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GENETICS
Genetics
• The study of heredity.
               heredity

• Gregor Mendel (1860’s) discovered the
  fundamental principles of genetics by breeding
  garden peas.
         peas
Genetics
• Alleles
  1. Alternative forms of genes.
  2. Units that determine heritable traits.
  3. Dominant alleles (TT - tall pea plants)
                                     plants
         a. homozygous dominant
  4. Recessive alleles (tt - dwarf pea plants)
                                       plants
         a. homozygous recessive
  5. Heterozygous (Tt - tall pea plants)
                                 plants
Phenotype
• Outward appearance
• Physical characteristics

• Examples:
         1.     tall pea plant
         2.     dwarf pea plant
Genotype
• Arrangement of genes that produces the
  phenotype
• Example:
  1. tall pea plant
           TT = tall (homozygous dominant)
  2. dwarf pea plant
           tt = dwarf (homozygous recessive)
  3. tall pea plant
           Tt = tall (heterozygous)
Punnett square
• A Punnett square is used to show the
  possible combinations of gametes.
                           gametes
Breed the P generation
• tall (TT) vs. dwarf (tt) pea plants

             T     T


         t


         t
tall (TT) vs. dwarf (tt) pea plants


       T     T


             Tt      produces the
t      Tt
                     F1 generation

t      Tt    Tt      All Tt = tall
                     (heterozygous tall)
Breed the F1 generation
• tall (Tt) vs. tall (Tt) pea plants

               T     t


          T


          t
tall (Tt) vs. tall (Tt) pea plants

     T     t
                      produces the
     TT    Tt         F2 generation
 T
                     1/4 (25%) = TT
     Tt    tt        1/2 (50%) = Tt
 t
                     1/4 (25%) = tt
                      1:2:1 genotype
                      3:1 phenotype
Monohybrid Cross
• A breeding experiment that tracks the inheritance
  of a single trait.

• Mendel’s “principle of segregation”
  a. pairs of genes separate during gamete
     formation (meiosis).

  b. the fusion of gametes at fertilization pairs
     genes once again.
Homologous Chromosomes

eye color locus                       eye color locus
B = brown eyes                        b = blue eyes




                                      This person would
                                      have brown eyes (Bb)



                  Paternal Maternal
Meiosis - eye color
                                     B

                    B                     sperm
                                     B

   Bb
                                         haploid (n)
                                     b
diploid (2n)        b

                                     b
        meiosis I       meiosis II
Monohybrid Cross
• Example:
  Example         Cross between two heterozygotes
                  for brown eyes (Bb)
BB = brown eyes
                            B      b      male
Bb = brown eyes                           gametes
bb = blue eyes
                     B
      Bb x Bb
                     b


                         female gametes
Monohybrid Cross

              B    b

                        1/4 = BB - brown eyed
          B   BB   Bb   1/2 = Bb - brown eyed
Bb x Bb                 1/4 = bb - blue eyed

          b   Bb   bb
                           1:2:1 genotype
                            3:1 phenotype
Dihybrid Cross
• A breeding experiment that tracks the inheritance
  of two traits.
• Mendel’s “principle of independent assortment”
  a. each pair of alleles segregates independently
     during gamete formation (metaphase I)
  b. formula: 2n (n = # of heterozygotes)
Independent Assortment
• Question: How many gametes will be produced
            for the following allele arrangements?


• Remember:       2n (n = # of heterozygotes)

  1.   RrYy
  2.   AaBbCCDd
  3.   MmNnOoPPQQRrssTtQq
Answer:
1. RrYy: 2n = 22 = 4 gametes
      RY Ry rY ry

2. AaBbCCDd: 2n = 23 = 8 gametes
     ABCD ABCd AbCD AbCd
     aBCD aBCd abCD abCD

3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
Dihybrid Cross
• Example:        cross between round and yellow
                  heterozygous pea seeds.

 R   = round                RrYy x RrYy
 r   = wrinkled
 Y   = yellow
                     RY Ry rY ry x RY Ry rY ry
 y   = green          possible gametes produced
Dihybrid Cross
     RY   Ry   rY   ry

RY


Ry


rY


ry
Dihybrid Cross
      RY    Ry      rY    ry
                                 Round/Yellow:     9
RY RRYY RRYy       RrYY   RrYy
                                 Round/green:      3
Ry RRYy     RRyy   RrYy   Rryy
                                 wrinkled/Yellow: 3

rY RrYY     RrYy   rrYY   rrYy   wrinkled/green:   1


ry   RrYy   Rryy   rrYy   rryy   9:3:3:1 phenotypic ratio
Test Cross
• A mating between an individual of unknown genotype
  and a homozygous recessive individual.
• Example: bbC__ x bbcc
  BB = brown eyes
  Bb = brown eyes
                                 bC      b___
  bb = blue eyes
                           bc
  CC = curly hair
  Cc = curly hair
  cc = straight hair
Test Cross

     • Possible results:

       bC    b___
               C                bC     b___
                                         c

bc    bbCc   bbCc   or     bc   bbCc   bbcc
Incomplete Dominance
• F1 hybrids have an appearance somewhat in
  between the phenotypes of the two parental
  varieties.
• Example: snapdragons (flower)
• red (RR) x white (rr)
                               R      R

     RR = red flower       r
     rr = white flower
                           r
Incomplete Dominance


      R   R

               produces the
r    Rr   Rr
               F1 generation

r    Rr   Rr   All Rr = pink
               (heterozygous pink)
Codominance
• Two alleles are expressed (multiple alleles)
                                      alleles
  in heterozygous individuals.
                   individuals
• Example: blood
  1.   type A    =   IAIA or IAi
  2.   type B    =   IBIB or IBi
  3.   type AB   =   IAIB
  4.   type O    =   ii
Codominance
• Example:          homozygous male B (IBIB)
                              x
                    heterozygous female A (IAi)

              IB        IB

       IA    I AI B    I AI B
                                 1/2 = IAIB
                                 1/2 = IBi
       i     I Bi       IBi
Codominance

• Example: male O (ii) x female AB (IAIB)


               IA     IB

         i    I Ai   IBi   1/2 = IAi
                           1/2 = IBi

         i    I Ai   IBi
Codominance
• Question:
  Question       If a boy has a blood type O and
                 his sister has blood type AB,
                 what are the genotypes and
                 phenotypes of their parents.

• boy - type O (ii) X girl - type AB (IAIB)
Codominance
• Answer:

      IA     i

I B I AI B          Parents:
                    genotypes = IAi and IBi
                    phenotypes = A and B
i            ii
Sex-linked Traits
• Traits (genes) located on the sex
  chromosomes



• Example:      fruit flies
     (red-eyed male) X (white-eyed female)
      red               white
Sex-linked Traits
                 Sex Chromosomes

                         fruit fly
                         eye color




XX chromosome - female          Xy chromosome - male
Sex-linked Traits
• Example:       fruit flies
     (red-eyed male) X (white-eyed female)
• Remember: the Y chromosome in males
  does not carry traits.

RR = red eyed                      XR        y
Rr = red eyed
rr = white eyed             Xr

Xy = male
                            Xr
XX = female
Sex-linked Traits


      XR      y

Xr   XR Xr   Xr y
                    1/2 red eyed and female
                    1/2 white eyed and male
Xr   XR Xr   Xr y
Population Genetics
• The study of genetic changes in populations.
                                  populations
• The science of microevolutionary changes in
  populations.
  populations
• Hardy-Weinberg equilibrium:
     the principle that shuffling of genes that occurs
     during sexual reproduction, by itself, cannot
     change the overall genetic makeup of a population.
• Hardy-Wienberg equation:         1 = p2 + 2pq + q2
Question:
• How do we get this equation?

Answer:   “Square” 1 = p + q
                          ↓
                    12 = (p + q)2
                          ↓
                    1 = p2 + 2pq + q2
Hardy-Wienberg equation
• Five conditions are required for Hardy-Wienberg
  equilibrium.
  1. large population
  2. isolated population
  3. no net mutations
  4. random mating
  5. no natural selection
Important
• Need to remember the following:

    p2 = homozygous dominant
    2pq = heterozygous
    q2 = homozygous recessive
Question:
• Iguanas with webbed feet (recessive trait) make
  up 4% of the population. What in the population
  is heterozygous and homozygous dominant.
                                     dominant
Answer:

1. q2 = 4% or .04   q2 = .04      q = .2

2. then use 1 = p + q
            1 = p + .2   1 - .2 = p    .8 = p
3. for heterozygous use 2pq
                   2(.8)(.2) = .32 or 32%
4. For homozygous dominant use p2
                         .82 = .64 or 64%
Hardy-Wienberg equation

          1 = p2 + 2pq + q2
•   64% = p2    = homozygous dominant
•   32% = 2pq   = heterozygous
•   04% = q2    = homozygous recessive
•   100%

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Genetics

  • 2. Genetics • The study of heredity. heredity • Gregor Mendel (1860’s) discovered the fundamental principles of genetics by breeding garden peas. peas
  • 3. Genetics • Alleles 1. Alternative forms of genes. 2. Units that determine heritable traits. 3. Dominant alleles (TT - tall pea plants) plants a. homozygous dominant 4. Recessive alleles (tt - dwarf pea plants) plants a. homozygous recessive 5. Heterozygous (Tt - tall pea plants) plants
  • 4. Phenotype • Outward appearance • Physical characteristics • Examples: 1. tall pea plant 2. dwarf pea plant
  • 5. Genotype • Arrangement of genes that produces the phenotype • Example: 1. tall pea plant TT = tall (homozygous dominant) 2. dwarf pea plant tt = dwarf (homozygous recessive) 3. tall pea plant Tt = tall (heterozygous)
  • 6. Punnett square • A Punnett square is used to show the possible combinations of gametes. gametes
  • 7. Breed the P generation • tall (TT) vs. dwarf (tt) pea plants T T t t
  • 8. tall (TT) vs. dwarf (tt) pea plants T T Tt produces the t Tt F1 generation t Tt Tt All Tt = tall (heterozygous tall)
  • 9. Breed the F1 generation • tall (Tt) vs. tall (Tt) pea plants T t T t
  • 10. tall (Tt) vs. tall (Tt) pea plants T t produces the TT Tt F2 generation T 1/4 (25%) = TT Tt tt 1/2 (50%) = Tt t 1/4 (25%) = tt 1:2:1 genotype 3:1 phenotype
  • 11. Monohybrid Cross • A breeding experiment that tracks the inheritance of a single trait. • Mendel’s “principle of segregation” a. pairs of genes separate during gamete formation (meiosis). b. the fusion of gametes at fertilization pairs genes once again.
  • 12. Homologous Chromosomes eye color locus eye color locus B = brown eyes b = blue eyes This person would have brown eyes (Bb) Paternal Maternal
  • 13. Meiosis - eye color B B sperm B Bb haploid (n) b diploid (2n) b b meiosis I meiosis II
  • 14. Monohybrid Cross • Example: Example Cross between two heterozygotes for brown eyes (Bb) BB = brown eyes B b male Bb = brown eyes gametes bb = blue eyes B Bb x Bb b female gametes
  • 15. Monohybrid Cross B b 1/4 = BB - brown eyed B BB Bb 1/2 = Bb - brown eyed Bb x Bb 1/4 = bb - blue eyed b Bb bb 1:2:1 genotype 3:1 phenotype
  • 16. Dihybrid Cross • A breeding experiment that tracks the inheritance of two traits. • Mendel’s “principle of independent assortment” a. each pair of alleles segregates independently during gamete formation (metaphase I) b. formula: 2n (n = # of heterozygotes)
  • 17. Independent Assortment • Question: How many gametes will be produced for the following allele arrangements? • Remember: 2n (n = # of heterozygotes) 1. RrYy 2. AaBbCCDd 3. MmNnOoPPQQRrssTtQq
  • 18. Answer: 1. RrYy: 2n = 22 = 4 gametes RY Ry rY ry 2. AaBbCCDd: 2n = 23 = 8 gametes ABCD ABCd AbCD AbCd aBCD aBCd abCD abCD 3. MmNnOoPPQQRrssTtQq: 2n = 26 = 64 gametes
  • 19. Dihybrid Cross • Example: cross between round and yellow heterozygous pea seeds. R = round RrYy x RrYy r = wrinkled Y = yellow RY Ry rY ry x RY Ry rY ry y = green possible gametes produced
  • 20. Dihybrid Cross RY Ry rY ry RY Ry rY ry
  • 21. Dihybrid Cross RY Ry rY ry Round/Yellow: 9 RY RRYY RRYy RrYY RrYy Round/green: 3 Ry RRYy RRyy RrYy Rryy wrinkled/Yellow: 3 rY RrYY RrYy rrYY rrYy wrinkled/green: 1 ry RrYy Rryy rrYy rryy 9:3:3:1 phenotypic ratio
  • 22. Test Cross • A mating between an individual of unknown genotype and a homozygous recessive individual. • Example: bbC__ x bbcc BB = brown eyes Bb = brown eyes bC b___ bb = blue eyes bc CC = curly hair Cc = curly hair cc = straight hair
  • 23. Test Cross • Possible results: bC b___ C bC b___ c bc bbCc bbCc or bc bbCc bbcc
  • 24. Incomplete Dominance • F1 hybrids have an appearance somewhat in between the phenotypes of the two parental varieties. • Example: snapdragons (flower) • red (RR) x white (rr) R R RR = red flower r rr = white flower r
  • 25. Incomplete Dominance R R produces the r Rr Rr F1 generation r Rr Rr All Rr = pink (heterozygous pink)
  • 26. Codominance • Two alleles are expressed (multiple alleles) alleles in heterozygous individuals. individuals • Example: blood 1. type A = IAIA or IAi 2. type B = IBIB or IBi 3. type AB = IAIB 4. type O = ii
  • 27. Codominance • Example: homozygous male B (IBIB) x heterozygous female A (IAi) IB IB IA I AI B I AI B 1/2 = IAIB 1/2 = IBi i I Bi IBi
  • 28. Codominance • Example: male O (ii) x female AB (IAIB) IA IB i I Ai IBi 1/2 = IAi 1/2 = IBi i I Ai IBi
  • 29. Codominance • Question: Question If a boy has a blood type O and his sister has blood type AB, what are the genotypes and phenotypes of their parents. • boy - type O (ii) X girl - type AB (IAIB)
  • 30. Codominance • Answer: IA i I B I AI B Parents: genotypes = IAi and IBi phenotypes = A and B i ii
  • 31. Sex-linked Traits • Traits (genes) located on the sex chromosomes • Example: fruit flies (red-eyed male) X (white-eyed female) red white
  • 32. Sex-linked Traits Sex Chromosomes fruit fly eye color XX chromosome - female Xy chromosome - male
  • 33. Sex-linked Traits • Example: fruit flies (red-eyed male) X (white-eyed female) • Remember: the Y chromosome in males does not carry traits. RR = red eyed XR y Rr = red eyed rr = white eyed Xr Xy = male Xr XX = female
  • 34. Sex-linked Traits XR y Xr XR Xr Xr y 1/2 red eyed and female 1/2 white eyed and male Xr XR Xr Xr y
  • 35. Population Genetics • The study of genetic changes in populations. populations • The science of microevolutionary changes in populations. populations • Hardy-Weinberg equilibrium: the principle that shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic makeup of a population. • Hardy-Wienberg equation: 1 = p2 + 2pq + q2
  • 36. Question: • How do we get this equation? Answer: “Square” 1 = p + q ↓ 12 = (p + q)2 ↓ 1 = p2 + 2pq + q2
  • 37. Hardy-Wienberg equation • Five conditions are required for Hardy-Wienberg equilibrium. 1. large population 2. isolated population 3. no net mutations 4. random mating 5. no natural selection
  • 38. Important • Need to remember the following: p2 = homozygous dominant 2pq = heterozygous q2 = homozygous recessive
  • 39. Question: • Iguanas with webbed feet (recessive trait) make up 4% of the population. What in the population is heterozygous and homozygous dominant. dominant
  • 40. Answer: 1. q2 = 4% or .04 q2 = .04 q = .2 2. then use 1 = p + q 1 = p + .2 1 - .2 = p .8 = p 3. for heterozygous use 2pq 2(.8)(.2) = .32 or 32% 4. For homozygous dominant use p2 .82 = .64 or 64%
  • 41. Hardy-Wienberg equation 1 = p2 + 2pq + q2 • 64% = p2 = homozygous dominant • 32% = 2pq = heterozygous • 04% = q2 = homozygous recessive • 100%