Question 1

2. Question 1 • In snapdragons (Antirrhinum), the phenotype for flower color is governed by two alleles – red (R) and white (W). Heterozygous individuals have pink flowers. Two pink individuals are crossed to produce 465 offspring. 1. Calculate how many of these offspring are expected to have the red phenotype. Round your response to the nearest whole number.

3. Question 1-Explanation • In snapdragons (Antirrhinum), the phenotype for flower color is governed by two alleles – red (R) and white (W). Heterozygous individuals have pink flowers. Two pink individuals are crossed to produce 465 offspring. 1. Calculate how many of these offspring are expected to have the red phenotype. Round your response to the nearest whole number. What are you being asked? You are being asked how many of the offspring produced by the mating of two pink snapdragons are expected to be red. What do you know? You know that there are two alleles, red (R) & white (W), and that heterozygotes (RW) have pink flowers. You know that two pink flowers mated. You know there are 465 offspring. What do you need to figure out? The expected phenotypic frequency from a cross of the parents (both are pink, so both are heterozygous (RW x RW). Do a monohybrid cross (Punnett square) This is an example of incomplete dominance

4. Question 1-Explanation R W • A cross between two pink snapdragons (RW x RW) should yield the following phenotypic ratio - 1:2:1 • 1 red (RR):2 pink (RW):1 white (WW). • 1 + 2 + 1 = 4 • So, 1/4 red, 2/4 pink, and 1/4 white = .25 red, .5 pink, & .25 white • 25% of the offspring are expected to be red • so the calculation would be .25 x 465 =116.25, which is rounded to the whole number 116. R R R R W W R W W W

5. Question 2 • In corn (Zea mays), purple kernels (R) are dominant to yellow kernels (r). Cobs from the offspring of a cross between a purple plant and yellow plant were used in a lab. A student counts 329 purple and 299 yellow kernels on one cob. 2. Calculate the chi-squared value for the null hypothesis that the purple parent was heterozygous for purple kernels. Give your answer to the nearest tenth.

6. Question 2-Explanation • In corn (Zea mays), purple kernels (R) are dominant to yellow kernels (r). Cobs from the offspring of a cross between a purple plant and yellow plant were used in a lab. A student counts 329 purple and 299 yellow kernels on one cob. 2. Calculate the chi-squared value for the null hypothesis that the purple parent was heterozygous for purple kernels. Give your answer to the nearest tenth. What are you being asked? You are being asked to calculate a chi square value to test it the purple parent was heterozygous. What do you know? • You know you are calculating a chi square, and the equation is on your Formula sheet. • You know that there are two alleles, purple (R) is dominant to yellow (r), • Since you are testing to see if the purple plant is heterozygous, you must solve for chi square as though the purple parent is heterozygotes (Rr). • So you know a purple (assumed heterozygote (Rr)) is crossed with a yellow plant. • You know the observed (‘O’ from eqn) numbers you will use (Opurple=329 & Oyellow=299) What do you need to figure out? • What genotype is the yellow parent plant? Since yellow is recessive, it must be rr. • You need to figure out the expected phenotypic frequency from crossing the purple (Rr) and yellow (rr), and then use those frequencies to determine the expected values (‘E’) of purple and yellow offspring.

7. R r Question 2-Explanation rr r Rr Rr rr r • A cross between a heterozygous purple corn plant (Rr) and a yellow corn plant (rr) would be expected to yield offspring that display a 1:1 ratio between purple and yellow kernels. Of the 628 total kernels (we got this number by adding our observed totals given to us in the problem 329 purple + 299 yellow = 628 total), it would be expected that 314would be purple and 314would be yellow. • 1 purple:1 yellow • 1+1=2. so ½ purple, and ½ yellow, or .5 purple & .5 yellow. • So, we know that we had 628 offspring, • based on the expected ratio, we would expect to have ½{628) =314 purple and ½(628)=314 yellow. • The chi-square value is calculated step by step on the next slide. The range of acceptable answers for this question should be 1.43-1.44. From formula sheet

8. Question 2-Explanation • A cross between a heterozygous purple corn plant (Rr) and a yellow corn plant (rr) would be expected to yield offspring that display a 1:1 ratio between purple and yellow kernels. Of the 628 total kernels (we got this number by adding our observed totals given to us in the problem 329 purple + 299 yellow = 628 total), it would be expected that 314would be purple and 314would be yellow. • 1 purple:1 yellow • 1+1=2. so ½ purple, and ½ yellow, or .5 purple & .5 yellow. • So, we know that we had 628 offspring, • based on the expected ratio, we would expect to have ½{628) =314 purple and ½(628)=314 yellow. • The chi-square value is calculated below. The range of acceptable answers for this question should be 1.43-1.44. From formula sheet (Opurple-Epurple) 2/Epurple + (Oyellow-Eyellow)2/Eyellow = (329 - 314)2/314 + (299 - 314)2/314 = (15)2/314 + (15)2/314 = 225/314 + 225/314 = 0.72 + 0.72 = 1.44

9. Question 3 • In a dog breed known as the Mexican Hairless, the “hairless” phenotype is a result of a mutation displaying an autosomal dominant pattern of inheritance. Homozygous recessive individuals (hh) display a “coated” phenotype. Inheriting two copies of the mutation (HH) is lethal during embryonic development. 3. In a cross between two dogs with the hairless phenotype, what proportion of puppies born is expected to be hairless? Give your answer in the form of a fraction.

10. Question 3-Explanation In a dog breed known as the Mexican Hairless, the “hairless” phenotype is a result of a mutation displaying an autosomal dominant pattern of inheritance. Homozygous recessive individuals (hh) display a “coated” phenotype. Inheriting two copies of the mutation (HH) is lethal during embryonic development. 3. In a cross between two dogs with the hairless phenotype, what proportion of puppies born is expected to be hairless? Give your answer in the form of a fraction. What are you being asked? You are being asked how many of the offspring produced by the mating of two hairless dogs are expected be hairless. What do you know? • You know hh gives ‘coated’ phenotype (so not hairless) • You know H is dominant to h, and that heterozygous dogs (Hh) are hairless. • You are told that homozygous dominant is lethal during the embryonic stage, so these puppies will not be born. • You know that you are crossing 2 hairless dogs. What do you need to figure out? • What genotype is the hairless dog? • Since a genotype of ‘hh’  ‘coated’ and ‘HH’ dead, the only way to be hairless is to be heterozygous, it must be Hh. • You need to figure out the expected phenotypic frequency (fraction) from crossing the 2 heterozygous (Hh x Hh).

11. Question 3-Explanation H h Not Born Hh H HH • Inheriting two copies of the hairless mutation is lethal in embryonic development; therefore, the parents must be heterozygous (Hh) for their hairlessness. • In the offspring, individuals with the (HH) genotype die before birth and are not calculated as a genotypic class. So we are left with: • 2 Hh :1 hh • 2+1=3; 2/3 Hh & 1/3 hh or 2/3 hairless & 1/3 ‘coated’ • This means that the proportion of hairless puppies born would be 2/3. Hh hh h

12. Question 4 • Wild-type fruit flies have red eyes (+). The “white-eyed” phenotype (w) is recessive and results from a mutation on the X chromosome. During a lab, students cross a white-eyed male with a homozygous red-eyed female. A red-eyed female and a red-eyed male from the F1 generation are then bred to produce 573 offspring. 4. How many of the offspring are predicted to be white-eyed males? Round your response to the nearest whole number.

13. Question 4-Explanation • Wild-type fruit flies have red eyes (+). The “white-eyed” phenotype (w) is recessive and results from a mutation on the X chromosome. During a lab, students cross a white-eyed male with a homozygous red-eyed female. A red-eyed female and a red-eyed male from the F1 generation are then bred to produce 573 offspring. 4. How many of the offspring are predicted to be white-eyed males? Round your response to the nearest whole number. What are you being asked? You are being asked how many of the offspring produced by the mating of a red eyed female and red eyed male from the F1generation are expected have red eyes AND be male. What do you know? • Wild type (+) = red eyes • White eyes are recessive (w). • The allele is X-linked, so we should designate the alleles as: • red = X+ (dominant) or • white = Xw(recessive) • P generation Cross between a White Eyed Male & a Homozygous red-eyed female • Cross of red-eyed female and a red-eyed male from the F1 generation produce 573 offspring. What do you need to figure out? Next slide

14. Question 4-Explanation What are you being asked? You are being asked how many of the offspring produced by the mating of a red eyed female and red eyed male from the F1generation are expected have red eyes AND be male. What do you know? • Wild type (+) = red eyes • White eyes are recessive (w). • The allele is X-linked, so we should designate the alleles as: • red = X+ (dominant) or • white = Xw(recessive) • P generation Cross between a White Eyed Male (XwY)& a Homozygous red-eyed female (X+ X+) • Cross of red-eyed female and a red-eyed male from the F1 generation produce 573 offspring. What do you need to figure out? P genetation Cross • What genotype do the F1 parents (male & female red eyed) have? (you need to cross the original parents (white-eyed male with a homozygous red-eyed female) • From the cross you now know that all F1 males are red eyed, and all F1 females are heterozygous • Using the above information: What is the expected frequency of white eyed males resulting from the F1 cross (red male with red eyed female.. X+ X+ Xw X+Xw X+Xw X+Y X+Y Y Continued on next slide

15. P genetation Cross (from last slide) X+ X+ Xw X+Xw X+Xw X+Y X+Y Y F1 Cross From the problem we know we are crossing a red eyed male with a red eyed female From crossing the P generation (above square) we know the genotypes of the F1 cross The male is X+Y And the female is X+Xw To figure out how many of the 573 offspring are males with white eyes, we must perform the cross and discover the phenotypic frequency of white eyed males. X+ Xw X+ X+X+ X+Xw X+Y XwY Y

16. Question 4-Explanation • To recap, the parental cross (X+X+x XwY) produced male offspring with the genotype X+Y and females with the genotype X+Xw. • If a male and female from the F1 are crossed,1/4, or 25% of the offspring will be white-eyed males. • So, 0.25 x 573 = 143.25 or 143 rounded to the nearest whole number.

17. Question 5 • A plant geneticist is investing the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Citrulluslanatus). Explosive rind is recessive and causes watermelons to burst when cut. Non-bitter watermelons are a result of the recessive genotype (susu). The geneticist wishes to determine if the genes assort independently. She performs a testcross between a bitter/nonexplosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced: • bitter/non-explosive – 88 • bitter/explosive – 68 • non-bitter/non-explosive – 62 • non-bitter/explosive – 81 5. Calculate the chi-squared value for the null hypothesis that the two genes assort independently. Give your answer to the nearest tenth.

18. Question 5-Explanation • A plant geneticist is investing the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Citrulluslanatus). Explosive rind is recessive and causes watermelons to burst when cut. Non-bitter watermelons are a result of the recessive genotype (susu). The geneticist wishes to determine if the genes assort independently. She performs a testcross between a bitter/nonexplosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced: • bitter/non-explosive – 88 • bitter/explosive – 68 • non-bitter/non-explosive – 62 • non-bitter/explosive – 81 5. Calculate the chi-squared value for the null hypothesis that the two genes assort independently. Give your answer to the nearest tenth. What do you not need to know from this problem You are calculating a chi square, which means you will need observed and expected values You have been given the observed values, you need to find the expected values, which means you need to find the expected phenotypic ratio, and then use it to determine the expected value for each phenotype. If these traits assort independently of one another, you should be able to use a normal dihybrid cross (or probability) to determine the likelyhood of each phenotype based on the parents. Cross on next slide

19. testcross between a bitter/nonexplosive hybrid (Susu/Ee) and a plant homozygous recessive for both traits (susu/ee) • Susu/Eex susu/ee • expected to produce offspring in a 1:1:1:1 phenotypic ratio. • 1+1+1+1=4; 1/4 bitter/non-explosive, 1/4 bitter/explosive, 1/4 non-bitter/non-explosive, & 1/4 non-bitter/explosive • So we add up all the offspring • bitter/non-explosive – 88 • bitter/explosive – 68 • non-bitter/non-explosive – 62 • non-bitter/explosive – 81 • And have 299 offspring. ¼(299)= 74.75 (Obitter/nonex-Ebitter/nonex) 2/Ebitter/nonex+ (Obitter/explosive-Ebitter/explosive)2/Ebitter/explosive+ (Ononbitter/nonex-Enonbitter/nonex) 2/Enonbitter/nonex + (Onon-bitter/explosive-Enon-itter/explosive)2/Enon-bitter/explosive • The chi-square value is calculated below. The answer for this question should be 5.7

20. How To Make a Punnett Square • Determine what alleles would be found in all of the possible gametes that each parent could produce. Male OspreyFemale Osprey Bb x Bb B b B b

21. Draw a table with enough spaces for each pair of gametes from each parent. Enter the genotypes of the gametes produced by both parents on the top and left sides of the table. How To Make a Punnett Square B b B b

22. How To Make a Punnett Square Fill in the table by combining the gametes’ genotypes. B b B b b b

23. How To Make a Punnett Square Fill in the table by combining the gametes’ genotypes. B b B b B b b b

24. How To Make a Punnett Square Fill in the table by combining the gametes’ genotypes. B b B b B b b b B b

25. How To Make a Punnett Square Fill in the table by combining the gametes’ genotypes. B b B b B b B B b b B b

26. How To Make a Punnett Square • Determine the genotypes and phenotypes of each offspring. • Calculate the percentage of each. • In this example, three fourths of the chicks will have large beaks, but only one in two will be heterozygous. B b B b B B B b B b b b