A heterozygous tall pea plant is allowed to self-fertilze. From the F1 progeny, a phenotypically tall plant was randomly chosen and mated to a phenotyplcally dwarf pea plant. What is the probability of producing any phenotypically dwarf plants from this mating? Multiple Choice 1/2 2/3 1/4 1/3
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- Trihybrid-Cross Pure-breeding parents RRGGWW rrggww F, progeny Phenotype Frequency (round) (yellow). (purple) Flower color Seed color R-G-W- Gamete formation R-G- (round) •(yellow). (white) RGW rgw Seed shape R-G-ww ... Fertilization R- (round) ·(green) (purple) R-ggW- ..... Trihybrid RrGgWw R-gg- (round) (green) (white) R-ggww F, Practice Problem What is the probability of getting a plant with smooth green seeds and white flowers? How about a plant with smooth yellow seeds and purple flowers? GENE302/2021BLACK 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…97 Assume that the trihybrid cross MM SS tt X mm ss TT is made in a plant species in which Mand S are dom- inant but there is no dominance between T and t. Consider the F2 progeny from this cross, and assume independent assortment. (a) How many phenotypic classes are expected? (b) What is the probability of genotype MM SS t? obib (c) What proportion would be expected to be 15 2nod bolleo homozygous for all genes? ablla Insninmobo.(om
- IN D Genetics Tutorial 3-Satish Sasikumar 30-12-2020.pdf In radishes, flower color may be red, purple, or white. The edible portion of the radish may be long or oval. When only flower color is studied, no dominance is evident, and red white crosses yield all purple. If these F1 purples are interbred, the F2 generation consists of 1/4 red: 1/2 purple: 1/4 white. Regarding radish shape, long is dominant to oval in a normal Mendelian fashion. (a) Determine the F1 and F2 phenotypes from a cross between a true-breeding red, long radish and a radish that is white and oval. Be sure to define all gene symbols at the start. (b) A red oval plant was crossed with a plant of unknown genotype and phenotype, yielding the following offspring: 103 red long: 101 red oval 98 purple long: 100 purple oval Determine the genotype and phenotype of the unknown plant.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…Height of the plant if a homozygous tall pea plant is crossed with a homozygous short pea plant, what are the possible genotype and phenotypes of the offspring? Legend: T- tall pea plant t- short pea plant Genotypes of the parents TT x tt Question 1. what is the genotypic ratio 2. what are the possible phenotypes of the offspring? 3. What is the phenotypic ratio? 4. what is the probability of having a short trait?
- The second parental (P2)cross using a Punnett square Male Female Gg gg G GG Gg Because the P1 and P2 crosses consider the inheritance of only one character, seed color, they are called monohybrid crosses. Activity 3. SELF- CHECK 1. Windows peak (S) is dominant over straight hairline (s). Given that the mother has heterozygous gene pair for window's peak and the father has straight hairline. a. What is the genotype of the mother? b. What is the genotype of the father? C. How many kinds of gametes can the mother produce? List them down. d. How many kinds of gametes can the father produce? List them down. e. Perform a cross using a Punnett square. f. Give the genotypic ratio of the cross g. Give the phenotypic ratio of the crossAn a cross between two true-breeding plants, Parent #1 with yellow, round peas and Parent #2 with green, wrinkled peas, 1. a. What is the genotype of each plant in the cross. What proportion of the F1 progeny will have the same phenotype as Parent #2. After you self the F1s, what proportion of the progeny will have the same phenotype as Parent #1. C. What proportion of the progeny will have the same genotype as Parent # 1. d. b.. In nature, the plant Plectritis congesta is dimorphic forfruit shape; that is, individual plants bear either winglessor winged fruits, as shown in the illustration.Wingless fruit Winged fruitPlants were collected from nature before floweringand were crossed or selfed with the following results:Number of progenyPollination Winged WinglessWinged (selfed) 91 1*Winged (selfed) 90 30Wingless (selfed) 4* 80Winged × wingless 161 0Winged × wingless 29 31Winged × wingless 46 0Winged × winged 44 0*Phenotype probably has a nongenetic explanation.Interpret these results, and derive the mode ofinheritance of these fruit-shaped phenotypes. Usesymbols. What do you think is the nongeneticexplanation for the phenotypes marked by asterisks inthe table?
- Monobrid crosses with complete dominance Genetic problem 1 Seeds can be round or wrinkled, use a genetic cross to show the genotype and the phenotype of fertilization generation when two heterozygous plant with round seeds are crossed.The allele for round seeds is dominant over the allele for round and for wrinkledCrossing Pea Plants: Mendels Study of Single Traits An unspecified characteristic controlled by a single gene is examined in pea plants. Only two phenotypic states exist for this trait. One phenotypic state is completely dominant to the other. A heterozygous plant is self-crossed. What proportion of the progeny of plants exhibiting the dominant phenotype is homozygous?Human females have two X chromosomes (XX); males have one X and one Y chromosome (XY). a. With respect to X-linked alleles, how many different types of gametes can a male produce? b. If a female is homozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele? c. If a female is heterozygous for an X-linked allele, how many types of gametes can she produce with respect to that allele?