Using the Fly Breeder tool in the Fly Biolab, we analyze the heritability of eight other traits. Based on the phenotype ratios observed, explain whether each trait is caused by a dominant or recessive allele, and whether the trait is autosomal or sex-linked. Support each conclusion by referring to a completed Punnett Square.
Trait 1: Purple Eyes
Parent Phenotypes: Male- Red Eye Female- Purple Eye
F1 Results (phenotype ratios): 100% red eyes
F2 Results (phenotype ratios): 75% red: 25% purple
Conclusion (Incl. Punnett Squares): ???
Trait 2: No Eyes
Parent Phenotypes: Male- Normal Female- No Eyes
F1 Results (phenotype ratios): 100% with eyes
F2 Results (phenotype ratios): 75% normal eyes: 25% no eyes
Conclusion (Incl. Punnett Squares): ???
Trait 1: Star Eyes
Parent Phenotypes: Male- Normal Female- Star Eyes
F1 Results (phenotype ratios): 100% star eyes
F2 Results (phenotype ratios): 75% star eyes: 25% normal
Conclusion (Incl. Punnett Squares): ???
Trait 1: Ebony Body
Parent Phenotypes: Male- Normal Female- Ebony
F1 Results (phenotype ratios): 100% normal
F2 Results (phenotype ratios): 75% normal: 25% ebony
Conclusion (Incl. Punnett Squares): ???
Trait 1: Yellow Body
Parent Phenotypes: Male- Normal Female- Yellow
F1 Results (phenotype ratios): -All Male Yellow -All Female Normal
F2 Results (phenotype ratios): 50:50
Conclusion (Incl. Punnett Squares): ???
***Need Punnett Squares***
Using the Fly Breeder tool in the Fly Biolab, we analyze the heritability of eight other...
1. What would be the phenotype for each of the following birds, including their gender? Color Gender Bb ZZi: - bb ZiZi: - BB ZZ: - bb ZW: - Bb ZiW: - 2. Using the phenotypes of the parents and of the progeny given in table 1, determine the genotype of each parent in the first nest. Male Female 3. Using the phenotypes of the parents and of the progeny given in table 2, determine the genotype of each parent...
can someone please explain this answer through punnett squares? Having troubles understanding the process. 141 In Drosophila, white eyes (w) and yellow body (y) are both recessive X-linked mutations. The wild type alleles, w+ and y+, control red eyes and dark body color, respectively. If a homozygous yellow body, red-eyed female is crossed with a dark body, white-eyed male, and F1 progeny are interbred, what will the phenotypes and ratios of the F1 and F2 be? osla 900 Answer: F1-females:...
It is about fruit fly monohubrid cross Fruit Fly Lab Worksheet Part A: using the site mate your two parental flies to obtain the Fı generation and answer questions 1 through 9. Questions: 1. What is the phenotype of the female generation fly? 2. What is the genotype of the female P generation fly? 3. What is the phenotype of the male P generation fly? 4. What is the genotype of the male P generation fly? Draw a Punnett square...
how do i do this table? how is question 13 done? Part II: More Practice with Punnett Squares #1 Data Table II: Note: When you record a ratio, whether it is genotypic or phenotypic ratio, always record the most dominant characteristic first, followed by the recessive. Scenario Genotype Genotype Ratio of Offspring Ratio of Offspring Phenotype of Parent 1 of Parent Genotype 2 Gg gg 2Gg: 2gg or 1:1 2 gray: 2 black or 1:1 #2 #3 #4 #5 #6...
I just need help with part 5 e. I dont know how to set up a punnett square that includes sex chromosomes with wing length as well as eye color. 5. A male fruit fly with red eyes and long wings is mated to a female with brown eyes and short wings. All of the F1 males have red eyes and short wings, and all of the F1 females have red eyes and long wings. The F1s are intercrossed to...
If an apterous female fruit fly with wild type (+) red eyes is crossed with a white eyed male with normal wild type wings, what is the resting F1 generation? Please show on a punnet square.Also what is the punnet dihybrid cross for the F2 generation. Note: the recessive white eyed allele is X-linked while the apterous (wingless) recessive allele is NOT x-linked. SO to summarize: male with recessive white eyes and with normal wild type wings and a recessive...
3) Dihybrid cross: The dark-bodied and rudimentary winged genes are on two different autosomal chromosomes. We will use this cross to determine the ratios of phenotypes in the F2 generation when there are two genes involved. 3.1) Use the following symbols to write out the cross: Male: 3 Female: 9 Dark body color (ebony) = e Wild-type body color: E or E+ Rudimentary wings- (vestigial wings= v) Wild-type wings: V or V+ 3.2) What would the cross look like if...
1)When Gregor Mendel conducted his genetic experiments with pea plants, he observed that a trait’s inheritance pattern was the same regardless of whether the trait was inherited from the maternal or paternal parent. Mendel made these observations by carrying out reciprocal crosses: For example, he first crossed a female plant homozygous for yellow seeds with a male plant homozygous for green seeds and then crossed a female plant homozygous for green seeds with a male plant homozygous for yellow seeds.Unlike...
Virtual Fly Lab Data, the above table are results of monohybrid crosses and dihybrid below. please analysis the statistical analysis using the F2 generations for the above to see if the results follow mendelian postulate for ratios 3:1 (monohybrid) and 9:3:3:1 (dihybrid) draw a conclusion based on statistical analysis thank you olders/1/messages/119807/AHnQ11nLSi2XMpkcArRiOsnko4:2tfullscreen 1 ws M Gmail female wt 10 male apterous 10 590 female wt 607 male wt F2 471 female wt 421 male wt 151 female AP 164 male...
Please show work. 2. You mate a female fly with a black body, purple eyes, and vestigial wing size to a wild- type male. These three alleles are located on chromosome II (autosome) in Drosophila. You testcross one of the Fi females to a male with all three mutations and record the results from the testcross in following table. Phenotype # of each phenotype Body color eye color wing size red red red red 900 le normal vestigial 90 50...