5. A cross between two wild-type flies gave progeny all of which were wild type. When vg e flies the following results were obtained: vg e they were test crossed to (a) 4 of the test crosses gave wild type, vestigal ebony, vestigal, and ebony in a 1:1:1:1 ratio. (b) 4 of the test crosses gave all wild type. (c) 4 of the test crosses gave vestigal and wild type in a 1:1 ratio. (d) 4 of the test crosses gave ebony and wild type in a 1:1 ratio. What are the genotypes of the original pair of wild-type flies?

5. A cross between two wild-type flies gave progeny all of which were wild type. When vg e flies the following results were obtained: vg e they were test crossed to (a) 4 of the test crosses gave wild type, vestigal ebony, vestigal, and ebony in a 1:1:1:1 ratio. (b) 4 of the test crosses gave all wild type. (c) 4 of the test crosses gave vestigal and wild type in a 1:1 ratio. (d) 4 of the test crosses gave ebony and wild type in a 1:1 ratio. What are the genotypes of the original pair of wild-type flies?

Name: 5. A cross between two wild-type flies gave progeny all of which were
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Description: 5. A cross between two wild-type flies gave progeny all of which were wild type. Whenvg ethey were test crossed toflies the following results were obtained:vg e(a) 4 of the test crosses gave wild type, vestigal ebony, vestigal and ebony in a 1:1:1:1

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter13: Genes, Chromosomes, And Human Genetics

Section: Chapter Questions

Problem 8TYK: Another gene in Drosophila determines wing length. The dominant wild-type allele of this gene...

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Another gene in Drosophila determines wing length. The dominant wild-type allele of this gene produces long wings; a recessive allele produces vestigial (short) wings. A female that is true- breeding for red eyes and long wings is mated with a male that has purple eyes and vestigial wings. F1 females are then crossed with purple-eyed, vestigial-winged males. From this second cross, a total of 600 offspring are obtained with the following combinations of traits: 252 with red eyes and long wings 276 with purple eyes and vestigial wings 42 with red eyes and vestigial wings 30 with purple eyes and long wings Are the genes linked, unlinked, or sex-linked? If they are linked, how many map units separate them on the chromosome?
In cats, the genotype AA produces tabby fur color; Aa is also a tabby, and aa is black. Another gene at a different locus is epistatic to the gene for fur color. When present in its dominant W form (WW or Ww), this gene blocks the formation of fur color and all the offspring are white; ww individuals develop normal fur color. What fur colors, and in what proportions, would you expect from the cross AaWw Aa Ww?
5. A strain of Neurospora with genotype A B is crossed with a strain with the genotype a b. Half the progeny are A B and half are a b. Explain how this is possible?
1.) Using the table above as a guide, determine the map distance between the linked genes for eye colour and wing type in the following experiment. You perform the following cross: PpVv × ppvv. You count 1000 offspring in the F1 generation. You find that 406 of the flies have normal eyes and normal wings, 398 have purple eyes and vestigial wings, 96 have normal eyes and vestigial wings, and 100 have purple eyes and normal wings. 2. From previous research findings, you know that the distance between the gene for eye colour and the gene for body colour is 12.2 map units, and the gene for body colour is 7.4 map units away from the gene for wing type. All three genes are on the same chromosome. Draw a chromosome map showing the relative distances between the three linked genes. 3.) You conduct the same cross again, but this time you get an almost exact 1:1 ratio of flies with normal eyes and normal wings to flies with purple eyes and vestigial wings. There are no recombinant types.…
1. In a cross between a black and a white guinea pig, all members of the F1 generation are black. The F2 generation is made up of approximately 3/4 black and 1/4 white guinea pigs. (a) Diagram this cross, showing the genotypes and phenotypes. (b) What will the offspring be like if two F2 white guinea pigs are mated?
8. In guinea pigs, the allele for black fur (B) is dominant over the allele for white fur (b). Similarly, the allele for straight hair (H) is dominant over the allele for having rough hair (h). Pure breeding rough haired guinea pigs with black fur were crossed with pure breeding straight haired guinea pigs with white fur. a. State the genotype and the phenotype of the F1 individuals produced as a result of this cross.b. Two F1 offspring were mated together. Calculate the expected ratio of phenotypes in the F2 generation.c. Show the completed Punnett square below.
1. In pea plants, purple flower color is dominant over white flower color. If you had a purple flowered plant, its genotype could be either PP or _______? A. Pp B. pp C. PP 2. To figure out which genotype your purple plant had, you could do a test cross and mate it with a __________ plant? A. White Flower B. Purple Flower 3. If any of the offspring were white, that would tell you that your mystery purple plant had the genotype __________? A. Pp B. PP C. pp
1) By decomposing the cross, MmNn X Mmnn, into single gene outcomes, the possibility of obtaining a double recessive outcome for the ‘N’ gene is: a) 50%! b) 100! c) 0%! d) 25%! 2) A 16-square Punnett is time-consuming to draw out. Dr. Thompson can easily solve this problem by decomposing her dihybrid crosses into separate monohybrid Punnett squares, which would also be quicker than using a forked line diagram. If so, how many individual monohybrid Punnett squares will she need to draw? a) Four Punnett squares. b) Two Punnet squares. c) One Punnett square. d) Sixteen Punnett squares since there are 16 possible outcomes.
1. If a purebred pea plant with purple flowers was crossed with a purebred pea plant with white flowers, all of the offspring's flowers ended up being purple. In this instancethe WHITE flower trait is considered to be what?
1) By convention, the recessive trait will determine the abbreviation used to track crosses if it is a mutant condition, since it stands out in contrast to the rest of the population. For example, if pea pods are typically green in color, a mutant condition might result in yellow pea pods; therefore, the lowercase letter ‘y’ would be used to depict the mutant state, yellow, while uppercase ‘Y’ would depict the wildtype (wt) condition, green. Remember that the mutant condition is not necessarily always recessive. 1 A) . If green is dominant wt (Y) and yellow is a recessive mutant condition (y), depict a Yy father mated with a YY mother in the Punnett. 1B) Considering the dominant allele, what colors are the parents in the cross above? Yy = _____________________ YY = _____________________ 1C) What colors are the offspring in the cross above? 1D) What is the phenotypic ratio of the offspring? 1E) What is the genotypic ratio of the offspring? (Remember, the genotypic ratio is based…
2. A maize plant heterozygous for three genes is crossed in a test cross. Are obtained the following results: Progeny phenotype ABC 304abc 301Abc 35aBC 33ABc 8abC 6AbC 172aBc 169 A. Determine the order and distance of the genesb. Determine the coefficient of coincidence and the interference.
2. Three different pure-breeding strains of corn that produce ears with white kernels were crossed to eachother. In each case, the F1 plants were all red, whileboth red and white kernels were observed in the F2generation in a 9:7 ratio. These results are summarized in the following table.F1 F2white-1 × white-2 red 9 red : 7 whitewhite-1 × white-3 red 9 red : 7 whitewhite-2 × white-3 red 9 red : 7 whitea. How many genes are involved in determiningkernel color in these three strains?b. Define your symbols and show the genotypes for thepure-breeding strains white-1, white-2, and white-3.c. Diagram the cross between white-1 and white-2,showing the genotypes and phenotypes of the F1and F2 progeny. Explain the observed 9:7 ratio.