Podgorski Genetics- Problem Set 2 PDF

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Answers to associated problem set 2 questions. Set assigned by Dr. Podgorski in his genetics Fall 2016 class. Disclaimer, some answers may not be completely correct! Not meant to be directly copied, best used to aid students in finding correct answers, or for checking their own answers. ...

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1. In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. The F1 are intercrossed to produce the F2. What is the predicted phenotypic ratio in the F2? A dihybrid cross with 2 heterozygotes results in a 9:3:3:1 phenotypic ratio. 9 with bitter fruit and yellow spots (B-S-) 3 with sweet fruit and yellow spots (bbS-) 3 with bitter fruit and no spots (B_ss) 1 with sweet fruit and no spots. (bbss) 2. In watermelons, bitter fruit (B) is dominant over sweet fruit (b), and yellow spots (S) are dominant over no spots (s). The genes for these two characteristics assort independently. A homozygous plant that has bitter fruit and yellow spots is crossed with a homozygous plant that has sweet fruit and no spots. If an F1 plant is backcrossed with the sweet, nonspotted parent, what phenotypes and proportions are expected in the offspring? A heterozygous and homozygous recessive cross will result in a 1:1:1:1 ratio. 1 with bitter fruit and yellow spots (BbSs) 1 with sweet fruit and yellow spots (bbSs) 1 with bitter fruit and no spots (Bbss) 1 with sweet fruit and no spots (bbss) 3. In mice, an allele for apricot eyes (a) is recessive to an allele for brown eyes (A). At an independently assorting locus, an allele for tan (t) coat color is recessive to an allele for black (T) coat color. A mouse that is homozygous for brown eyes and black coat color is crossed with a mouse having apricot eyes and a tan coat. The resulting F1 are intercrossed to produce the F2. In a litter of eight F2 mice, what is the probability that two pups will have apricot eyes and tan coats? P= .074 Binomial expansion P=8!/(2!*6!)*(1/16)^2 (15/16)^6= .074 chance of obtaining apricot eyes and tan of 2 out of 8 children. 4. In Drosophila, yellow body color is due to an X-linked gene that is recessive the gene for gray body. a. A homozygous gray female is crossed with a yellow male. The F1 are intercrossed to produce F2. Give the genotypes and phenotypes, along with their expected proportions, of the F1 and F2 progeny. F1 Generation: 100% Gray Phenotype - Males: 100% X+Y (Gray) - Females: 100% X+ Xy (Gray) F2 generation: ¾ Gray and ¼ Yellow Phenotype - Males: 50% X+Y (Gray) and 50% XyY (Yellow) - Females: 50% X+X+ (Gray) and 50% X+Xy (Gray) b. A yellow female is crossed with a gray male. The F1 are intercrossed to produce F2. Give the genotypes and phenotypes, along with their expected proportions, of the F1 and F2 progeny.

F1 Generation: ½ Gray and ½ Yellow Phenotype - Males: 100% XyY ( Yellow) - Females: 100% X+Xy (Gray) F2 Generation: ½ Gray and ½ Yellow Phenotype - Males: 50% X+Y (Gray) and 50% XyY (Yellow) - Females: 50% X+Xy (Gray) and 50% XyXy (Yellow) c. A yellow female is crossed with a gray male. The F1 females are backcrossed with gray males. Give the genotypes and phenotypes, along with their expected proportions, of the backcross progeny. F1 Generation: ½ Gray and ½ Yellow Phenotype - Males: 100% XyY ( Yellow) - Females: 100% X+Xy (Gray) F2 generation: ¾ Gray and ¼ Yellow Phenotype - Males: 50% X+Y (Gray) and 50% XyY (Yellow) - Females: 50% X+X+ (Gray) and 50% X+Xy (Gray) 5. The Talmud states that if a woman bears two sons who die of bleeding after circumcision, and additional sons she has should not be circumcised. Furthermore, the Talmud states that the sons of her sisters must not be circumcised, whereas the sons of her brothers should be. Is this religious law consistent with sound genetic principles? Explain your answer. Yes. We assume the gene is an X-linked recessive trait so we know that both the woman and her sister could carry the gene, which can be passed to their sons. The brother can only give a Y chromosomes to his sons, so whether he has it or not, he cannot pass to his sons. The brothers sons should be safe as long as the wife of the brother is not a carrier of the gene. (Same situation as color blindness) 6. Individuals of the following genotypes are crossed: AaBbCcX+X+ x AaBBccX+Y, where a, b, and c represent alleles of autosomal genes. What is the probability of obtaining genotype aaBbCcX+X+ in the first offspring? ¼ * ½ * ½ * ½ = 1/32 ¼ chance of aa, ½ of getting Bb, ½ of getting Cc, and ½ of getting X+X+ 7. Women who are heterozygous for an X-linked recessive trait sometimes exhibit mild expression of the trait. In contrast, in Drosophila females heterozygous for an X-linked recessive trait, expression of the recessive trait is not seen. What might cause this difference in expression between human and fly females? Humans experience X-Chromosome inactivation which causes one of the X chromosomes in females to be mostly deactivated leaving 15-20% genes active. Inactivation is completely random, causing about 50% of cells to express the X-linked recessive trait. In these cells there is still a chance that the 15-20% active part of the inactive chromosome could hinder the inactive gene. Overall, X-chromosome inactivation causes the X-linked recessive gene to be expressed in some human cells creating mild symptoms of the trait. The reason this is not seen in Drosophila is simply because there is no X-Chromosome inactive in flie.

8. Occasionally, a mouse X chromosome is broken into two pieces that each become attached to different autosomes. When this occurs, only one of the pieces of the X chromosome never undergo X inactivation. Propose a plausible explanation for this observation. The xist gene located on the X-chromosome must be on only one of the pieces of the chromosome and not on the other. Because the xist gene is separated from the other half of the chromosome the other half can never be deactivated. 9. In Drosophila, forked bristles are caused by a recessive X-linked allele, and brown eyes are caused by a recessive autosomal allele. In a cross of a female heterozygous at both loci with a brown eyed, forked bristle male, what proportion of their daughters will have brown eyes and forked bristles? ¼ of the daughters will have forked bristles and brown eyes, They can only receive a recessive allele for eye color from their father, as well as forked bristles. And half of them will get either allele from their mother. 10. You cross a homozygous female rat with pink toe pads and pointy ears to a homozygous male rat with black toe pads and round ears. The ear-shape gene is X-linked, while the toe pad color gene is autosomal. Round ears and pink toe pads are dominant traits. If the F1 are crossed to produce F2 progeny, what proportion of the F2 will be black-padded, pointy-eared males? 1/16 of the progeny of the F2 generation will be XbY rr, or black-padded, pointy-eared males. ⅛ of the males will show the black-padded, pointy-eared phenotype. 11. In Drosophila, yellow body is due to an X-linked gene (Xy) that is recessive to the gene for gray body (X+). A homozygous gray female is crossed with a yellow male. The F1 are intercrossed to produce the F2. What are the predicted phenotypes of males and females and their ratios in the F2? Females: 100% are gray bodies with the genotypes with 50% X+Xy and 50% X+X+. Males: 50% gray bodies with a X+Y phenotype, and 50% yellow bodies with XyY phenotype. 12. In chickens, baldness is due to a Z-linked recessive gene. A bald rooster is mated with a normal hen. The F1 from this cross are interbred to F1 females: 100% ZbW (bald) F1 males: 100% Z+Zb (normal) F2 females: 50% Z+W(normal) and 50% ZbW(bald) F2 males: 50% Z+Zb(normal) and 50% ZbZb(bald) 13. Anhidrotic ectodermal dysplasia is an X-linked recessive disorder in humans characterized by small teeth, no sweat glands, and sparse body hair. The trait is usually seen in men, but women who are heterozygous carriers of the trait often have irregular patches of skin with few or no sweat glands (see the illustration). Explain why the distribution of patches of skin lacking sweat glands differs among the females depicted in the figure, and even among identical twins.

The differences in the distribution of patches of skin lacking sweat glands could be due to X- chromosome inactivation. X-chromosome inactivation occurs in the embryo. The cells randomly inactivate one of the X-chromosomes and the inactivation is passed onto the daughter cells after mitotic division. (This example is similar to the patchiness of calico cats) Because the inactivation is random, even though twins will have identical genotypes, the distribution of the patches of skin will be different. 14. Allozymes are enzymes produced by different alleles of the same gene and can be detected using gel electrophoresis as proteins that migrate to different positions on the gel. Extracts from an SfSs heterozygous Morning Glory produce two different bands on a gel. When separated on a gel, what form of dominance do allozymes exhibit? Explain. At this level of observation, these allozymes express co-dominance, because both enzymes are produced and observed. 15. Considering the same alleles as the problem above, imagine that the Sf allele is wild type and produces a functional enzyme and the Ss allele is a loss-of-function mutant allele. When assaying levels of enzyme activity (enzyme activity is the rate at which a reaction is catalyzed) present in extracts from lysed cells, which form of dominance would these alleles exhibit? Explain. The cells would represent incomplete dominance, the amount of enzyme activity present would be intermediate to the levels of an SfSf homozygote and an SsSs homozygote, because 50% of the cells express Sf functional enzymes, and 50% produces Ss loss of function enzymes. 16. Considering the same alleles as the problems above, imagine that the Sf allele encodes a plant growth hormone. One copy of the Sf allele produces far more growth hormone than needed for normal growth but does not cause the plants to grow beyond their normal height. The Ss allele is a loss-of-function mutant allele. What form of dominance would these alleles exhibit? Explain. Complete dominance, because the phenotype of the heterozygote, S fSs, would be indistinguishable to a homozygote, SfSf, plant. In other words, their heights would be the same, because only one functional allele is needed to achieve maximum potential height. 17. When two plants with yellow leaves were crossed, progeny were produced in a ratio of 2 yellow: 1 green. When two green leaf strains were crossed, all the progeny had green leaves. When yellow and green leaf strains were crossed, progeny were produced in a 1 yellow: 1 green ratio. Explain these results. This is an example of a lethal allelism because the F1 generation produced a 2:1 ratio from two heterozygous parents. Genotypic ratios were ⅔ Yy and ⅓ yy with any offspring that had the genotype YY being unable to survive. This was also confirmed when a testcross was performed between yellow leaf plants (Y_) and a green leaf plants (yy). The offspring where ½ yellow (Yy) and ½ green(yy), concluding that the yellow leaf plants are heterozygous and none are homozygous dominant.

18. The yellow mouse coat color allele first studied by Cuénot is a recessive lethal allele. What form of dominance does the yellow allele exhibit for coat color? Explain your reasoning. The yellow allele expresses complete dominance or incomplete dominance for coat color in the case of a heterozygote. We can’t determine what type of dominance due to the recessive lethality of the gene. 19. A variety of poppy having lacerate leaves was crossed with a variety that has normal leaves. All the F1 had lacerate leaves. Two F1 plants were interbred to produce the F2. Of the F2, 249 had lacerate leaves and 16 had normal leaves. The following calculations prove that the 249:16 ratio is equivalent to a 15:1 ratio. Phenotype Observed Expected (O-E) (O-E)2 (O-E)2/E Lacerate 249 248.4 0.6 0.36 0.00145 Normal 16 16.7 -0.7 0.49 0.02934 x2=0.03079 Degree of freedom = n-1 = 2-1 = 1 P...