 5
Adenine = C5H5N5
Pillar of Islam = 5
Namaz/Azan = 5
Panjantan e Pak = 5
Ribs on Left Side = 12
Ribs on Right Side = 12
Thoracic Vertebrae = 12
Pair of Cranial Nerves = 12
Light Reactions = 12
Water Used = 12
Non Cyclic Reactions = 12
ATP Cyclic Reactions = 12
NADH Cyclic Reactions = 12
Photolysis = 12
O atom produced = 12
Duodenum length inch = 12
One foot have inches = 12
Heart Chakra Petals = 12
Max wind speed hurricane = 12
Months of Year = 12
Footballs Players = 12
Hours = 12
Imam = 12
Sure-e-Yusuf = 12
Facial Bones of Man = 14
Phalanges of Left Hand = 14
Phalanges of Right Hand = 14
Phalanges of Left Foot = 14
Phalanges of Right Foot = 14
Pea Plant Chromosomes = 14
Woodlouse legs = 14
Catalan Number = 14
Companion Pell Number = 14
Atomic Wt of Nitrogen = 14
Electrons of f = 14
Holy Helpers = 14
Muqatta (Quran) = 14
Masoomen (Infallibles) = 14
5 12
14
Dr. Tahir Abbas Baloch
