From the following data set of a testcross with a heterozygous female, which of the following map distances are correct? b C 128 at b* c 387 b c+ c C+ o o o o o o 9 7 389 b+c+ 132 a+b+ b a+ cand bis 27.85 m.u. none of the choices are correct cand bis 24.71 m.u. cand bis 26.24 m.u.
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17. Please solve and make sure all relevant calculations of m.u. take into consideration double crossovers!
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- An individual has the following genotype. Gene loci (A) and (B) are 15 m.u. apart. What are the correct frequencies of some of the gametes that car be made by this individual? Bl a O A. Ab = 7.5%; AB = 42.5% B. ab = 25%; aB = 50% O C. AB = 7.5%; aB = 42.5% O D. aB = 15%; Ab = F0% E. aB = 70%; Ab = 15% Reset Selection OMark for Review What's This?Punnett Square for a Dihybrid Cross to Investigate Coat Colour in Mice be Вс BbCc Answer: D Ova BBCC BbCc bc) bb Cc (BC) BbCC D BbCc Bbcc X BBCC bbcc BBcc bbcc Bb Cc BC O Sperm BbCC Bb Cc Bbcc bc BBCC bbcc Вс BbCc Legend Black Brown White Coat colour in mice is controlled by the interaction of two genes. Three phenotypes result: black coat, brown coat, and white coat. The expected phenotypic ratio resulting from a cross between a bbCc female mouse and BbCc male mouse, is black: brown : white Enter only the number values in the order black brown white (Do not include spaces, commas, etc.)ISlate edu/ d2l/le/content/5003190/viewContent/44248878/View Google Tranx 4 My Drive-G X 4. Suppose that a parent Drosophila is e ca* ca The gamete frequency is as follows: e'ca e ca 16% e'ca е са 31% 14% 29% a. Circle the recombinant gametes. b. What is the map distance between the ebony and claret genes?
- BLACK VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (bl+/blt.vg+/vg+) with a pure-breeding black, vestigial (bl/bl vg/vg) to produce an F1 generation that is all wild-type (bl+/bl vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (bl+/bl vg+/vg) with tester males, which are black, vestigial (bl/bl;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number 440 Wild-type 394 108 135 Black, vestigial Vestigial Black Copy the table into your notes and derive the dihybrid gametes following the example in the previous section. The columns in blue (phenotypes and numbers of offspring) are what you can see and count. The genotypes of the testcross offspring (orange) must be deduced from the phenotypes and knowing that the tester contributed bl vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete…Dihybrid Cross Practice In a breed of dog called a Doberman, black fur is dominant to brown fur and floppy ears are dominant to straight ears. These letters represent the genotypes and phenotypes of the dogs: EE = floppy ears Ee = floppy ears bb = brown fur ee = pointed ears BB = black fur %3D Bb = black fur 1. A female dog with the genotype BBee is crossed with a male dog with the genotype bbEe The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring. Ве Ве Ве Ве bE How many out of 16 have: black fur and floppy ears? be black fur and pointed ears? brown fur and floppy ears? bE brown fur and pointed ears? beuh ec CV + + 10.5 SC 9.1 scute bristles echinus eyes 9.2 ct + crossveinless wings Table 1: phenotype wild-type tapdance feet crossveinless wings tapdance & crossveinless cut wings 15.9 vermilion eyes V + + 66.8 Drosophila X chromosome Use the map provided above for problems 1 & 2. Problem 1: 11.2 10.9 garnet eyes M A new gene is being investigated in fruit flies. The recessive allele of this gene (t) causes the flies' feet to grow tiny tapdance shoes, while the dominant allele (t*) permits wild-type feet to develop. Preliminary studies indicate that this new gene is located on the X-chromosome. You decided to perform a two-point testcross to determine its position relative to the well-established crossveinless wings gene (cv). You cross a female heterozygous for both genes with a testcross male fly and obtain the male offspring results shown in table 1, below. Using this information, answer the following questions: # male offspring 13 405 401 11 forked bristles a) is the original…
- Miniature wings, X in Drosophila melanogaster result from an X-linked allele that is recessive to the allele for long wings, 9 + X. Match the genotypes for each parent in the crosses. Male parent phenotye long miniature miniature 111 long long Female parent phenotype long long long miniature long m m X X Male offspring phenotypes 231 long, 250 miniature 610 long 410 long, 417 miniature 753 miniature 625 long m X Y Female offspring phenotypes 560 long 632 long 412 long, 415 miniature 761 long 630 long Answer Bank ++ X X Male parent genotype + X Y Female parent genotype + X X m 00What would justify the following ratio appearing after phenotyping the outcome of a crossing trial: 8.9: 2.9: 3.2:1? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a b C d Obviously this represents independent assortment based on crossing dihybrid heterozygotes. Obviously this represents gene linkage based on test crossing a dihybrid heterozygote. Obviously this represents the results of a trihybrid test cross. Obviously this represents independent assortment based on crossing monohybrid heterozygotes.In fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.4
- What is the diploid genotype after mating these two strains? Parent 1: MATα msh6Δ::kanMX leu2-3,112 ura3-1 trp1-1 his3-11,15 + pRS415 (msh6- K337T and LEU2 gene) Parent 2: Mata his3∆ leu2∆ lys2∆ ura3∆ TRP1 MSH6A three point cross is carried out and the following 1000 offspring are recovered. PHENOTYPES NUMBER OBSERVED CP+ 5 CPR 330 C++ 115 C+R 70 +++ 385 ++R 10 +PR 70 +P+ 15 - A. B. C. D. E. What is the correct gene map for these genes? (A-E) F. G. H. I. J. What is the interference value? (F-J) A. P----20----C----10----R B. P----20----R----10----C C. C----20----R----10----P D. R----20----C----10----P E. C----20----P----10----R…In Drosophila (fruit flies), jammed wings (J), daughterless (da), curly wings (Cy), star eyes (S), and a black body (b) are determined by genes located on the same chromosome. Gene Combination Recombination Frequency J and Cy 34.9 J and da 1.7 S and Cy 4.8 Cy and b 42.4 S and da 38 b and S 47.2 What is the map unit distance between b and da?Answer map unitsRecord your answer as a value rounded to one decimal place.