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1 Warm-Up Blood types are determined by a single gene with several alleles. The allele encoding the Type A phenotype (I A ) is dominant to the allele encoding the Type O phenotype (i). Determine the phenotype and genotype distributions resulting from a cross between a Type O individual and a heterozygous Type A individual.

2 Yesterday s Picture

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4 Mendel s Law of Segregation: During meiosis, each allele is segregated to different gametes. 1m 1p 2p 2m m p 3p 3m allele a allele A 4m 4p 2n n = 4

5 Mendel s Law of Segregation: During meiosis, each allele is segregated to different gametes. Mendel s Law of Independent Assortment: During meiosis, two genes segregate independently from each other 1m 2p 1p 2m m p 3p 3m allele a allele A 4m 4p m allele F p allele f 2n n = 4

6 Mendel s Law of Segregation: During meiosis, each allele is segregated to different gametes. Mendel s Law of Independent Assortment: During meiosis, two genes segregate independently from each other a F one possibility A f

7 Mendel s Law of Segregation: During meiosis, each allele is segregated to different gametes. Mendel s Law of Independent Assortment: During meiosis, two genes segregate independently from each other because of random alignment during meiosis. m p 1m 2p 3p 1p 2m 3m allele a allele A 4p 4m p m 2n n = 4 allele f allele F

8 Mendel s Law of Segregation: During meiosis, each allele is segregated to different gametes. one possibility Mendel s Law of Independent Assortment: During meiosis, two genes segregate independently from each other because of random alignment during meiosis. a F A f another possibility a f A F

9 Mendel s Law of Segregation: During meiosis, each allele is segregated to different gametes. Mendel s Law of Independent Assortment: During meiosis, two genes segregate independently from each other because of random alignment during meiosis. c a B A b C For n different genes on n different chromosomes, there are 2 n possibilities of gamete combinations. d D 2 4 = 16 possible gametes: a B c d a b c d a b C d a b c D a B C d a B C D a b C D a B c D A B c d A b c d A b C d A b c D A B C d A B C D A b C D A B c D

10 CTQ #1 In pea plants, the q gene is present on a separate chromosome from the r gene. Draw a somatic pea plant cell (2n) heterozygous for the q gene and the r gene. Label the maternal and the paternal chromosomes which contain the q gene and the maternal and paternal chromosomes which contain the r gene, and label the dominant Q, recessive q, dominant R, or recessive r on the appropriate chromosomes. Draw FOUR possible gametes (1n) produced through meiotic division of this cell depending on the alignment of the chromosomes, and indicate the genotype of each gamete. (LO 3.12)

11 A dihybrid or multi-hybrid cross is used to predict genotype and phenotype probabilities with multiple genes.

12 A dihybrid or multi-hybrid cross is used to predict genotype and phenotype probabilities with multiple genes. For each gene, perform a monohybrid cross and multiply the probabilities. PpSs x PpSs gene 1 gene 2 gene 1 gene 2 parent 1 parent 2 S s P p

13 A dihybrid or multi-hybrid cross is used to predict genotype and phenotype probabilities with multiple genes. For each gene, perform a monohybrid cross and multiply the probabilities. PpSs x PpSs gene 1 gene 2 gene 1 gene 2 parent 1 parent 2 P p S s P PP Pp 1# 4 S 1# 4 SS Ss 1# 4 1# 4 p Pp pp s Ss ss 1# 1 4 # 4 1# 1 4 # 4

14 A dihybrid or multi-hybrid cross is used to predict genotype and phenotype probabilities with multiple genes. For each gene, perform a monohybrid cross and multiply the probabilities. PpSs x PpSs gene 1 gene 2 gene 1 gene 2 parent 1 parent 2 Genotype Distribution: PP: ¼ Pp: ¼ + ¼ = ½ pp: ¼ Phenotype Distribution: Purple: ¼ + ¼ + ¼ = ¾ White: ¼ Genotype Distribution: SS: ¼ Ss: ¼ + ¼ = ½ ss: ¼ Phenotype Distribution: Smooth: ¼ + ¼ + ¼ = ¾ wrinkled: ¼

15 A dihybrid or multi-hybrid cross is used to predict genotype and phenotype probabilities with multiple genes. For each gene, perform a monohybrid cross and multiply the probabilities. PpSs x PpSs gene 1 gene 2 gene 1 gene 2 parent 1 parent 2 Genotype Distribution: PP SS: ¼ x ¼ = PP Ss: ¼ x ½ = PP ss: ¼ x ¼ = Pp SS: ½ x ¼ = Pp Ss: ½ x ½ = ¼ Pp ss: ½ x ¼ = pp SS: ¼ x ¼ = pp Ss: ¼ x ½ = pp ss: ¼ x ¼ = Phenotype Distribution: purple, smooth: ¼ = purple, wrinkled: + = white, smooth: + = white, round:

16 A dihybrid or multi-hybrid cross is used to predict genotype and phenotype probabilities with multiple genes. For each gene, perform a monohybrid cross and multiply the probabilities. example x Bbww Tan body (B), white eyes (w) bbww Black body (b), red eyes (W)

17 CTQ #2 Determine the phenotype distributions from the cross between an individual heterozygous for two genes, A and B (AaBb), and an individual homozygous dominant for A and homozygous recessive for B (AAbb). (LO 3.14)

18 Closure If two genes are on the same chromosome, explain why a dihybrid cross would not follow Mendel s Law of Independent Assortment. (LO 3.15)

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