Podgorski Genetics- Problem Set 3 PDF

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Summary

Answers to associated problem set 3 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.

What is (or are) the possible inheritance pattern(s) for the characteristic in this pedigree? a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant e. All of the above are possible.

2.

This pedigree is for red-green colorblindness, an X-linked recessive trait. What is the chance that the first child born to II-2 and II-3 will be colorblind? ¼ because there is a ½ chance that the II-2 is heterozygous and that cross would result in ½ of the children being color-blind

3.

This pedigree shows the inheritance of sickle cell disease, a rare autosomal recessive disorder. What is the chance that the first child born to II-2 and II-3 will have sickle cell disease?1/16 because there is a ½ chance that II-2 and II-3 are heterozygous and the heterozygous cross would result in a ¼ chance their children would have sickle-cell disease.

4.

The pedigree below is of an autosomal dominant disease. If individual IV-7 married a person who was unaffected with the disease, what is the probability that their first child would have the disease? ½ because IV-7 is heterozygous and the spouse would be homozygous recessive, resulting in a ½ chance of being affected by the disease

5.

What is the most likely mode of inheritance in the following pedigree? Explain briefly. Autosomal dominant, there is no bias between the sexes (not sex-linked) and it does not skip generations

6.

What is the most likely mode of inheritance in the following pedigree? Explain briefly. Autosomal recessive, not bias towards the sexes (no sex-linked), skips generations, and consanguinity

often results in the reappearance of affected children

7.

What is the most likely mode of inheritance in the following pedigree? Explain briefly. X-linked dominant. In generations II and III the affected father passes the trait to all of his daughters, and the affected females pass the trait to half of their offspring (regardless of sex) and overall the daughters are more likely to be affected.

8. Joe is colorblind. Both his mother and father have normal vision, but his mother’s father is colorblind. All Joe’s other grandparents have normal color vision. Joe has three sisters—Patty, Betsy, and Lora— all with normal color vision. Joe’s oldest sister, Patty, is married to a man with normal color vision; they have two children, a 9-year-old color-blind boy and a 4-year-old with normal color vision. Assume color blindness in this family is an X-linked recessive trait. a. If Patty and her husband have another child, what is the probability that the child will be a color-blind boy? There is a ¼ chance they will have a colorblind boy. Patty is X+ Xb and her husband is X+Y, its a 50% chance they will have a boy and a 50% chance that boy will have a XBY phenotype. ½*½ =¼ b. If Lora marries a man with normal color vision, what is the chance that their first child will be colorblind? There ⅛ chance her child will be color blind. There is a 50% chance laura has a XB allele from her mother, and another 25% chance her first child will be a boy with color blindness. ½ * ¼ = 1/8

9. In a snail species, an autosomal allele that causes a banded shell is recessive to the allele for an unbanded shell. Alleles at a second locus determine whether the shell is yellow, a recessive phenotype, or brown. A banded yellow snail (bbyy) is crossed with a homozygous unbanded brown snail (BBYY). The F1(BbYy) are then testcrossed to a banded yellow snail. a. What will the results of the test cross be if the two genes are linked with no crossing over? BbYy x bbyy cross will result in 50% progeny BbYy (unbanded brown), and 50% bbyy (banded yellow), because all the alleles will nonrecombinant/paternal alleles. (BY and by) b. What will the results of the test cross be if the two genes assort independently? This will result in a 1:1:1:1 ratio, 25% BbYy(unbanded brown), 25% Bbyy(unbanded yellow), 25% bbYy(banded brown), 25% bbyy(banded yellow).

c.

What will the results of the test cross be if the two genes are linked and 20 cM apart? There will be both recombinant and nonrecombinant progeny, with nonrecombinant phenotype occurring 80% of the time, and the recombinant phenotype occurring 20% of the time.

10. If there is exactly one crossover in each meiosis, what fraction of gametes will be recombinant gametes? 50% of the gametes will be recombinant because in crossing one pair of sister chromatids will be involved in crossing over which will result in 2 recombinant chromatids and 2 nonrecombinant chromatids. 11. In tomatoes, tall (D) is dominant over dwarf (d), and smooth fruit (P) is dominant over pubescent (p) fruit, which is covered with fine hairs. A farmer has two tall and smooth tomato plants, which we will call plant A and plant B. The farmer crosses plants A and B with the same dwarf and pubescent plant and obtains the following numbers of progeny: Progeny of _ Plant A Plant B Dd Pp 122 2 Dd pp 6 82 dd Pp 4 82 dd pp 124 4 a. What are the genotypes of plant A and plant B? Both plants are heterozygous and their following phenotypes are: Plant A - DP/dp Plant B - Dp/dP. b. What is the map distance between the loci that determine height of the plant and pubescence? Plant A recombinant percentage of 3.9%, plant B had a recombinant percentage of 3.5%. The average between the two percentage is 3.72%, giving an approximate map distance of 3.7cM. c. Explain why different proportions of progeny are produced when plant A and plant B are crossed with the same dwarf pubescent plant. The tomato plants have linked alleles (DP and dp), the other genotypes are a result of a small percentage of crossing over. Plant A is Cis and plant B is Trans Configuration.

12. In cucumbers, heart-shaped leaves (hl) are recessive to normal leaves (Hl), and having numerous fruit spines (ns) is recessive to having few fruit spines (Ns). The genes for leaf shape and number of spines are located on the same chromosome; mapping experiments indicate that they are 32.6 map units apart. A cucumber plant having heart-shaped leaves and numerous spines is crossed with a plant that is homozygous for normal leaves and few spines. The F 1 are crossed with plants that have heart-shaped leaves and numerous spines. Assuming no need to correct for double crossovers, what phenotypes and proportions are expected in the progeny of this cross? HlhlNsns: normal leaves, few spines: 33.7% hlhlnsns: heart-shaped, numerous spines: 33.7% Hlhlnsns: normal leaves, numerous spines: 16.3% hlhlNsns: heart-shaped, few spines: 16.3%

13. In mice, apricot eye (a) is recessive to normal eye (A) and black coat (b) is recessive to normal coat (B). The eye and coat color genes are 10 cM apart on mouse chromosome 3. In a cross of AB/ab x Ab/aB what fraction of progeny are predicted to have apricot eyes and black coats? 2.25%, from the AB/ab parent, there is a 45% chance of inheriting the ab gamete and there is a 5% of inheriting the ab gamete from the Ab/aB parent. 45% x 5%= 2.25% 14. In Drosophila, the held out-wings (HOW) gene is at 4 cM on chromosome 2 and the curved wing (CW) gene is at 75.5 cM on the same chromosome. Held-in and straight wings are dominant to both held-out and curved wings. In a test cross of a HOW+ CW+/HOW- CW- fly, what percentage of the offspring are expected to have held-out, straight wings? Justify your answer. 35.75% will have heldout, straight wings (HOW HOW CW+ CW). HOW+ HOW CW+ CW = 14.25% HOW+ HOW CW CW= 35.75% HOW HOW CW+ CW= 35.75% HOW HOW CW CW= 14.25% Because the genes are 71.5 cM apart, causing the recombinant genes to be 35.75% each, leaving 14.25% non-recombinants 15. In house flies, curved wings (cu) are recessive to normal wings (cu+) and protruding eyes (pr) are recessive to normal eyes (pr+). Both traits are encoded by autosomal genes. A fly has the genotype cu+ cu pr+ pr with the genes in repulsion. What will be the most common set of alleles of these genes in gametes produced by this fly? The possible allele combinations are: cu+ pr & cu pr+. For repulsion to occur, there would have to be a dominant allele and recessive allele on each gamete 16. The human clotting factor X and neuraminadase genes linked on chromosome 12. Imagine that you measure the genetic map distance between these genes in females and determine this to be 16 cM. Assuming two-fold lower rate of recombination in males relative to females, what map distance would you find if you repeated this study in males? Justify your prediction. 8 cM, because males would have half the recombination rate as females, they would have 8% recombination frequency, because females are 16% due to 1 cM=1% frequency and 16cM=16% frequency. 16%/2=8%. Thus males would be 8%=8cM. 17. The human clotting factor X and neuraminadase genes are linked on chromosome 12. Imagine that you measure the genetic map distance between these genes in females and determine this to be 16 cM with the genes separated by 16 mB (one mB is one million base pairs of DNA). If you repeated this study in males, what would you estimate the spacing of these genes to be in mB? 8 million base pair, because there is 8% recombination in males, 1% recombination is equal to 1 cM, which is equal to 1 million base pairs, 8x1 million= 8 million base pairs apart. The physical map distance for males and females would be the same, while the genetic map is different. Recombination occurs twice as often in females, because the males have a smaller map distance, it is more likely to be correct. 18. R. W. Allard and W. M. Clement determined recombination rates for a series of genes in lima beans. The following table lists paired recombination rates for eight of the loci (D, Wl, R, S, L, Ms, C, and G) that they mapped. On the basis of these data, draw a series of genetic maps for the different chromosomes on which these genes are located, indicating the distances between the genes. Keep in mind that these rates are estimates of the true recombination rates and that some error is associated with each estimate. An asterisk beside a recombination frequency indicates that the recombination frequency is significantly different from 50%.

D--Wl---------------------R 2.1 38 D to R 40.1 (39.3 is underestimated) L1---------S 26.9 C----Ms 14.7 Linkage group 1: D, Wl, R Linkage group 2: L1, S Linkage group 3: C, Ms Linkage group 4: G The different linkage groups could be spaced far apart on the same chromosome or could be on different chromosomes. 19. Species I is diploid (2n = 8) with chromosomes AABBCCDD; related species II is diploid (2 n = 8) with chromosomes MMNNOOPP. Individuals with the following sets of chromosomes represent what types of chromosome mutations? a. AAABBCCDD: Trisomy A b. MMNNOOOOPP: Tetrasomy O c. AABBCCDDAAABBBCCCDDD : Autopentaploidy d. AABBCCDDMMNNOOPP : Allotetraploidy or Amphidiploid 20. A species has 2n = 16 chromosomes. How many chromosomes will be found per cell in each of the following mutants in this species? a. monsomic: 15 = (2n-1) b. autotriploid: 24 = (2n+n) c. autotetraploid: 32 = (2n+2n) d. trisomic: 17 = (2n+1) e. nullisomic: 14 = (2n-2) f. autopentaploid: 40 = (2n+3n) g. tetrasomic: 18 = (2n+2)...