Create dihybrid crosses from the following information: 39. A heterozygous tall double eye individual with a short single eyed individual. Draw your chart below. a. What are the two parents genotypes: b. What are their children's genotypes: С. What are their children's phenotypes:

Create dihybrid crosses from the following information: 39. A heterozygous tall double eye individual with a short single eyed individual. Draw your chart below. a. What are the two parents genotypes: b. What are their children's genotypes: С. What are their children's phenotypes:

Name: Create dihybrid crosses from the following information:39. A heterozy
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: Create dihybrid crosses from the following information:39. A heterozygous tall double eye individual with a short single eyed individual. Draw your chartbelow.What are the two parents genotypes:Xa.b.What are their children's genotypes:С.What are the

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter6: Cytogenetics: Karyotypes And Chromosome Aberrations

Section: Chapter Questions

Problem 2QP: Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the...

See similar textbooks

Similar questions

Given the karyotype shown at right, is this a male or a female? Normal or abnormal? What would the phenotype of this individual be?
25. a)What is “dihybrid cross”? b) Consider two traits color and height. For color, let Y=yellow be dominant over y=green; for height, let T=tall be dominate over t=short. i. Set up the P1 generation ii. enumerate the possible gametes in the sex cells iii. Put the possible gametes along the top and sides of a Punnett square. iv. Hence, or otherwise, find a) the proportion (in percentage) of the F1 are yellow and tall with genotype YyTt b) The probability of obtaining, the genotype YyTt
Some mice are born with a yellow coat controlled by the gene Y. Examine the setup and outcome of the following crosses, which produce two generations: Po: yellow mouse X gray mouse F1 :5 yellow mice, 4 gray mice P: yellow mouse from F1 X yellow mouse from F1 F2 : 7 gray mice, 15 yellow mice with 9 dead Analyze the information provided to answer these questions: • What type of cross is the first cross? • Which trait is dominant? • What is the genotype of the yellow mouse in the first cross? • What is the genotype of the gray mouse in the second cross? Why are 9 mice dead in F2?
For the cross: PpAa x PpAa P = purple flowers (Dominant) p = white flowers A = axial flowers (Dominant) a = terminal flowers a. What are the possible gamete classes that can form from these parents? b. What are the expected offspring genotype classes and ratios/proportions/fractions which will result from the cross? c. What are the expected offspring phenotype classes and ratios/proportions/fractions which will result from the cross? 2. Predict ratios/proportions/fractions of genotypes and phenotypes of the following crosses. T = tall stem t = dwarf stem P = purple flowers p = white flowers G = green pods g = yellow pods A = axial flowers a = terminal flowers R = round peas r = wrinkled seeds A. ttPp x Ttpp B. GgRr x ggRr C. PpGg x ppgg
DIHYBRID CROSS Heterozygous Yellow and heterozygous round seed crossed with homozygous yellow and heterozygous round seed. Find the following: 1. Alleles of both parents (given problems) 2. Genotype 3. Phenotype ratio 4. Write the punnet square -Please answer this on the paper and explain the answer step by step. Thank you asap
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.
Monohybrid Cross: 1) In dogs, wire hair (S) is dominant to smooth (s). a. In a cross between a homozygous wire-haired dog with a smooth haired dog what are the genotypes and phenotypes of the F1 generation? b. If a brother and sister from the F1 generation are crossed what is the expected ratio of wire-haired to smooth-haired dogs produced from this cross?
+ 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…
. a. A mouse cross A/a ⋅ B/b × a/a ⋅ b/b is made, and inthe progeny there are25% A/a ⋅ B/b, 25% a/a ⋅ b/b,25% A/a ⋅ b/b, 25% a/a ⋅ B/bExplain these proportions with the aid of simplifiedmeiosis diagrams.b. A mouse cross C/c ⋅ D/d × c/c ⋅ d/d is made, and inthe progeny there are45% C/c ⋅ d/d, 45% c/c ⋅ D/d,5% c/c ⋅ d/d, 5% C/c ⋅ D/dExplain these proportions with the aid of simplifiedmeiosis diagrams.
If a homozygous tall pea plant is crossed with a homozygous shortpea plant, what are the possible genotype and phenotype of theoffspring? Legend: T- tall pea plant t- short pea plant Genotypes of the parents: TT x tt Question to answer ff: 1. What is the genotypic ratio? 2. What are the possible phenotypes of the offspring? 3. What is the phenotypic ratio?
38 Which cross would produce phenotypic ratios that would best illustrate the Law of Dominance? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. TT X tt TT X Tt tt X tt TTX TT 身

3.) A Heterozygous tall heterozygous red flowered plant is crossed with a Homozygous short white flowered plant. What is the genotype of the heterozygous tall heterozygous red flowered plant? TtRr What is the genotype of the homozygous short white flowered plant? ttrr Fill in the genotype in the parentheses and FOIL each parent’s genotypes: heterozygous tall heterozygous red flowered plant ( homozygous short white flowered plant (