Lec 2

• Mamallian Cells have the primary function to reproduce and grow into multicellular organisms
• Cell division and differentiation

  1. Mitotic process

  2. Retaining exact copies of parental genome
• Gametogenesis


  1. Spermatogenesis

  2. Oogenesis

  3. Meiotic process

  4. Reduce from 2N to N chromosomes

  5. Formation of haploid cells from diploid parental cells
• Spermatogenesis


  • Spermatogonia divide by mitosis to produce more spermatogonia or differentiate into spermatocytes
  • The walls of the seminiferous tubules consist of diploid spermatogonia, stem cells that are the precursors of sperm
  • Meiosis of each spermatocyte produces 4 haploid spermatids, which differentiate into sperms
• Oogenesis


  • Diploid stem cells called oogonia divide by mitosis to produce more oogonia and primary oocytes
  • The primary oocyte grows much larger and completes the meiosis I, forming a large secondary oocyte and a small polar body
  • Only if fertilization occurs will meiosis II ever be completed
  • Completion of meiosis II converts the secondary oocyte into a fertilized egg or zygote (and also a second polar body)
• Cytogenetics


  The study of the molecular and genetic features of genes observed in parallel with cytological features of chromosomes and chromosomal DNA using microscopy
• Clinical cytogenetics


  The branch of cytogenetics concerned with relationship between chromosomal abnormalities and pathologic conditions
• Cytogenetic Lab


  Standard chromosomal analyses for diagnosis of congenital anomalies and mental retardation; leukemia, lymphoma and other hematological disorders; and solid neoplasm, mainly in humans
• Karyotype of Human Chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes
• Swamp Buffalo Karyotype: No. of 2N chromosomes = 48, 1,5 metacentric chromosomes, 2,3,4 submetacentric chromosomes, 5 -23 acrocentric chromosomes, X, Y acrocentric chromosomes
• Chromosome no. of some domestic animal species

  • Cat, Felis catus: 38
  • Pig, Sus scrofa domesticus: 38
  • Rabbit, Oryctolagus cuniculus: 44
  • Swamp buffalo, Bubalus bubalis: 48
  • Water buffalo, Bubalus bubalis: 50
  • Sheep, Ovis aries: 54
  • Goat, Capra hircus: 60
  • Cattle, Bos indicus/taurus: 60
  • Horse, Equus caballus: 64
  • Dog, Canis familaris: 78
  • Jungle fowl, Gallus gallus: 78
  • Ducks, Anas platyrhynchos: 80
• 1. locus Inheritance or Dihybrid Crossing

  • 2 loci with 2 alleles at each locus
  • Locus A with 2 alleles: A, a
  • Locus B with 2 alleles: B, b
• Dihybrid cross or genotype = AaBb
• Possible gametes in a dihybrid cross: AB, Ab, aB, ab
• Dihybrid crossing = crossing between individuals differing at 2 loci
• Punnett Square for dihybrid crossing of AaBb male x AaBb female


  Genotypic ratio: 1AABB, 1AAbb, 2AABb, 2AaBB, 4AaBb, 2Aabb, 2aaBb, 1aaBB, 1aabb
• Branching System for dihybrid crossing of AaBb x AaBb

  Genotypic ratio: 1AABB, 1AAbb, 2AABb, 2AaBB, 4AaBb, 2Aabb, 2aaBb, 1aaBB, 1aabb
• The Foil Method for dihybrid crossing of AaBb x AaBb

  Genotypic ratio: 1AABB, 1AAbb, 2AABb, 2AaBB, 4AaBb, 2Aabb, 2aaBb, 1aaBB, 1aabb
• Genotypic ratio for dihybrid crossing of AaBb x AaBb: 1 AABB, 1 AAbb, 2 AABb, 2 AaBB, 4 AaBb, 2 Aabb, 2 aaBb, 1 aaBB, 1 aabb
• Phenotypic ratio for dihybrid crossing of AaBb x AaBb: 9 A_B_, 3 A_bb, 3 aaB_, 1 aabb when A is dominant over a and B is dominant over b
• Crossing of AABB x AaBb using Branching system

  Genotypic ratio: 1 AABB: 1 AABb: 1 AaBB: 1 AaBb
• Crossing of AABB male x AaBb female using Punnett Square


  Genotypic ratio: 1 AABB: 1 AABb: 1 AaBB: 1 AaBb
• Crossing of AaBb x Aabb genotypes

  1. Genotypic ratio: 1 AABb, 1 AAbb, 2 AaBb, 2 Aabb, 1aaBb, 1 aabb

  2. Phenotypic ratio: 3 A_B_: 3 A_bb : 1 aaBb : 1 aabb
• Crossing of AaBb male x Aabb female

  Genotypic ratio: 1 AABb, 1 AAbb, 2 AaBb, 2 Aabb, 1aaBb, 1 aabb
• Branching system of arriving at offspring genotypic ratio for crossing of AaBb x Aabb

  Offspring genotypic ratio: 1 AABb : 1 Aabb : 2 AaBb : 2 Aabb : 1 aaBb : 1 aabb
• In horses black (B) is dominant over chestnut (b) and trotting gait (T) is dominant over pacing gait (t). A heterozygous black trotter F1 stallion (BbTt) is mated to several chestnut pacer mares (bbtt). What will be the appearance of the resultant offspring?
• Crossing of BbTt stallion x bbtt mares

  Genotypic ratio & Phenotypic ratio: 1:1:1:1
• aabb


  Phenotypic ratio: 3 A_B_: 3 A_bb: 1 aaBb: 1 aabb
• Crossing of AaBb male x Aabb female

  1. Female Ab

  2. Female ab

  3. Male AB
• Offspring genotypes


  • AABb
  • AAbb
  • AaBb
  • Aabb
  • aaBb
  • aabb
• Genotypic ratio


  1 AABb, 1 AAbb, 2 AaBb, 2 Aabb, 1 aaBb, 1 aabb
• Branching system of arriving at offspring genotypic ratio

  1. Crossing of AaBb x Aabb genotypes

  2. Locus A >> possible gametes 1AA:2Aa:1aa

  3. Locus B >> possible gametes 0BB:1Bb:1bb
• Offspring genotypic ratio


  • 1 AABb:1 Aabb:2 AaBb:2 Aabb:1 aaBb:1 aabb
• Black (B) is dominant over chestnut (b) and trotting gait (T) is dominant over pacing gait (t). A heterozygous black trotter F1 stallion (BbTt) is mated to several chestnut pacer mares (bbtt).


  1. Stallion gametes BT, Bt, bT, bt at 1:1:1:1 ratio

  2. Mare gamete bt at 1
• Offspring genotypes and phenotypes

  • BbTt (black trotter)
  • Bbtt (black pacer)
  • bbTt (chestnut trotter)
  • bbtt (chestnut pacer)
• Genotypic ratio


  1 BbTt: 1 Bbtt: 1 bbTt: 1 bbtt
• Phenotypic ratio


  1 black trotter: 1 black pacer: 1 chestnut trotter: 1 chestnut pacer
• Assume two loci, S and Y, each with 2 alleles: S and s alleles for the S locus and Y and y alleles for the Y locus, affect 2 traits, respectively: A and B. S is dominant over s and Y is dominant over y. Which of the following genetic crosses would be predicted to give a phenotypic ratio of 9:3:3:1?

  
  

  A. SsYY x SSYy

  B. SSYY x ssyy

  C. SsYy x SsYy

  D. Ssyy x ssYY

  E. ssYY x ssyy
• Mendelian ratio

  A statistical ratio that fits the expected genetic inheritance pattern
• Testing of Fit to Mendelian Ratio

  1. Hypothesis that the phenotypic ratio obtained from a specific mating fits a Mendelian ratio

  2. Chi-Square Formula X2 = Σ (Observed value –Expected value)2/Expected value

  3. Degrees of freedom (df) = n-1 where n is the number of classes

  4. Significant level p < 0.05 or 0.01