Heredity Heredity, Meiosis & Independent Assortment Classwork 1. Describe what is meant by the terms: dominant and recessive traits. 2. In genetics we often use the terms homozygous and heterozygous. Describe the meaning of the terms and provide an example of each. 3. Describe the term alleles in terms of the sequence of DNA. 4. Geneticists often use the terms genotype and phenotype in the description of an organism’s characteristics. Describe the meaning of each and provide an example. 5. Mendel used a testcross to determine the genotype of an unknown plant. Describe the process of performing a testcross. Describe the possible results. 6. In pea plants the allele T produces tall plants and allele t produces dwarf. A tall plant is crossed with a dwarf plant, producing offspring with equal numbers of both tall and dwarf plants. What are the genotypes of the parents? 7. Brown eyes are dominant to blue eyes. A brown eyed couple, both of whom had one blue eyed parent and one brown eyed parent are expecting their first child. What is the probability that the child will have blue eyes? 8. Describe Mendel’s law of independent assortment. 9. One way of predicting the outcome of a cross is to use a Punnett square. Describe what is meant by a monohybrid cross and a dihybrid cross. Show examples of each. 10. People who are able to taste PTC paper have a dominant allele. A homozygous recessive person cannot taste the PTC paper. Brown eyes are dominant to green eyes. Neither trait is sex-linked. A brown-eyed male taster marries a female green-eyed nontaster. Describe the kinds of children and proportions they could expect if one of the father’s parents was a green eyed non-taster. 11. Describe the disease known as Huntington disease and explain how it is inherited. Homework 12. Brown eyes are dominant to blue eyes. Two brown-eyed parents have four blue-eyed children. What is the chance that their next child will have brown eyes? What is the chance it would have blue eyes? 13. In jack rabbits the allele for smooth fur (S) is dominant to the allele for rough fur (s). A smooth fur parent mates with a rough fur parent and produce offspring with the proportion ½ smooth and ½ rough. If two of the smooth offspring are mated what would be the ratio of their offsprings’ phenotypes? 14. A novice gardener decides to cross lima bean plants that are true breeding for green pods with another lima bean plant that is true breeding for yellow pods. After collecting the seeds and planting, she notices all the F1 have yellow seed pods. Which plant is the dominant phenotype? Show a Punnett square to prove you answer. 15. The ability to roll your tongue in a complete circle is controlled by a dominant gene. Two parents that can roll their tongue have a child that cannot roll her tongue. Explain why she cannot. List the genotypes and phenotypes for the parents and the child. 16. Suppose the homozygous recessive genotype is lethal for a particular trait. What ratios would you expect to be living in the following crosses? Dd x DD and Dd x Dd 17. A researcher observes that when two blue-flowered plants are mated together they always produce blue flowers. When two pink-flowered plants are mated together 75 % are pink and 25% are blue. Which flower color is dominant? What is the genotype of the pink flowered plants that were mated together?

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18. You have two rose plants the each produce different color petals, one red and one yellow. You want to cross them for next year’s flower show. How could you determine which plant has the dominant flower color? 19. A grey rabbit is crossed with a Chinchilla rabbit and produces 3 grey rabbits and 1 white rabbit. Determine the genotypes of the parents. 20. In chickens, pea combs are produced by a dominant allele, P. The recessive allele, p, produces a single comb. Feathered legs are produced by a dominant allele, F. Assuming no gene linkage, a farmer performs the following crosses with four chickens: #1, 2, 3, and 4. Using the results below, determine the genotypes of chickens #1, 2, 3, and 4.

Rooster #1 #1 #1

Hen #2 #3 #4

Results All feathered and all pea comb 75% feathered, all pea comb 56.25% feathered, pea comb; 18.75% featherless pea comb; 18.75 feathered, single-comb; 6.25% featherless, single comb

21. In humans, the gene for brown hair, B, is dominant to the gene for blond hair, b. A brown-haired man, whose mother had blond hair, and father had brown hair marries a woman having blond hair, both of whose parent were brown-haired. Determine the genotype of the man. Determine the possible genotypes for the children produced from this marriage. 22. Sweet corn comes in two varieties, white and yellow kernels. Yellow kernels, Y is dominant to white kernels, y. What kinds of gametes world be produced by the parent pants in the following crosses? A. YY x Yy B. Yy x Yy C. Yy x yy What Mendel Didn’t Know Classwork 23. Describe incomplete dominance and provide an example. 24. Four o’clock flowers show incomplete dominance. Red flowers are homozygous for the dominant allele R, white flowers are homozygous for the recessive allele, r. Heterozygotes have pink flowers. Determine the expected offspring from the following crosses: RR x Rr, RR x rr, and Rr x Rr. 25. Describe the genetic terminology of pleiotropy and polygenic inheritance. 26. Cats produce many different coat colors. Tabby fur color can be produced by AA or Aa, black is aa. Another gene pair is epistatic to the gene for fur color. When present in its dominant form (WW or Ww), this gene blocks the formation of fur color resulting in offspring with white fur (ww offspring have normal fur). Determine the proportions of offspring from the following cross: AaWw x AaWw. (Assume no gene linkage). 27. Explain why people with type O blood are known as universal donors. Likewise, explain why people with type AB blood are known was universal recipients. 28. Red-green color blindness is a sex-linked recessive trait. A man with normal color vision and a color-blind woman have a son. Determine the probability that their son will be color-blind. If they have a daughter, what is the probability that she will be color-blind?

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29. Use the pedigree below to determine the probability that a son of the 3rd generation female will be color-blind if she marries a normal male. What if she marries a color-blind male? Explain your answer.

30. Describe what is meant by linkage, relating to genes. 31. Discuss the relationship between frequency of recombination between genes and their positions on the chromosomes. Homework 32. Describe polygenic inheritance and explain how it differs from the inheritance of traits by a single gene. 33. Describe epistasis and provide an example. 34. A woman with normal color vision whose father was color-blind marries a man with normal color vision whose father was also color-blind. Determine the proportion of probable offspring. 35. Suppose you are working on a genetics lab in your college biology class. You cross the assigned parental generation of Drosophila flies. You notice that the F1 generation has twice as many females as males. Explain what might be the cause of the unusual ratio of sexes. Show a genetic cross that supports your explanation. 36. In Drosophila, normal wings are dominant to vestigial wings. Wing version is not sexlinked. However, the gene for eye color is sex linked. Wild type flies have red eyes, which is dominant to black eyes. A red eyed male with normal wings is mated with a black-eyed female with vestigial wings. Determine the expected phenotypic ratio in the F1 generation. 37. Explain what is meant by recombination in terms of chromosomes and chromatids. 38. Place the following genes that are located on the same chromosome in the correct order. There are three genes: X, Y, and Z. In a genetic recombination experiment, 12% recombination is found between genes X and Y. Eight percent recombination is found be X and Z, and 3% recombination between Y and Z. 39. Determine the sequence map of the following data showing crossing-over frequency.

Genes B and A C and D C and A B and D C and B www.njctl.org

Frequency 5% 8% 9% 4% 4% PSI AP Biology

Heredity

Probability, Statistics & Genetics Classwork 40. When performing a chi-squared analysis a null hypothesis is necessary. Explain what a null hypothesis is and why it is necessary. 41. A group of biology students wanted to see if there was a color preference towards the blue trays in class. They set up 4 separate piles of trays consisting of blue, green, orange and yellow. Using chi-squared analysis, can they reject the null hypothesis that these trays are selected at random?

Tray Color Blue Green Orange Yellow

Number of Trays Chosen 22 17 54 27

42. The laws of probability can simplify many genetic problems. A 6-sided die is tossed 5 times. Determine the probability of rolling a 4, five times in a row. 43. A woman has the following combination of alleles for the genes H, I, J, K, L, M and N: HhIIJjkkLlMmNn. How many different kinds of gametes can be produced? Homework 44. A biology student is trying to determine if his genetic corn was derived from a true dihybrid cross. He obtained the following data regarding two traits. Use the chi-square test to determine if this is a true dihybrid cross.

Purple and smooth Purple and wrinkled Yellow and smooth Yellow and wrinkled

Observed 283

Expected

X2

81 79 24 467 total

45. A researcher believes she has invented a new drug to lessen tooth decay. Two groups totaling 1000 participants were asked to rinse with an unknown solution after brushing twice a day for 1 year. One group was given a placebo (group A, 500 members) and the other the drug (group B, 500 members). Both groups were assessed for cavities. Use the data to complete the chi-squared analysis to determine if there is a significant difference between the placebo and the new drug.

Group A Group B

Observed cavities 227 186

X2

46. Suppose we have 4 gene pairs, each of which affect a different characteristic. Tt, Gg, Hh, and Mm are the alleles for these genes. Determine the probability of obtaining the genetic combination of TtGGHhmm from the parental cross of TfGgHhMm x TTGgHhMm

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47. Using the same genes as the previous question, determine the probability of obtaining the genetic combination of TtGGHhmm from the parental cross of ttGGhhMm x TTGGHhmm 48. Describe the advantage to an organism reproducing through sexual reproduction.

Free Response

Calculated distances and percent crossovers. Shaw, K. & Miko, I. (2008) Chromosome theory and the Castle and Morgan debate. Nature Education 1(1):142

1. If traits occur together 50% of the time, then they also occur apart 50% of the time. If traits occur together more often than they do separately, then they show crossing-over events less than 50% of the time, and the traits are predicted to be more closely associated on the chromosome. This crossing-over percentage is therefore a measure of their degree of linkage on the same chromosome. Specifically, for genes on the same chromosome, the following statements hold true: Crossing-over rate < 50% = Linkage (for genes and traits on same chromosome) Crossing-over rate = 50% = No linkage for genes and traits on separate chromosomes) A. H. Sturtevant, an undergraduate student working with Morgan, crossed Drosophila and looked at the linkage of 5 traits (P,R,M,C,O) around a single locus or trait B. The percent crossovers and calculated distance from B are shown in the above table. a. Draw a plausible genetic map using the crossover percent data in above table. b. Describe why these traits would not follow Mendelian phenotypic ratios between homozygous Parental genotypes or F1 generation crosses.

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Tall

Short

Expected

300

100

Observed

305

95

Lobo, I. (2008) Genetics and statistical analysis. Nature Education 1(1):109

2. When Mendel crossed a pure breed Tall with a pure breed Short pea plant, the progeny was always Tall, but when he crossed the Tall progeny- he got a 3 Tall:1 Short ratio. For four hundred plants, the expected and observed numbers for a heterozygous cross is shown in the above table. a. Perform a Chi-square test for the above data. The null hypothesis for this test will be that observed data will differ from the expected 3:1 Mendel ratios by chance at a significance level of 95%. Does your test confirms or rejects the null hypothesis? Show your calculations. b. If you did a test cross with the parent generation and F1 generation, how would probabilities of the offspring differ?

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http://en.wikipedia.org/wiki/Mitochondrial_DNA Miko, I. (2008) Non-nuclear genes and their inheritance. Nature Education 1(1):135

3. Right: In humans, mitochondrial DNA is the smallest chromosome, coding for 37 genes and containing approximately 16,600 base pairs. Most of the genes are involve in respiration or protein synthesis. In most species, including humans, mtDNA (mitochondrial DNA) is inherited solely from the mother. Left: Mitochondria that have wildtype mtDNA are shown in red (lighter shade); those having mutant mtDNA are shown in blue (darker shade). a. In a heterozygous population of cells or tissue, how would the distribution of mutant alleles present in the mitochondrial DNA differ from the distribution of mutant alleles present in chromosomal DNA? b. Would how the distribution of mutations in mtDNA be affected in mitosis versus meiosis?

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4. In 1909, not long after Mendel's principles of inheritance became well accepted, Carl Correns noticed some strange patterns of inheritance in fouro'clock plants, Mirabilis jalapa. He crossed female plants (seed) of three different phenotypes with male plants (pollen) having the same three phenotypes: white, green, and variegated stem and leave colors. Correns’s results are shown in the table on the left. Miko, I. (2008) Non-nuclear genes and their inheritance. Nature Education 1(1):135

a. Identify the dominant phenotype, and describe whether or not the sex of the plant matters. b. How do Correns’s results differ from the expected Mendelian patterns of inheritance?

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5. Microsatellites are repeated sequences that usually consist of two, three or four nucleotides (di-, tri-, and tetranucleotide repeats respectively), and can be repeated 3 to 100 times, with the longer loci generally having more alleles due to the greater potential for slippage, a type of replication error. Microsatellites, such as CA nucleotide repeats, are very frequent in human and other genomes and can be present every few thousand base pairs. If human DNA is cut with one or more enzymes, the fragments will be of varying lengths for individuals; therefore, theses difference to can be used to identify a specific person with near perfect accuracy. The diagram above shows an electronic readout for the electrophoresis of DNA fragments from a potential father, mother, and child. The numbered peaks represent specific DNA fragments that are specific to the respective individual. Adams, J. (2008) Paternity testing: blood types and DNA. Nature Education 1(1):146

a. How do you know the child is the son of this father and mother? b. Using the following chromosome representations or models, show how meiosis would produce the gametes that through fertilization generated the genetic combination seen in the DNA electrophoregram. 9.3

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6

7

9

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Heredity-Answer Key

1. Dominant traits are alleles that show up more often compared to the recessive trait. The dominant allele trait can show up in the homozygous dominant and heterozygous. The recessive trait will only show up in the homozygous recessive. 2. Heterozygous means two different alleles, such as Rr. Homozygous means two of the same alleles such as RR or rr. 3. An allele is alternate for of a gene. Alleles can be either dominant or recessive and are represented as a capital letter or lowercase letter respectively. 4. Genotype is the genetic makeup of an organism. Phenotype is the outward appearance of the organism. 5. A test cross is performed by mating a plant that is homozygous recessive with an unknown individual. If the organism is homozygous recessive all the offspring will look like the known recessive. If the results are half like one parent and half like the other parent, than the unknown is heterozygous. If the traits exhibited are like the unknown individual than it is homozygous dominant. 6. Tt x tt 7. Both parents would be heterozygous, therefore the chance would be ¼. 8. The law of independent assortment is when different genes that control different traits separate independently in the formation of gametes. 9. Monohybrid cross involves genetic crosses that involve a singe trait from both parents. A dihybrid cross involves two different traits. 10. The children would be expected to be ¼ brown eyed tasters, ¼ brown eyed non-tasters, ¼ green eyed tasters and ¼ green-eyed nontasters. 11. Huntington’s Disease is a degenerative brain disorder, that eventually takes away a persons ability to walk, think, talk and reason. Huntington’s disease is a result of a dominant allele. 12. Chance of brown eyes is ¾, blue eyes is ¼. 13. ¼ SS, ½ Ss, ¼ ss therefore the phenotypes would be ¾ smooth and ¼ rough 14. Yellow would be dominant. 15. Both parents are heterozygous and the child received the recessive allele from each parent. Parent’s genotype Tt, phenotype tongue roller: child genotype tt and phenotype non-tongue roller. 16. ½ DD and ½ Dd for the first cross. ¼ DD and ½ Dd for the second cross. ¼ will die. 17. Pink is dominant. Parents are both heterozygous for pink. 18. Complete a test cross. Cross two red and two yellow to determine if both are homozygous. Cross the red with the yellow to create the heterozygous. Cross the one whose color does not show up with the heterozygous, if the result is half red and half yellow you know the yellow is recessive and the red is dominant. 19. The grey rabbit is Ggw and the chinchilla is Ggw 20. Chicken #1 – FfPp, Chicken #2, FFPP, Chicken # 3 FfPP, Chicken #4 – FfPp 21. The man would be Bb. Therefore, the possible offspring would be ½ Bb and ½ bb. 22. A. ½ YY, ½ Yy B. ¼ YY, ½ Yy, ¼ yy C. ½ Yy, ½ yy 23. The effects of both alleles can be detected in a diploid heterozygous with two different alleles of a gene. The heterozygous gives a different phenotype than homozygous dominant or homozygous recessive. One example is Red four o’clock flower being mated with a white four o’clock gives a pink four o’clock. 24. ½ will have red flowers; RR, ½ will have pink flowers; Rr. Next cross: all will have pink flowers Rr. Last cross ¼ will have red RR, ½ will have pink Rr, and ¼ will have white rr.

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25. Pleiotropy involves a situation where a single gene influences more than one phenotypic characteristic. Polygenic inheritance involves a pattern of inheritance in which the interaction of two or more similar genes determines the phenotype. 26. White, tabby and black in 12:3:1 ratio 27. Type O blood type lack the A and B antigens they can be given to any one. Type AB blood does no have the neither anti-A nor anti-B antibodies, therefore they can receive any blood because they don’t have the antibodies to reject them. 28. For the son, 100%. For the daughter none will be colorblind. 29. The chance is ½ for both. 30. Genes that are located on the same chromosome and are near one another tend to be inherited together. 31. Genes that are closer together on the same chromosome have a lesser chance of recombination. 32. In polygenic inheritance the same phenotype is controlled by two or more different genes with various alleles. Inheritance by a single gene is controlled by one set of alleles. 33. Epistasis is where the alleles of one gene overrides the expression of one or more different genes. In guinea pigs hair color produces black or brown. The expression of the B – black gene is controlled by a different gene C for color. For any color, black or brown to be expressed they must have a CC or Cc, otherwise the guinea pig will be albino. 34. All females will have normal vision, ½ males will be color blind, ½ normal color vision. 35. A recessive allele carried on the X chromosome is lethal in males. XLXl x XLY = ¼ XLX , ¼ X LXl, XLY, ¼ XlY (lethal) 36. 3/8 red eyes/ normal, 3/8 black eye/ normal, 1/8 red /vestigial, 1/8 black /vestigial 37. Recombination occurs when homologues on each chromatid cross over, break of and exchange places. 38. X – 8% -- Z -3% -Y ----------12%------39. ADBC 40. A null hypothesis is no difference between what is being tested and what is accepted and any difference is due to chance. 41.

Observed 22 17 54 27 Total = 120

Blue Green Orange Yellow

Expected 30 30 30 30 Chi = 27.2

Df = 4-1 = 3. At 0.05 we see a p value of 7.82 our value of 27.2 is much greater so we reject our null hypothesis meaning there is a color preference. 42. 1/6 x 1/6 x 1/6 x 1/6 x 1/6 = 1/7776 43. 32 different combinations. 44.

Purple and smooth Purple and wrinkled Yellow and smooth www.njctl.org

Observed #

Expected #

283 81

262 88

(observedexpected)2/expected 1.68 0.56

79

88

0.92

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Yellow and wrinkled

24

29

0.86

467 total

Total =4.03

Df = n – 1 or 4 – 1 = 3 Chi-square = 7.82 our value of 4.03 is less indicating that there is a good fit between our data and the expected values. 45.

Observed with cavities 227 186

Group A Group B

Expected with cavities 250 250

Sum of (observedexpected)2/expected 2.12 16.38 18.50

Df = n-1 probability at 0.05 would be 3.84; since our value is greater we reject the null hypothesis. 46. ½ x 1/4 x ½ x ¼ = 1/64 47. 1x1x1/2x1/2 = ¼ 48. Sexual Reproduction provides more opportunity for genetic variability in the offspring. There is also a greater chance of producing more fit offspring do to sexual cycles limiting when a species can produce offspring. 1. Recombination Frequency

a. b. Since these genes and the traits they contribute to are situated on the same chromosomes, the traits would not be able to independently assort and segregate during meiosis; therefore, the number of possible alleic gametes and combinations between these genes would be diminished and the phenotypic and genotypic ratios would differ from the ratios expected from Mendelian inheritance. Learning Objectives: LO 3.12 The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring. [See SP 1.1, 7.2] 2. Chi Squared N=1

(305-300)2/300

.083

a. Since χ2=,333 and is well below the allowed value for χ2 of 3.841 at a significance level of 95%, we Degrees of Total (χ2) .333 must confirm the null freedom= hypothesis that the N-1= (2-1)= 1 differences between the expected and observed results are due to fluctuations in the data that can be accounted for by chance. N=2

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(95-100)2/100

.25

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b. A test cross involves the testing of a putative homozygous or heterozygous dominant individual with a known homozygous recessive. The progeny of a homozygous recessive individual with a homozygous dominant genotype would give heterozygous dominant phenotype and genotypes. . The progeny of a homozygous recessive individual with a heterozygous dominant genotype would give half heterozygous dominant phenotype and genotypes and half homozygous recessive. The test cross allows you to discriminate between homozygous and heterozygous genotypes. Learning Objectives: LO 3.12 The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring. [See SP 1.1, 7.2] LO 3.14 The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data sets. [See SP 2.2] 3. mtDNA a. Since mitochondria do not independently assort and segregate, like chromosomes, their distribution and thus their allelic distribution will be more random than alleles in chromosomes. Due to random segregation of mitochondria during cell division, some tissues may become enriched in cells with a large number of mutant, possibly defective, alleles that lead to organ failure or disorders. b. The distribution of mitochondria would be random in either case and not be affected by the type of cell division. But in the gametogenesis of some organisms, like humans, all four cells from meiosis do not mature into oocytes or egg cells; therefore, if the cell with more mutant mitochondria is selected then the consequences of those mutations would be more evident in the offspring that is derived from the oocyte. Learning Objectives: LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. [See SP 6.5] LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. [See SP 6.3] LO 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. [See SP 1.2] 4. Non Mendelian Inheritance a. The dominant phenotype depends on the plant that provides the ovule that becomes the seed after fertilization with pollen. Whatever phenotype the seed plant is, the progeny show that phenotype, or in the case of variegated- all three phenotypes. b. Mendelian inheritance involves the contribution of two alleles to one trait, with one allele being dominant over another allele. The flower and stem color in Mirabilis jalapa only matters the contribution of the seed plant without regard to pollen plant phenotype. At first, the inheritance looks Mendelian until the sex of the plants is reverse and the dominant allele become recessive and vice versa. Learning Objectives: LO 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. [See SP 6.5]

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LO 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. [See SP 6.3] LO 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. [See SP 1.2] 5. Paternity a. The child has DNA fragments with microsatellites that are derived from both the mother and the father (fragments 6 from the mother and fragment 7 from the father). Sexually reproducing organisms contribute equally to the genetic makeup of the offspring. b.

6 6

9.3 9.3

Meiosis I

6

6

Meiosis I

9.3

6

9 9

7 7

9.3

Meiosis II

6

9 9

7 7

Meiosis II 9.3

9.3

7

7

9

9

Allelic combination that yielded the observed results

Learning Objectives: LO 3.12 The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring. [See SP 1.1, 7.2]

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