Week 7 Discussion Questions SP24

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Apr 3, 2024

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Discussion Questions Week 7 Questions SP23 1. In tomatoes, the D gene controls plant height, and the H gene controls fruit skin texture. The D allele produces tall plants and d allele produces dwarf phenotype. The H allele produces smooth fruits and h allele produces hairy fruits. A gardener mates two tall, smooth heterozygotes (Plant A & Plant B) with a dwarf, hairy tester plant and gets the following offspring. a. What are the genotypes of the A and B plants? (If you think the genes are linked, write the genotypes to reflect that. A genotypes: DH/dh B genotypes: Dh/dH b. Are the D and H genes linked? If so, what is the map distance between the loci? Round to the nearest whole number. A plant: (6+5)/256 *100 = 4.2 ๏ƒ  4 mu B plant: (3+4)/172 * 100 = 4.1 ๏ƒ  4 mu c. Which heterozygous plant, A or B, has the repulsion/trans arrangement of the alleles for the two traits? Plant B has the repulsion arrangement (Dh/dH) 2. In the snail Cepaea nemoralis , an autosomal allele causing a banded shell, u , is recessive to the allele for an unbanded shell, U . Genes at a different locus determine the background color of the shell. The allele for a yellow shell, b , is recessive to the allele for a brown shell, B . A banded, yellow snail is crossed with a homozygous unbanded, brown snail. The F1 progeny are then crossed with banded, yellow snails in a testcross. a. Write out the genotypes of the parent and F1 generation for the cross described. P 1 : uu bb P 2 : UU BB b. What is the phenotype of the F1 snails? The phenotype would be unbanded and brown c. Predict the percentage of F2 progeny with each phenotype if the two genes are 1) linked with no recombination, 2) assort independently, and 3) are 10 map units (m.u.) apart. Write the expected percentage for each phenotype for each of the scenarios in the table. Relationship of U and B loci Phenotypes of F2 Snails Banded, yellow Unbanded, brown Banded, brown Unbanded, yellow 1) Linked with no recombination 50 50 0 0 2) Independently Assorting 25 25 25 25 3) Linked with and 10 m.u. apart 45 45 5 5
Discussion Questions Week 7 Questions SP23 3. A geneticist is using a three-point testcross to map three linked Drosophila recessive mutations called s , z , and n , where s is associated with abnormally slow movement, z is associated with zigzag movement pattern, and n is associated with narrow wings. The geneticist first crosses homozygous narrow flies ( nn ) to homozygous slow, zigzag flies (ss zz ). Next, he testcrosses the F1 progeny to slow, zigzag, and narrow parents and obtains the results in the table. NOTE: The first parent is homozygous for wild type alleles of the other two genes ( s + s + z + z + ). The second parent is homozygous for the wild type alleles of the wing gene ( n + n + ). a. Identify the Parental (P), single crossover (SCO) and double crossover (DCO) classes among the progeny shown. Write it alongside the table. b. Based on the data obtained, what is the order of the three genes? The order will be s- n-z c. Calculate the distances between each pair of genes. n-s: (219+241+8+5)/2143 *100% = 22.1 mu n-z: (209+223+8+5)/ 2143 *100% = 20.8 mu s-z: 22.1+20.8 = 42.8 d. Draw a simple genetic map showing the order and distances of the three genes on the chromosome. e. Calculate the coefficient of coincidence and interference for the region. 13 / (0.221)(0.207)(2143)= .87 P/NCO P/NCO SCO 1 SCO 1 SCO 2 SCO 2 DCO DCO
Discussion Questions Week 7 Questions SP23 4. Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver , qu , and determines that it is an autosomal recessive mutation. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings , vg . She crosses a fly homozygous for quiver and wild type wings with a fly homozygous for vestigial wings and wild type for behavior. She then uses the resulting F 1 females in a testcross. She obtains the flies from this testcross. Test the hypothesis that the genes  quiver  and  vestigia l assort independently by calculating the chi-square value,  ๐‘‹ 2 , for this hypothesis. Use the Chi-square Test for Independence (using a contingency table) . Provide the  ๐‘‹ 2  to one decimal place. Then look up the p-value for 1 degree of freedom, and interpret it P value indicates that the difference between the observation and the expected number of progeny would be statistically significant. So in this case we would be rejecting the this. 5. In Drosophila melanogaster , black body ( b ) is recessive to gray body ( b + ), purple eyes ( pr ) are recessive to red eyes ( pr + ), and vestigial wings ( vg ) are recessive to normal wings ( vg + ). The loci encoding these traits are linked, with the map distances shown in the image. The interference among these genes is 0.5. A fly with a black body, purple eyes, and vestigial wings is crossed with a fly homozygous for a gray body, red eyes, and normal wings. The female progeny are then crossed with males that have a black body, purple eyes, and vestigial wings. If a total of 1000 progeny are examined, determine the expected number of offspring that result from each crossover event. (Problem continued on next page.) Phenotype Number of flies ๐‘ฃ๐‘” +   ๐‘ž๐‘ข + 99 vg qu 97 vg  ๐‘ž๐‘ข + 224 ๐‘ฃ๐‘” +  qu 230
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Discussion Questions Week 7 Questions SP23 Double crossovers (DCO): (0.06)(.13)(1000)(0.5)= 3.9 which rounds up to 4 Single crossovers (SCO) between b and pr : (0.06) (1000) - 4 =56 (28 per type) Single crossovers (SCO) between pr and vg : (0.13) (1000)-4 =126 (63 per type) Nonrecombinant offspring: 1000 - (4+56+126) =814 (407 per type)