Week 5 – Homework – Answer Key
Due Feb. 23, 2013
A total of 20 points are possible for this homework
1. A black guinea pig is crossed with an albino guinea pig, producing 12 black offspring. When the very same albino is crossed with another black guinea pig, 7 black and 5 albinos are obtained. Explain this genetic outcome by writing out the genotypes for the parents, gametes, and offspring in both crosses.
First Cross: The fact that all F1 offspring are black suggests that the parents of the first cross were genotype BB x bb (where B=black and b=albino). The gametes produced by the black parent would have carried the B allele, while those produced by the albino parent carried the b allele. The F1 offspring of such a cross would be Bb, and
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Tan could also be autosomal dominant. Under these circumstances the tan female parent would be heterozygous (Tt) and the wt male would be homozygous recessive (tt). A cross between these two would still produce 50:50 normal colored / tan offspring, of both sexes.
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A Third solution: tan and curved are both sex linked.
C = wt wings, c = curved wings (recessive)
T = tan body, t = wt body color (recessive)
XcT / Xct
XCT / Y
Female Parent fly is
Male P fly is
This works, too!
+2 points for any workable solution
4. Clear wing, Black eye, and Hairless (c, b, and h) are linked, recessive traits carried on chromosome 3 of the Narragansett quahaug fly. A clear-winged, black-eyed fly is crossed with a hairless fly.
A) Their F1 offspring were 97 wild type quahaug flies. What is the genotype of these F1 flies??
[4 points]
Genotype:
Cc Bb Hh
or, [cbH / CBh], etc…
(heterozygous for all 3 traits)
+2 points for genotype
10 of these flies were then mated with flies that exhibit all three traits (clear-winged, black-eyed, and hairless). They produced a total of 1,000 F2 offspring: clear-wings, black eyes hairless clear wings black eyes, hairless wild type clear wings, black eyes, hairless black eyes clear wings, hairless
cbH
CBh
cBH
Cbh
CB cbh CbH cBh 415
411
61
58
22
26
4
3
• Use these data to construct a map of these three genes in two steps.
• Use these data to construct a map of these three genes, showing your work along the way. The map should show the order of the
6. Gather data: On the DESCRIPTION tab, click Reset. Set DD and dd to any values you like. Fill in the initial values in the table below, and then run the Gizmo for five generations. Record the allele and genotype percentages for each generation in the table below.
Introduction: The intention of this lab was to gain a better understanding of Mendelian genetics and inheritance patterns of the drosophila fruit fly. This was tasked through inspecting phenotypes present in the dihybrid crosses performed on the flies. An experimental virtual fly lab assignment was also used to analyze the inheritance patterns. Specifically, the purpose of our drosophila crosses is to establish which phenotypes are dominant/recessive, if the traits are inherited through autosome or sex chromosomes and whether independent assortment or linkage is responsible for the expressed traits.
Apply your understanding of how alleles assort and combine during reproduction to evaluate a scenario involving a monohybrid cross.
In our experiment, we created three crosses; DxG,BxD, and BxG. In crosses BxG and BxD we found a rare mutant fly. This unusual mutant had a misshapen abdomen, deformed wings, and was pale in color. While the mutant was rare, due to the fact that it popped up in both those populations, we hypothesized that this genetic mutant had already previously been discovered and named.
Table 2 shows the phenotypes of the F1 flies produced by crossing P1 wild-type males and P1 no-winged females. The results of that cross was that there was fifty nine wild-type females and forty one males. Therefore there was a total of one hundred wild-type flies produced from crossing P1 true breeding wild-type males and P1 true breeding virgin
* How is it possible for an offspring to exhibit a recessive trait if neither parent exhibited that
Purpose The purpose of this experiment is to determine whether the fruit flies were dominant/recessive or linked/non-linked. The traits I chose for this activity was the fruit fly with vestigial wings and purple eyes, the other fruit fly I chose was a normal fly, also called wild type. While writing out my plan for this activity I thought it would be interesting to test a female mutant and the wild type male, the ratios I came up for this experiment was 2:2 and the mutant allele being recessive to the wild type. In this case, the words recessive and dominant means, if the child born from the parents inherits more of the genes and traits from let us say the father then the alleles of the father is dominant over the mother’s genes.
There were eight different phenotypes among the progeny. The highest phenotypic frequency was the w+m+f+ at 40% of the progeny. The lowest was the w+mf+ with only 2 % of the progeny (Table 3). The sum of the recombinant frequencies between genes, table 4, was used to determine the gene distance. The recombinant frequency was determined by counting the number of individuals whose genes differed from that of the parental type. For example, how many individuals white eye gene, and miniature wing gene, differed from both wild-type or both mutants. Recombination occurred between the white and miniature gene 33 times. Recombination occurred between the miniature and the forked genes 31 times. Recombination occurred between the white and forked genes 44 time. Double recombination occurred 10 times. Therefore, genes w and f are 64 m.u. apart, m and w are 33 m.u. apart, and m and f are 31 m.u. apart (Figure
Now mate a mutant F1 female fly with a mutant F1 male fly. Out of the 50 F2 progeny, what percentage of flies are wild type and what percentage are mutant
Bloom's Level: 1. Remember Learning Outcome: 06.01.03 Predict the outcome of crosses involving genetic variation in chloroplast genomes. Section: 06.01 Topic: Extranuclear Inheritance: Chloroplasts
You are also provided with a heterozygous female, and a homozygous recessive male for a genetic cross. In this particular female, all the dominant alleles are on one chromosome, and the recessive counterparts are on the other homologous chromosome. Due to a chromosomal condition, in the female no recombination occurs between the M and N loci. Normal recombination occurs between the L and M loci. Diagram this cross, and show the genotypes and frequencies of all offspring expected from this cross.
A vial was set up for the F1 cross, such as was done in Week One. The F1 generation was anesthetized with FlyNap. The number of male and females were observed in Table 1. Five males and five females were added to the new culture to create the F2 generation.
This table helps show all the possible genotypes from one set of parents. The table shows that the genotypes purple and starchy are dominant, and the genotypes yellow and sweet are recessive.(stallsmith)
Drosophila melanogaster are a great species for students to learn the process of Mendelian inheritance. They reproduce rapidly and have distinct phenotypes that are easily observable under a microscope. The experiment involved anesthetizing, observing and categorizing these flies based on their wild type and mutant phenotypes to figure out the mutant phenotypes mode of inheritance. We hypothesized that for the mutant vestigial wings phenotype, the mode of inheritance was autosomal dominant. Based on our chi-square value of .57 and p-value of .05, we failed to reject our hypothesis. Comparing our data to other research we learned that the actual mode of inheritance of the vestigial wing mutant phenotype is autosomal recessive. The error could have been from the fact that we did not have a sufficient number of flies to analyze and therefore gave us an inaccurate ratio of wild type to mutant phenotypes.
Based upon observation of the F1 generation, we hypothesize that the inheritance of the white-eye (W) mutation is sex-linked and recessive wild type.