In the F2 generation, 425 flies with long wings and 94 with a curly wing phenotype were observed. Suppose the calculated x2 value is 0.15. Usir the distribution chart below, what is the x2 range? What is the p-value range? Based on this information, do you accept or reject the null hypothesis? If the chart does not appear click here Percentage Points of the Chi-Square Distribution Degrees of Probability of a larger value of x

In the F2 generation, 425 flies with long wings and 94 with a curly wing phenotype were observed. Suppose the calculated x2 value is 0.15. Usir the distribution chart below, what is the x2 range? What is the p-value range? Based on this information, do you accept or reject the null hypothesis? If the chart does not appear click here Percentage Points of the Chi-Square Distribution Degrees of Probability of a larger value of x

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter15: Genomes And Genomics

Section: Chapter Questions

Problem 2QP: Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To...

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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?
. If you intercrossed F1 heterozygotes of the formA b / a B in mice, the phenotypic ratio among the F2progeny would vary with the map distance betweenthe two genes. Is there a simple way to estimate themap distance based on the frequencies of the F2phenotypes, assuming rates of recombination areequal in males and females? Could you estimatemap distances in the same way if the mouse F1heterozygotes were A B / a b?
A wild-type fruit fly (heterozygous for gray body color andnormal wings) is mated with a black fly with vestigial wings.The offspring have the following phenotypic distribution: wildtype, 778; black vestigial, 785; black normal, 158; gray vestigial,162. What is the recombination frequency between these genesfor body color and wing size? Is this consistent with the resultsof the experiment in Figure 15.9?
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.
Do a Punnett square for a cross between a homozygous red-eyed female (XR XR ) and a white-eyed male (Xr Y). What would be the genotypes of the female F1 flies? ____ phenotype? ______ What would be the genotypes of the male F1 flies? ___ phenotype? _______ Do a Punnett square for the F2 results if one of the F1 females mated with one of the F1 males. *To show results of sex-linked traits: Divide phenotypes into males and females first, keep results as fractions (or %). List the phenotypes, with the corresponding genotype (in parentheses) next to each phenotype. Males: Females: A sex-linked trait in humans: hemophilia = blood clotting disorder. Normal clotting (XN) is dominant to hemophilia (Xn). Do a Punnett square for a normal clotting (homozygous dominant) woman and a man with hemophilia.
In a vial of Drosophila, a research student noticedseveral female flies (but no male flies) with bag wingseach consisting of a large, liquid-filled blister insteadof the usual smooth wing blade. When bag-wingedfemales were crossed with wild-type males, 1/3 of theprogeny were bag-winged females, 1/3 were normalwinged females, and 1/3 were normal-winged males.Explain these results.
Summer squash exist in long, spherical, or disk shapes. When atrue-breeding long-shaped strain was crossed to a true-breedingdisk-shaped strain, all of the F1 offspring were disk-shaped. Whenthe F1 offspring were allowed to self-fertilize, the F2 generationconsisted of a ratio of 9 disk-shaped to 6 round-shaped to 1 longshaped. Assuming the shape of summer squash is governed by twodifferent genes, with each gene existing in two alleles, propose amechanism to account for this 9:6:1 ratio
. A snapdragon plant that bred true for white petals wascrossed with a plant that bred true for purple petals, andall the F1 had white petals. The F1 was selfed. Among theF2, three phenotypes were observed in the followingnumbers:white 240solid purple 61spotted purple 19Total 320a. Propose an explanation for these results, showinggenotypes of all generations (make up and explain yoursymbols).b. A white F2 plant was crossed with a solid purple F2plant, and the progeny werewhite 50%solid purple 25%spotted purple 25%What were the genotypes of the F2 plants crossed?
In Drosophila, a cross was made between females expressing thethree X-linked recessive traits, scute bristles (sc), sable body (s),and vermilion eyes (v), and wild-type males. All females were wildtype in the F1, while all males expressed all three mutant traits.The cross was carried to the F2 generation and 1000 offspringwere counted, with the results shown in the following table. Nodetermination of sex was made in the F2 data. Question:Calculate the coefficient of coincidence; does this represent positive or negative interference? Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156sc + v 46+ s + 30sc s + 10+ + v 14
In maize trisomics, n + 1 pollen is not viable. If adominant allele at the B locus produces purple colorinstead of the recessive phenotype bronze and a B b btrisomic plant is pollinated by a B B b plant, whatproportion of the progeny produced will be trisomicand have a bronze phenotype?
What is the expected number of double crossovers based on the map? What is observed number of double crossover’s based on the data in the table? Calculate the interference using the formula.. Interpret the interference in terms of whether a crossover inhibits or stimulates an additional crossover in the same interval of the chromosome?
The a, b, and c loci are all on different chromosomesin yeast. When a b+ yeast were crossed to a+ b yeastand the resultant tetrads analyzed, it was found thatthe number of nonparental ditype tetrads was equal tothe number of parental ditypes, but there were no tetratype asci at all. On the other hand, many tetratypeasci were seen in the tetrads formed after a c+ wascrossed with a+ c, and after b c+ was crossed withb+ c. Explain these results.