Part 2: Monohybrid Crosses. 1. In minks, coat color is controlled by a single gene.
The allele for a brown (B) coat is dominant to the allele for silverblue (b) coats.
ANSWER KEY
Basic Genetics Practice Problems Part 1: Mendel’s Works Mendel made the following crosses with pea plants. Complete the Punnett Squares and answer the following questions.
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R = red flowers r = white flower He crossed a red flowered R plant with a white flowered r plant. His results were 126 red flowered plants and 122 white flowered plants. Which of the Punnett squares above best show the parents and offspring that could give these results? Punnett square #4 best shows the parents and offspring from this cross. He crossed a red flowered plant with a white flowered plant. His results were 307 red flowered plants and 0 white flowered plants. Which of the Punnett squares above best show the parents and offspring that could give these results? Punnett square #1 best shows the parents and offspring from this cross. He crossed a red flowered plant with a red flowered plant. His results were 306 red flowered plants and 110 white flowered plants. Which of the Punnett squares above best show the parents and offspring that could give these results? Punnett square #3 best shows the parents and offspring from this cross. He crossed a red flowered plant with a red flowered plant. His results were 300 red flowered plants and 0 white flowered plants. Which of the Punnett squares above best show the parents and offspring that could give these results? Punnett square #2 best shows the parents and offspring from this cross. It is possible that Punnett square #3 could have produced these results as well and just by chance, none of the offspring were white.
Part 2: Monohybrid Crosses 1.
In minks, coat color is controlled by a single gene. The allele for a brown (B) coat is dominant to the allele for silverblue (b) coats. B = brown coats b = silverblue coats a. A homozygous brown mink was crossed with a silverblue mink. There were 9 offspring in the F1 generation. What color were they? All 9 offspring would be brown. They each inherited the dominant all from their brown parent. Cross: BB x bb B B b Bb Bb b Bb Bb
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Two of the offspring from the F1 generation above were mated. What would the ratio of brown to silverblue offspring be in the F2 generation? In the F2 generation, the ratio of brown minks to silverblue minks would be 3:1. Cross: Bb x Bb B b B BB Bb b Bb bb i. What fraction of the F2 generation will be homozygous brown? ¼ of the F2 generation would be brown (BB or Bb). ii. Heterozyous? ½ of the F2 generation would be heterozygous (Bb). iii. Homozygous silverblue? ¼ of the F2 generation would be silverblue (bb).
2.
In domestic cats, the gene for Tabby stripes (T) is dominant over the gene for no stripes (t). T = Tabby stripes t = stripeless a. If a breeder crossed a heterozygous Tabby cat with a stripeless cat, what percent of the F1 generation would have tabby stripes? Fifty percent of the F1 generation would have Tabby stripes. Cross: Tt x tt T t t Tt tt t Tt tt b.
There is a recent demand in pet stores for Tabby cats. As a breeder, which two cats from the F1 generation above would you cross to produce the most Tabby kittens? Why? To increase your chances of producing Tabby cats, it would be best to breed the heterozygous Tabbies from the F1 generation. Cross: Tt x Tt Cross: Tt x tt Cross: tt x tt T t T t t t T TT Tt t Tt tt t tt tt t Tt tt t Tt tt t tt tt
3.
If you had a Tabby cat and wanted to know if this cat was Homozygous or Heterozygous, what type of cat could you breed your Tabby with to prove its genotype? Explain your answer. You could perform a test cross and breed the Tabby cat with a stripeless cat. If there are any stripeless cats in the F1 generation, you can conclude that your Tabby was heterozygous. Cross: Tt x tt Cross: Tt x Tt T t T T t Tt tt t Tt Tt t Tt tt t Tt Tt
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In squash, the allele for white fruits is dominant over the allele for yellow fruits. If a white-fruited plant is crossed with a yellow-fruited plant and all of the offspring are yellow, what are the possible genotypes of the P generation and F1 generation? Trait: Fruit color Alleles: W = white fruit w = yellow fruit
Since all of the offspring are yellow, their genotypes must be homozygous recessive (ww). The only way that a white-fruited plant can produce a yellow plant is if it is carrying the allele for yellow fruit, so the white-fruited plant must be heterozygous (Ww) Cross: W? x ww W ? Parent Genotypes: Ww x ww F1 Generation Genotypes: ½ Ww: ½ ww w Ww ww w Ww ww 5.
In humans, normal skin pigmentation is due to a dominant gene, albinism to its recessive allele. N = normal pigmentation n = albino (no pigmentation) a. A normal man marries an albino woman. Their first child is an albino. What are the genotypes of these three persons? The woman and the first child are homozygous recessive (nn). The man must be heterozygous. If he was homozygous dominant, he would not be able to have an albino child. b. If they have more children, what would they probably be like, and in what ratio? If they have more children, they have a 50% chance that they could be albino and a 50% chance that they could have normal skin pigmentation. Cross: Nn x nn N n n Nn nn n Nn nn
Part 3: Dihybrid Crosses 1.
A dihybrid cross involves a study of inheritance patterns for organisms differing in two traits. Mendel studied dihybrid crosses and determined that different traits of pea plants, such as flower color and seed shape, were inherited independently. In pea plants, S is the allele for the dominant, spherical shape characteristic; s is the allele for the recessive, dented shape characteristic. Y is the allele for the dominant, yellow color characteristic; y is the allele for the recessive, green color characteristic. Traits: seed shape & flower color Y = yellow peas Alleles: S = spherical shaped peas y = green peas s = dented shaped peas a. A pea plant is heterozygous for both seed shape and seed color. What will be the distribution of these two alleles in this plant's gametes? Genotype: SsYy Possible Gametes: SY, Sy, sY and sy b. What would the expected genotypic and phenotypic ratio of the offspring of a SsYy x ssyy test cross? Cross: SsYy x ssyy Possible Gametes: (SY, Sy, sY or sy) x sy SY Sy sY sy sy SsYy Ssyy ssYy ssyy Genotypic ratios: 1 SsYy: 1Ssyy: 1ssYy: 1 ssyy Phenotypic ratios: 1 spherical, yellow: 1 spherical green: 1 dented yellow: 1 dented green c.
What are the expected parental genotypes if the phenotypic ratio in the F1 generation is 9 spherical, yellow: 3 spherical, green: 3 dented, yellow: 1 dented, green? Possible genotypic ratios for given phenotypes: 9 S_Y_: 3 S_yy: 3 ssY_: 1 ssyy Cross: Unknown SY Sy sY sy SY SSYY SSYy SsYy SsYy Sy SSYy SSyy SsYy Ssyy sY SsYY SsYy ssYY ssYy
sy
SsYy Ssyy ssYy ssyy
The parents must both be heterozygous for pea shape and pea color. 2.
In summer squash, white fruit color (W) is dominant over yellow fruit color (w) and disk-shaped fruit (D) is dominant over sphere-shaped fruit (d).. If a squash plant true-breeding for white, disk-shaped fruit is crossed with a plant true-breeding for yellow, sphere-shaped fruit, what will the phenotypic and genotypic ratios be for: Traits: Fruit color and shape Alleles: W = white fruit D = disk-shaped fruit w = yellow fruit d = sphere-shaped fruit a. The F1 generation? Parental cross: WWDD x wwdd Possible Gametes for each parent: WD & wd WD WD WD WD wd WwDd WwDd WwDd WwDd wd WwDd WwDd WwDd WwDd wd WwDd WwDd WwDd WwDd wd WwDd WwDd WwDd WwDd Genotypic ratio: 100% WwDd Phenotypic ratio: 100% white, disk-shaped fruit b. The F2 generation? F1 cross: WwDd x WwDd Possible Gametes for each parent: WD, Wd, wD or wd WD Wd wD wd WD WWDD WWDd WwDD WwDd Wd WWDd WWdd WwDd Wwdd wD WwDD WwDd wwDD wwDd wd WwDd Wwdd wwDd wwdd Genotypic ratio: 1/16 WWDD: 2/16 WWDd: 2/16 WwDD: 4/16 WwDd: 1/16 WWdd: 2/16 Wwdd: 1/16 wwDD: 2/16 wwDd: 1/16 wwdd Phenotypic ratio: 9 white, disk-shaped fruit: 3 white, sphere-shaped fruit: 3 yellow disk-shaped fruit: 1 yellow sphere-shaped fruit
3.
In horses, a black coat is dependent upon a dominant gene (B), and chestnut color upon its recessive allele (b). The trotting gait is due to a dominant allele, (T), the pacing gait to its recessive allele, (t). A stud is heterozygous for both a black coat and for trotting. He is mated to a mare that has a chestnut coat and is heterozygous for trotting. What is the probability that they may produce each of the following colts? Traits: Coat color & Gait Alleles: B= black, b = chestnut & T = trotting, t = pacing Parents: Stud (BbTt) x Mare (bbTt) Possible Gametes: Stud (BT, Bt, bT or bt) & Mare (bT or bt) BT Bt bT bt bT BbTT BbTt bbTT bbTt bt BbTt Bbtt bbTt bbtt a.
Black and trotting? (Genotypes BBTT or BbTt) 3/8 offspring b. Black and pacing? (Genotypes BBtt or Bbtt) 1/8 offspring c. Chestnut and trotting? (Genotypes bbTT or bbTt) 3/8 offspring d. Chestnut and pacing? (Genotypes bbtt) 1/8 offspring
