Create a Punnett Square representing the cross between the brown and albino mouse for each of the two possible brown mouse genotypes: Punnett Square 1: albino mouse x [possible brown mouse genotype 1] Punnett Square 2: albino mouse x [possible brown mouse genotype 2]
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Create a Punnett Square representing the cross between the brown and albino mouse for each of the two possible brown mouse genotypes:
Punnett Square 1: albino mouse x [possible brown mouse genotype 1]
Punnett Square 2: albino mouse x [possible brown mouse genotype 2]
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- A standard three-point mapping is conducted for recessive mutations in autosomal genes purple eye (pr), curved wing ( c) and black body (b). Their wild type alleles are also used for genetic mapping. An F1 Drosophila female heterozygous for purple eye (pr), curved wing (c) and black (b) is crossed to a triply homozygous mutant male. The observed numbers and phenotypes of the offspring are as follows: 360 pr c b 380 pr+ c+ b+ 104 pr c+ b 96 pr+ c b+ 30 pr c b+ 20 pr+ c+ b 6 pr c+ b+ 4 pr+ c b PROVIDE THE FOLLOWING: A) State the order of genes on this chromosome. B) Calculate map distances between the gene pairs: pr-c, pr-b, c-b. Show calculations, state the number of map units and which gene pairs they refer to.The image attached shows a parental cross that is homozygous wild female x white male. F1 were intercrossed to produce the F2 generation as indicated below: Wild Female: 416 Wild Male: 192 White Female: 0 White Male: 192 Use the data provided to prepare a formal table for a Chi-square analysis.Miniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X*). Give the genotypes of the parents in the following cross: Male parent Female parent Male offspring Female offspring 231 long, 250 miniature Long Long 560 long O male: Xm/Y and female X* /X* male: X* / Y and female Xm /x O male: X* /Y and female X* /X* male: Xm/Y and female Xm /x+
- Miniature wings (Xm) in Drosophila result from an X-linked allele that is recessive to the allele for long wings (X*). Give the genotypes of the parents in the following cross: Male parent Female parent Male offspring Female offspring Long Miniature 750 miniature 761 long O male: X* / X* and female X™ /x+ O male: X*/Xt and female Xm /xm O male: X*/ Y and female Xm /xm O male: Xm/ Y and female Xm /xmGiven the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?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.
- Given the following testcross data for corn in which the genes for fine stripe (f), bronze aleurone (bz) and knotted leaf (Kn) are involved: PhenotypeNumber Kn + +451 Kn f +134 + + + 97 + f bz436 Kn + bz18 + + bz119 + f +24 Kn f bz86 Total:1,365 Maize geneticists tend to be like Drosophila geneticists utilizing a + to indicate the wild type allele (which also is dominant) and a lower case letter for the mutant allele (which is recessive). The first thing I suggest in attacking this problem is to use the typical convention of upper case letters for dominant alleles and lower case letters for recessive alleles as follows: F = wild type, f = fine stripe; B = wild type, b = bronze aleurone K = wild type, k = knotted leaf Then we can rewrite the phenotypes in a “language” that is more easily understood: PhenotypeNumber k F B451 k f B134 K F B97 K f b436 k F b18 K F b119 K f B24 k f b86 Total:1,365 a) Determine the sequence (order) of the…In the fruit fly Drosophila melanogaster, the following genes and mutations are known:Wing size: recessive allele for tiny wings t; dominantallele for normal wings T.Eye shape: recessive allele for narrow eyes n;dominant allele for normal (oval) eyes N.For each of the four following crosses, give thegenotypes of each of the parents.Male FemaleWings Eyes Wings Eyes Offspring1 tiny oval × tiny oval 78 tiny wings, oval eyes24 tiny wings, narrow eyes2 normal narrow × tiny oval 45 normal wings, oval eyes40 normal wings, narrow eyes38 tiny wings, oval eyes44 tiny wings, narrow eyes3 normal narrow × normal oval 35 normal wings, oval eyes29 normal wings, narrow eyes10 tiny wings, oval eyes11 tiny wings, narrow eyes4 normal narrow × normal oval 62 normal wings, oval eyes19 tiny wings, oval eyesThe F1 from a cross of AABB x aabb is testcrossed, resulting in the following phenotypic ratios: AB 308 Ab 190 ab 292 aB 210 (1) what are the genotypes for the listed phenotypes? (2) why aren't the phenotypes in a 1:1:1:1 ratio? (3) what is the frequency of recombination between genes A and B?
- 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.Cross Cross A Cross A Cross B Cross B Phenotype F1 generation F2 generation F1 generation F2 generation Male red eyes 132 150 0 99 Female red eyes 135 295 110 101 Male white eyes 0 147 105 93 Female white eyes 0 0 0 95 Using “+” to indicate the wildtype red-eyed allele and “w” to indicate the mutant white-eyed allele, state the genotypes of the following: Wildtype red-eyed and white-eyed parental flies from cross A and cross B. Males and females from the F1 generation flies from cross A and cross B Males and females, F2 generation flies from cross A and cross B.a) Based upon the results of these crosses, is the dwarfism trait in Andean condors autosomal or sex-linked? Once again, please explain the logic behind your conclusion