What more can be learned from ordered tetrad analysis that cannot be learned from unordered tetrad analysis? i) Determine the total number of asci with first division segregation and second division segregation pattern. ii) Calculate the distance between the gene and the centromere.
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- Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?A series of three-point testcrosses is made to determine the genetic map order of seven linked allele pairs: A/a, B/b, G/g, H/h, Q/q, R/r, and Y/y.From each cross between a triply heterozygous parent listed below, two recombinant classes were noticed as the least frequent among all 8 progeny classes, and are listed at the right in the table. A. For each testcross write the genotype of the F1 heterozygous parent. F1 Parental Phenotype Least frequent F2 Phenotype 1.AHB&ahb AHb & ahB 2.RYh&ryH RYH & ryh 3.BhY&bHy Bhy & bHY 4.qYB&Qyb qYb & QyB 5.AbQ&aBq Abq & aBQ 6.ghR&GHr ghr & GHR B. Write the unified map order of these genes, showing your reasoning.
- In a cross in Drosophila, a female heterozygous for the autosomallylinked genes a, b, c, d, and e (abcde/ + + + + +) was testcrossedwith a male homozygous for all recessive alleles (abcde/abcde).Even though the distance between each of the loci was at least3 map units, only four phenotypes were recovered, yielding thefollowing data: Phenotype No. of Flies+ + + + + 440a b c d e 460+ + + + e 48a b c d + 52 Total = 1000 Why are many expected crossover phenotypes missing? Can anyof these loci be mapped from the data given here? If so, determinemap distances.An organism of the genotype AaBbCc was testcrossed to a triplyrecessive organism (aabbcc). The genotypes of the progeny are inthe following table.AaBbCc 20 AaBbcc 20aabbCc 20 aabbcc 20AabbCc 5 Aabbcc 5aaBbCc 5 aaBbcc 5 a.) Assuming simple dominance and recessiveness in each genepair, if these three genes were all assorting independently,how many genotypic and phenotypic classes would result inthe offspring, and in what proportion? b.) Answer part (a) again, assuming the three genes are sotightly linked on a single chromosome that no crossovergametes were recovered in the sample of offspring. c.) What can you conclude from the actual data about thelocation of the three genes in relation to one another?From a series of two-point crosses, the following mapdistances were obtained for the syntenic genes A, B,C, and D in peas:B ↔ C 23 m.u.A ↔ C 15 m.u.C ↔ D 14 m.u.A ↔ B 12 m.u.B ↔ D 11 m.u.A ↔ D 1 m.u.Chi-square analysis cannot reject the null hypothesis of no linkage for gene E with any of theother four genes.a. Draw a cross scheme that would allow you todetermine the B ↔ C map distance.b. Diagram the best genetic map that can be assembled from this data set.c. Explain any inconsistencies or unknown features inyour map.d. What additional experiments would allow you toresolve these inconsistencies or ambiguities?
- A homozygous strain of corn that produces yellow kernels is crossed with another homozygous strain that produces purple kernels. When the F1 are interbred, 197 of the F2 are yellow and 153 are prurple. Give the genotypes of the yellow and purple F2 and propose a genetic model that explains the inheritance of these kernel colors in corn.. In mice, the following alleles were used in a cross: W = waltzing gait w = nonwaltzing gait G = normal gray color g = albino B = bent tail b = straight tail A waltzing gray bent-tailed mouse is crossed with a nonwaltzing albino straight-tailed mouse and, over several years, the following progeny totals are obtained: waltzing gray bent 18 waltzing albino bent 21 nonwaltzing gray straight 19 nonwaltzing albino straight 22 waltzing gray straight 4 waltzing albino straight 5 nonwaltzing gray bent 5 nonwaltzing albino bent 6 Total 100 a. What were the genotypes of the two parental mice in the cross? b. Draw the chromosomes of the parents.c. If you deduced linkage, state the map unit value or values and show how they were obtained.E. W. Lindstrom crossed two corn plants with green seedlings and obtained the following progeny: 3583 green seedlings, 853 virescentwhite seedlings, and 260 yellow seedlings (E. W. Lindstrom. 1921. Genetics 6:91–110). a. Give the genotypes for the green, virescent-white, and yellow progeny. b. Explain how color is determined in these seedlings. c. Is there epistasis among the genes that determine color in the corn seedlings? If so, which gene is epistatic and which is hypostatic?
- In poultry, the genotype–phenotype relationships forcomb shape are R/– P/–, walnut; R/– p/p, rose, r/r P/–, pea;and r/r p/p, single. What will be the comb characters ofthe offspring of the following crosses? a. A walnut crossed with a single b. A rose crossed with a walnut c. A rose crossed with a pea d. A walnut crossed with a walnut Note- a, b, c, are already solved, posting this one to get the answers of d sub topic.. A true-breeding strain of Virginia tobacco has dominantalleles determining leaf morphology (M), leaf color(C), and leaf size (S). A Carolina strain is homozygousfor the recessive alleles of these three genes. Thesegenes are found on the same chromosome as follows:M C S6 m.u. 17 m.u.An F1 hybrid between the two strains is now backcrossedto the Carolina strain. Assuming no interference:a. What proportion of the backcross progeny willresemble the Virginia strain for all three traits?b. What proportion of the backcross progeny willresemble the Carolina strain for all three traits?c. What proportion of the backcross progeny will havethe leaf morphology and leaf size of the Virginiastrain but the leaf color of the Carolina strain?d. What proportion of the backcross progeny will havethe leaf morphology and leaf color of the Virginiastrain but the leaf size of the Carolina strain?A cross was performed using Drosophila melanogaster involving a female known to be heterozygous for both ebony body and sepia eyes and a male known to be homozygous for both of these recessive traits. The following data was produced from the cross. Test these data to determine if they are significantly different from the expected phenotypic ratio. Remember to use the 5% level of significance Wild eye Wild body – 102, Wild eye Ebony body – 94, Sepia eye Wild body – 100, Sepia eye Ebony body – 93. Your answer should include the hypothesized cross in genotypes, the Chi-squared value, the critical value and whether you reject or do not reject.
