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.
Page 2
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
Suppose the feather color of a bird is controlled by two alleles, D and d. The D allele results in dark feathers, while the d allele results in lighter feathers.
* What recessive features do you exhibit? Do you have any relatives who exhibit those same features?
_____ In swine, when a pure-breeding red is crossed with a pure-breeding white the F1 are all red. However, the F2 shows 9 red, 1 white and 6 of a new color, sandy. The Sandy phenotype is most likely determined by
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.
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.
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
The results of this cross was that there were thirty eight wild-type females and thirty five wild-type males. Therefore there were seventy three wild-type flies. There were sixteen no-winged mutant males and eleven no-winged mutant females. Therefore there was a total of twenty seven no-winged flies produced in this cross. The observed phenotypic ratio of wild-type flies and no-winged mutant flies was 2.7:1 (winged: no-winged).The predicted phenotypic ratio if the no-winged mutation was autosomal recessive would be 3:1 (winged: no-winged). The χ2 value obtained for this cross was 0.213. The p value that was obtained for this cross was
The two recessive alleles are both on the same chromosome. Genes A and B completely follow Mendel’s principles of inheritance; genes B and C are physically connected together and never are separated from each other at any time during any cell division cycle or fertilization event. Draw below the gamete genotypes that this individual could produce.
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.
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
This paper will discuss how phenotypes are driven by DNA inheritance in offspring by the genes known as alleles. Each parent provides an allele although on allele is hidden in one parent and can be passed on to his or her offspring. The information presented in the monohybrid cross that shows where one parent has the hidden allele of a gene that carries the trait of orange eyes that disappears and reemerges later in the genetic make-up of an offspring. The inheritance in the offspring is driven by the DNA of that individuals parents and the genetic traits one is born with are decided by the dominant and recessive genes of his or her parents (www.nature.com, 2014).
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.
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.
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)
The pairs of alternative traits examined segregated among the progeny of a particular cross, some individuals exhibiting one traits, some the other