Background to Problems 13-14 Consider a single egg produced by a diploid female in which n = 3. If there was a single crossover between EACH of the three paired homologs during Prophase I of the meiosis that produced this egg (meaning there were 3 crossovers in all), calculate the following:
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- PROBLEM SET #1;: The parental genotypes for a series of crosses are wild-type male fruit flies mated to females with white eyes (wh) and miniature (min) wings. The phenotypes of the F1 generation were wild-type females, and males with white eyes, and miniature wings. These flies were allowed to mate with each other and produced the following offspring: Red eyes, long wings White eyes, miniature wings Red eyes, miniature wings White eyes, long wings 770 716 401 318 Total 2205 A. Are these genes linked? Why or why not? Your answer should diagram a cross to support your decision. *Picture name - Tradescantia spathacea meiotic cell. HPO (400x)Shown below are photomicrographs of Rhoeo tradescantia cells undergoing meiosis. Answer the following question for each of the photomicrographs: Identify the cytogenetic abnormality observed (ex. ring, chain, laggard, bridge). Identify the meiotic stage in which these aberrations are observed (as shown in the photomicrograph). Explain how these aberrations are formed and relate to the possible causal mutation(s). Will this result to sterile and/or fertile gametes? Explain.7-5. In corn a pair of genes determines leaf shape and another pair determines pollen shape. A ragged-leafed plant with round pollen was crossed to a ragged-leafed plant with angular-pollen, and the resultant progeny were classified as follows: Class Phenotypes 186 ragged-leaf round-pollen 11 IV 174 ragged-leaf angular-pollen 57 smooth-leaf round-pollen Total 63/480 smooth-leaf angular-pollen (a) Using alphabetical letters of your choice, designate the genes for the different leaf and pollen characters. (b) On the basis of the symbols given in (a) 1 provide genotypes two parents. (c) (forth)/(nbers) According to your hypothesis what would you have expected for each of the four classes of progeny? (d) After re viewing Chapter 8, use the chi-square method (p. 132) to test your hypothesis and indicate whether you accept or reject it.
- This problem leads you through the derivation of acorrected equation for RF in yeast tetrad analysis thattakes into account double crossover (DCO) meioses. A yeast strain that cannot grow in the absence ofthe amino acid histidine (his−) is mated with a yeaststrain that cannot grow in the absence of the aminoacid lysine (lys−). Among the 400 unordered tetrads resulting from this mating, 233 were PD, 11 wereNPD, and 156 were T.a. What types of spores are in the PD, NPD, andT tetrads?b. Are the his and lys genes linked? How do you know? c. Using the simple equation RF = 100 × [NPD +(1/2)T]/total tetrads, calculate the distance in mapunits between the his and lys genes.d. If you think about all the kinds of meiotic eventsthat could occur (refer to Fig. 5.24), you can seethat the calculation you did in part (c) may substantially underestimate RF. What kinds of meioses(NCO, SCO, or DCO) generated each of the tetradtypes in this cross? e. What incorrect assumptions does the simple RFequation…QUESTION:- In Drosophila, assume that the gene for scute bristles (s) is located at map position 0.0 and that the gene for ruby eyes (r) is at position 25.0. Both genes are located on the X chromosome and are recessive to their wild-type alleles. A cross is made between scute-bristled females and ruby-eyed males. Phenotypically wild-type F1 females were then mated to homozygous double mutant males, and 1000 offspring were produced. Give the phenotypes and frequencies expected..1:30 Genetics problem on Typ...icular diploid genotype 2 Genetics problem on Types of gametes and their gamete genotype s from a particular diploid genotype Please use the formula to solve the problem Types of gametes=2 to the power of heterozygous genes (2 eggene Examples Diploid genotype of gamete type AABB 2 to the power 0-1 AABb 2 to the power 1-2 AaBb ABBER AABbc ABbCe AABCCDD ADBBCCDO ABDCCDD AaBbCxDD AnBbCcDd 2 to the power 2=4 7 7 7 7 ? ? ? ? Genotype of gametes Print Layout AB AB, Ab AB, Ab, ab, ab 2 7 ? ? If this genotype AaBBCCDdEE is crossed with this genotype AabbccddEę, what is the probability of an offspring of this aaBbssDdEe. Please show how you have calculated the probability of this offspring. AD Read Aloud B Headings ||
- Progeny of triploid tomato plants often contain parts of an extra chromosome, in addition to the normal complement of 24 chromosomes . Mutantswith a part of an extra chromosome are referred to as secondaries. James and Margaret Lesley observed that secondaries arise from triploid (3 n), trisomic (3 n + 1), and double trisomic (3 n + 1 + 1) parents, but never from diploids (2 n). Give one or more possible reasons that secondaries arise from parents that have unpaired chromosomes but not from parents that are normal diploids.Homozyogous wild type male Drosophila PPQQRRSSTTUUVV (all linked) are irradiated to induce the formation of chromosomal deletions and then mated with homozygous recessive females. In several of these matings a unique pattern of pseudodominance could be correlated with the loss of specific polytene chromosome bands, as shown in the table... Cross Chromosomal Bands Deleted Pattern of Pseudodominance PqRs TuV 1 2 3 4 5 5-6 1-2 3-5 6-8 1-3 PQRSTUV PqrSTUV PQRSTUV pQrSTUV In which chromosomal band is gene R located? A. 3 OB. 4 OC. 2 D.8 O E. none of these answers are correct(Problem 65a) In the plant Arabidopsis, the loci for pod length (L, long; I, short) and fruit hairs (H, hairy; h, smooth) are linked 16 m.u. apart on the same chromosome. The following crosses were made: (i) L H/L H × 1 h/l h F1 (ii) L h/L h × 1 H/I H F1 If the F1's from cross i and cross ii are crossed, what proportion of the progeny are expected to be I h/l h? Oa. 16.00% Оb. 8.00% Oc. 4.00% Od. 3.36% Oe. 1.28%
- 10:10 E ← sisAndMeiosis.docx FREE EDIT PDF FREE CONVERT PDF TO WORD 2. Why are chromosomes important? 3. How are meiosis I and meiosis II different? 1. What is the state of DNA at the end of meiosis I? What about at the end of meiosis II? 4. Why do you use non-sister chromatids to demonstrate crossing over? 7. Identify two ways that meiosis contributes to genetic recombination. 10. P FREE PDF FILLER 5. What combination of alleles could result from a crossover between BD and bd chromosomes? 8. Why is it necessary to reduce the number of chromosomes in gametes? 6. How many nuclei are present at the end of meiosis II? How many chromosomes are in each? a. Sperm Cell b. Egg Cell AP_5 c. Daughter Cell from Mitosis ON 5G I 9. Blue whales have 44 chromosomes in every cell. Determine how many chromosomes you would expect to find in the following: d. Daughter Cell from Meiosis II COUS ra PAGE 10... FREE CONVERT JPG TO PDF X OeScience Labs, 2016Shown below are photomicrographs of Rhoeo tradescantia cells undergoing meiosis. Answer the following question for each of the photomicrographs: Identify the cytogenetic abnormality observed (ex. ring, chain, laggard, bridge). Identify the meiotic stage in which these aberrations are observed (as shown in the photomicrograph). Explain how these aberrations are formed and relate to the possible causal mutation(s). Will this result to sterile and/or fertile gametes? Explain.16 - The phenotype of vestigial (short) wings (vg) in Drosophila melanogaster is caused by a recessive mutant gene that independently assorts with a recessive gene for hairy (h) body. Assume that a cross is made between a fly the is homozygous for normal wings with a hairy body and a fly with vestigial wings that is homozygous for normal body hair. The wild-type F1 flies were crossed among each other to produce 1024 offspring. Which phenotypes would you expect among the 1024 offspring, and how many of each phenotype would you expect? a) Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (256), vestigial (256), hairy (256), and vestigial hairy (256). b) O Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (576), vestigial (192), hairy (192), and vestigial hairy (64). C) O Phenotypes: wild, vestigial, hairy, and vestigial hairy; Numbers expected: wild (192), vestigial (256), hairy (64), and vestigial hairy (192). d) All vestigial…