In rabbits, gray hair (G) is dominant to white hair (g), and black eyes (B) are dominant to red eyes (b). These two traits are independent of each other. In other words, a female rabbit with the genotype GgBb may produce eggs with the alleles GB, Gb, gB, or gb. To predict the probability of these sorts of crosses, we will make a dihybrid Punnett Square. 1. What are the phenotypes (descriptions) of rabbits that have the following genotypes: Ggbb ggbb GgBb 2. A male rabbit with the genotype GGbb is crossed with a female rabbit with the genotype ggBb The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring.

In rabbits, gray hair (G) is dominant to white hair (g), and black eyes (B) are dominant to red eyes (b). These two traits are independent of each other. In other words, a female rabbit with the genotype GgBb may produce eggs with the alleles GB, Gb, gB, or gb. To predict the probability of these sorts of crosses, we will make a dihybrid Punnett Square. 1. What are the phenotypes (descriptions) of rabbits that have the following genotypes: Ggbb ggbb GgBb 2. A male rabbit with the genotype GGbb is crossed with a female rabbit with the genotype ggBb The square is set up below. Fill it out and determine the phenotypes and proportions in the offspring.

Name: In rabbits, gray hair (G) is dominant to white hair (g), and black ey
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Description: In rabbits, gray hair (G) is dominant to white hair (g), and black eyes (B) are dominant to red eyes (b). Thesetwo traits are independent of each other. In other words, a female rabbit with the genotype GgBb may produceeggs with the alleles GB, Gb,

Concepts of Biology

1st Edition

ISBN:9781938168116

Author:Samantha Fowler, Rebecca Roush, James Wise

Publisher:Samantha Fowler, Rebecca Roush, James Wise

Chapter8: Patterns Of Inheritance

Section: Chapter Questions

Problem 2ACQ: Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are...

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Figure 8.10 In pea plants, purple flowers (P) are dominant to white (p), and yellow peas (Y) are dominant to green (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? How many squares would you need to complete a Punnett square analysis of this cross?
Hemophilia and color blindness are both recessive conditions caused by genes on the X chromosome. To calculate the recombination frequency between the two genes, you draw a large number of pedigrees that include grandfathers with both hemophilia and color blindness, their daughters (who presumably have one chromosome with two normal alleles and one chromosome with two mutant alleles), and the daughters sons. Analyzing all the pedigrees together shows that 25 grandsons have both color blindness and hemophilia, 24 have neither of the traits, 1 has color blindness only, and 1 has hemophilia only. How many centimorgans (map units) separate the hemophilia locus from the locus for color blindness?
Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?
In autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. 1. What is the genotype of the Fl? 2. Show the types of gametes the Fl's may be expected to form and derive the expected proportion of each. Show your solution. 3. What phenotypic ratio of green to red is expected if: a. the Fl's are intercrossed? Show your solution. b. the Fl's are crossed with red plants? Show your solution. c. If the G locus were 50 or more map units from the centromere, what types and proportions of gametes would the Fl be expected to produce? Derive the expected F2 phenotypic ratio.
+ ec +/Y + + w/Y y ec +/Y + ec +/y ec w ++ w/y ec w у ес +у ес и Determine the order in which the three loci y, ec, and w Occur on the chromosome and prepare a linkage map. 7.22 A cross involving X-linked genes was made between yellow, bar, vermilion female fies and wild males, and the F1 females were crossed with y B v males. The following phenotypes were obtained when 1000 progeny were exam- ined: Dra ord ma the 7.2 546 244 160 50 + + + + Bv y Bv y+ + y+v y B+ and an and and and +B + re + + v ge Determine the order in which the three loci occur on the chromosome and prepare a linkage map. 7.23 Female Drosophila heterozygous for ebony (e"le), scarlet (st*/st), and spineless (ss*/ss) were testcrossed, and the following progeny were obtained: PROGENCY PHENOTYPES NUMBER ir Wild type Ebony Ebony, scarlet Ebony, spineless Ebony, scarlet, spineless Scarlet 67 8. 68 347 78 368 Scarlet, spineless Spineless (a) Are these genes linked? Justify your answer. (b) Write the genes given on a…
In autotetraploid Chinese primrose (Primula sinensis L.), the gene controlling stigma color is very near the centromere of the chromosome carrying it. The allele G for green stigma is dominant to g for red stigmas. A homozygous green autotetraploid strain is crossed with a homozygous red autotetraploid strain. 1. What phenotypic ratio of green to red is expected if: a. the Fl's are intercrossed? Show your solution. b. the Fl's are crossed with red plants? Show your solution. c. If the G locus were 50 or more map units from the centromere, what types and proportions of gametes would the Fl be expected to produce? Derive the expected F2 phenotypic ratio.
In dogs, dark coat color is dominant over albino, andshort hair is dominant over long hair. Assume that theseeffects are caused by two independently assorting genes.Seven crosses were done as shown below, in which D andA stand for the dark and albino phenotypes, respectively,and S and L stand for the short-hair and long-hairphenotypes.Number of progenyParental phenotypes D, S D, L A, S A, La. D, S × D, S 88 31 29 12b. D, S × D, L 19 18 0 0c. D, S × A, S 21 0 20 0d. A, S × A, S 0 0 29 9e. D, L × D, L 0 31 0 11f. D, S × D, S 45 16 0 0g. D, S × D, L 31 30 10 10Write the genotypes of the parents in each cross. Use thesymbols C and c for the dark and albino coat-color allelesand the symbols H and h for the short-hair and long-hairalleles, respectively. Assume parents are homozygousunless there is evidence otherwise.
The A & B antigens in humans may be found in water-soluble forms in secretions, including saliva of individuals with the genotypes E/E& E/e, but not individuals with the genotype e/e. Thus, the population contains ‘secretors’ & ‘nonsecretors’. A cross was performed between 2 heterozygous secretors, one with type O blood and the second with type AB blood. What phenotypic ratios are expected in their offspring? A) 1/4 type A : 1/4 type B : 1/2 non-secretor B) 3/8 type A : 3/8 type B : 2/8 non-secretor C) 1/2 type A : 1/2 type B D) 3/4 type AB : 1/4 non-secretor
Yellow guinea pigs crossed with white ones always produce cream-colored offspring. Two cream guinea pigs, when crossed, produce yellow, cream, and white offspring in the ratio of 1 yellow : 2 cream : 1 white. What principle of genetics is involved in this cross? (1 point) 2. The shape of radishes may be long, round, or oval. The following results were obtained in the different possible crosses: a. long x oval gave ½ long and ½ oval b. oval x round gave ½ oval and ½ round c. long x round gave all oval d. oval x oval gave ¼ long, ½ oval, and ¼ round Explain these results. Hint: Show genotypes of each cross) (2 points). a. b. c. d. 3. In human blood types, what are the genotypes of the following parents? (2 points). Phenotypes of ParentsPhenotypes of OffspringGenotypes of parents ABABO A x AB½ 0½ 0_______ x _______ A x AB½ ¼¼0_______ x _______ A x A¾00¼_______ x _______ A x O½00½_______…
The F1 from a cross of AABB x aabb is testcrossed, resulting in the following phenotypic ratios: AB 308 Ab 190 ab 292 aB 210 (1) what are the genotypes for the listed phenotypes? (2) why aren't the phenotypes in a 1:1:1:1 ratio? (3) what is the frequency of recombination between genes A and B?
Fruit flies having one allele for curly wings (Cy) and one allele for normal wings (Cy+) develp curly wings. Fruit flies homozygous for normal wings (Cy+) develop normal wings. When two curly-winged flies were crossed, 204 curly-winged and 98 normal-winged flies. In fact, all crosses between curly-winged flies produce nearly the same curly-winged to normal-winged ratio. How would you explain these results? The Cy allele is epistatic to the Cy+ allele. The Cy+ allelle shows partial dominance over the Cy allele. The Cy allele is dominant for the curly phenotype and is lethal when homozygous. The Cy+ allele is dominant over the Cy allele and is lethal when homozygous. The Cy+ and Cy alles are co-dominant.
Fruit flies with one allele for curly wings (Cy) andone allele for normal wings (Cy+) have curly wings.When two curly-winged flies were crossed, 203curly-winged and 98 normal-winged flies wereobtained. In fact, all crosses between curly-wingedflies produce nearly the same curly : normal ratioamong the progeny.a. What is the approximate phenotypic ratio in theseoffspring?b. Suggest an explanation for these data.c. If a curly-winged fly was mated to a normalwinged fly, how many flies of each type would youexpect among 180 total offspring?