Presentation on theme: "Solving Genetic Problems"— Presentation transcript:
1 Solving Genetic Problems Example: A red flowered plant is fertilized with pollen from a plant of unknown phenotype. 62 seeds are collected, planted the next year and grow into the F1 offspring. 16 of these offspring have white flowers and the rest have red flowers. What was the phenotype and genotype of the unknown parental plant?
3 (1) Use What You Already Know Flower color is controlled by a single gene.The dominant form of this gene is often written RThe recessive form of this gene, written r.Plants with the genotype RR or Rr have red flowers.Plants with the genotype rr have white flowers.
4 (2) Use the Information in the Question Phenotypes and Genotypes of parents (P generation)?Genotypes and Phenotypes of Offspring(F1 generation)
5 (2) Use the Information in the Question Phenotypes and Genotypes of parents (P generation)Parental Phenotypes ... one red flowered, one unknownParental Genotypes ... unknown, but the red flowered parent must have contained at least one R gene, with a genotype of RR or Rr.Genotypes and Phenotypes of Offspring(F1 generation)The data given was for 62 F1 offspring.Phenotypes: 16 white flowers, 46 red flowers.Genotypes: all 16 white flowered plants = rr. All red flowered plants = R?
6 (3) Reduce all raw data to ratios Heterozygous x Heterozygous Aa x Aa, which gives a ratio of 3:1 among the offspring.Heterozygous x Homozygous recessive Aa x aa, which gives a ratio of 1:1 among the offspring.Heterozygous x Heterozygous (two factors) AaBb x AaBb, which gives a ratio of 9:3:3:1 among the offspring
7 (3) Reduce all raw data to ratios What type of ratio is provided in the example we are currently solving ? ? ? ? ? ? ? ? When compared to the known ratios above, the cross is most likely _____________ ______________ or ? x ?.
8 (3) Reduce all raw data to ratios In the example we have worked on:There are 62 plants in the F1 generation and 16 have white flowers.Therefore 46 plants have to have red flowers(62-16).This gives us a ratio of 46 red flowers to 16 white flowers.If we divide both sides by 16, this can be reduced to red flowers to 1 white flower or approximately 3:1.When compared to the known ratios above, the cross is most likely Heterozygous x Heterozygous or Rr x Rr.
9 (4) Diagram and Fill in the Blanks Parents: R? x ?? (the genotype of the second parent, the male, is the whole point of the problem)NOTE: In the Punnett Square a genotype of rr has been entered for one quarter of the offspring; why?R?R?rr
10 The data said that 16 out of 62 F1 offspring had white flowers. White flowered plants can only have the genotype rr.The raw numbers were reduced to the ratio of 3:1 in step 3.So one quarter of the F1 offspring (one box in the Punnett Square) must hold the genotype rr.Each of the other three offspring must have at least one R gene in order to be red.This was also given in the data and deduced in step 3.Now fill in the blanks.
11 If it is not obvious from the Punnett Square, work backwards If it is not obvious from the Punnett Square, work backwards. Since some of the offspring have white flowers (genotype rr), the gametes that formed them must have been r. So the only arrangement that would give this result is...Rr?R?rr
12 Now most of the gametes are filled in along the sides of the Punnett Square, some of the missing genotypes can also be filled in...Rr?R?Rrrr
13 This only leaves one gamete and one F1 genotype unresolved This only leaves one gamete and one F1 genotype unresolved. There are only two choices for the missing gamete; R or r. Putting a recessive, r gamete in place of the question mark does not give the right answer…RrRrrr
14 Punnett 5 rightso the missing gamete must be R. So the complete Punnett Square is...ANSWER: The genotype of the unknown male plant must have been Rr (a fact deduced from the two gametes this plant must have provided for the Punnett Square to work). The phenotype of this plant must have been - red flowers.RrRRRrrr
15 Your Turn Do Questions 1-4 Teachers: you can use data from 4th SW SFA #7, to decide if students should do SA (Straight Ahead) or UH (Up Hill).After students finish, you can go over answers several ways:Have students discuss with their table or neighbor table.Teacher go over as whole class orHave different groups volunteer to show class.
16 EOC PracticeIn fruit flies, a single gene controls wing phenotype. The diagram below shows the phenotypes for long wings and vestigial wings in fruit flies.Two fruit flies that have long wings are crossed. Of the 95 offspring produced, 73 have long wings. The other 22 have vestigial wings.Which of the following conclusions about the inheritance of long wings and vestigial wings is best supported by the results of this experiment?A. The alleles for long wings and vestigial wings are sex-linked.B. The alleles for long wings and vestigial wings are codominant.C. The allele for long wings is dominant and the allele for vestigial wings isrecessive.D. The allele for long wings is recessive and the allele for vestigial wings isdominant.Answer is C
17 EOC Practice2. In sheep, the allele for white wool (W) is dominant, and the allele for black wool (w) is recessive. A farmer has mated two Suffolk sheep for a few years. These matings have resulted in six offspring, four with white wool and two with black wool. One parent has white wool and the other has black wool. Which of the following could be the genotypes of the parent sheep? A. WW and Ww B. WW and ww C. Ww and Ww D. Ww and wwAnswer is D