A white-eyed female fruit fly X"XW is crossed with a red-eyed male XW*Y. Select all of the viable offspring that could be produced when nondisjunction occurs in the female. V red-eyed females (XWtXw) O white-eyed females (X"X"Y) O red-eyed females (XW*x O white-eyed males (X"YY) O red-eyed males (XW*O) O red-eyed females (XW*x"Y)
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- PURPLE VESTIGIAL DIHYBRID CROSS In the parental generation, you mate a pure-breeding wild-type female (put/pu+;vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (pu*/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (pu*/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Genotype Tester Gamete Dihybrid Gamete Number Wild-type 437 417 77 59 Purple, vestigial Vestigial Purple Copy the table into your notes and derive the dihybrid gametes following the example in the first 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 pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete contribution…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…A true breeding male fly with eosin eyes (CCXw-eY) is crossed to a red-eyed female who is heterozygous for both the cream (C) and eosin eyes (Xw-e) allele. What will be the phenotypic ratio of their offspring?
- PURPLE VESTIGIAL DIHYBRID TESTCROSS In the parental generation, you mate a pure-breeding wild-type female (put/put.vg+/vg+) with a pure-breeding purple, vestigial (pu/pu;vg/vg) to produce an F1 generation that is all wild-type (put/pu;vg+/vg). Note that the F1 flies are all dihybrid. Next, you mate several F1 dihybrid females (put/pu;vg+/vg) with tester males, which are purple, vestigial (pu/pu;vg/vg). The offspring of this dihybrid testcross are: Phenotype Wild-type Purple, vestigial Vestigial Purple Genotype Tester Gamete Dihybrid Gamete Number 437 417 77 59 Copy the table into your notes and derive the dihybrid gametes following the example in the first 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 pu vg gametes. Finally, you can deduce the dihybrid gametes (green) by subtracting the tester gamete…Three genes yellow (y) white (w) and cut (ct) are under investigation for possible linkage on the X chromosome. Pure breeding yellow body/white eye/normal wing female flies are crossed with pure breeding black body/red eye/cut wing males. The females in the F1 are black/red/normal, the males are all yellow/white/normal.The F2 males and females are crossed, but ONLY the male flies are are scored (see table). Phenotype Male offspring y + ct 9 + w + 6 y w ct 90 + + + 95 + + ct 424 y w + 376 A. Is the gene order as written (y-w-ct) the same as the gene order on the chromosome? B.What is the mapping distance between yellow and white? C.What is the between cut and white? D.What is the mapping distance between yellow and cut?Fruit flies can have straight wings (S) or curly wings (s), and they can have be female XX or male XY. (A) For a standard monohybrid cross (Ss ´ Ss), what proportion of the offspring will have the genotype ss? (Express the proportion as a simple fraction) (B) For the following cross (SsXX ´ SsXY), what proportion of the offspring will have the genotype Ss? (Express the proportion as a simple fraction) (C) What proportion will have the genotype XX? (Express the proportion as a simple fraction) (D) What proportion will have the genotype SsXX? (Express the proportion as a simple fraction)SHOW YOUR WORK
- In the following cross, imagine that you have a female fly that has two Xs and one Y due to a nondisjunction event in her mother's germ cells. Draw out what the possible gametes are for both the female and the male and also a Punnett square showing the genotypes, phenotypes, and sex of the possible flies as a result of this cross. You do not need to provide the probabilities of each of these. Red-eyed wi C Ở Red-eyed wt XX Y X Y Meiosis(a) DdGGww x DDGgWw ow being (b) ddggWw x DdGgww (c) DdGgWw x DdggWw (d) DdGgWW x DdGgWW de 10. What phenotypes are expected and in what proportion from each of the following crosses? (Assume D, G and Ware dominant over d, g and w, respectively.) mo to esc Ils ovit 101 SUO hamnow brid Jesus 1860 88 HOW bildicion baild am lemon a Bal 101 mem nem lemonHere are the progeny of this cross: (Note that the categories are not in any particular order.)Fly type # of prog. Phenotype symbols Categorywt eyes black body wt wings 97 grn+ blk crv+Green eyes black body curved wings 709 ParentalGreen eyes wt body wt wings 9Green eyes black body wt wings 162wt eyes wt body wt wings 727wt eyes black body curved wings 12wt eyes wt body curved wings 179Green eyes wt body curved wings 105Total = 2000 9.) Write the phenotype symbols in the right-hand column. The first one has been done for you.10.) Next to that, label all fly categories as parental (NCOs), SCOs, and DCOs. One has been donefor you.11.) After each SCO/DCO label, write which gene got “unlinked” in these offspring.12.) Put these three genes into a genetic map in the proper order.13.) Calculate the genetic distance between the genes and label the map with these distances.14.) Calculate the cross-over interference15.) Return to questions #1-6 above. For question 6, you gave your opinion, but…
- A heterozygous round seeded plant (Rr) is crossed with a homozygous round seeded plant (RR). What percentage of the offspring will be homozygous (RR)? _type P = n! (p)* (q)"* х! (n - х)! Practice Problem: You cross a true-breeding pea plant with red flowers to a true-breeding pea plant with white flowers. All of your offspring have red flowers. Which gene is dominant? Why? What is the genotype of your offspring? You then cross the offspring to each other. What ratio do you expect? Why? You count 1000 plants and look at their flowers. Your results are as follows: 740 red 260 white Does this follow a simple Mendelian inheritance pattern? Why or why not? DADT 2 MEA SUDI ND D LUT IONSAssume that the trihybrid cross AABBrr x aabbRR is made in a plant species. Assume that A and B are dominant alleles, but there is no dominance effect of alleles at the R locus. a) How many different gametes are possible in the F1generation? What are the genotypes of these gametes? b) What is the probability of the parental aabbRR genotype in the F2 progeny? c) What proportion of the F2 progeny would be expected to be homozygous for all three genes?