We have crossed true-breeding flies today. Wild type to mutant. Their phenotypes were wt and vg se. (Let’s not worry about males and females for now, just assume both genes are autosomal, which is normal in a Mendelian experiment). The results of a cross are phenotypes. I will be asking you for numbers or a ratio of phenotype numbers you expect to see in the offspring from this cross. To answer this, you’ll need to start with the genotypes of parents, then write their gametes, then write the offspring genotypes, then get the phenotypes based on them. Predict the results of this cross: In case both mutations are recessive; In case both mutations are dominant; In case vestigial wings are recessive and the sepia eyes are dominant.

We have crossed true-breeding flies today. Wild type to mutant. Their phenotypes were wt and vg se. (Let’s not worry about males and females for now, just assume both genes are autosomal, which is normal in a Mendelian experiment). The results of a cross are phenotypes. I will be asking you for numbers or a ratio of phenotype numbers you expect to see in the offspring from this cross. To answer this, you’ll need to start with the genotypes of parents, then write their gametes, then write the offspring genotypes, then get the phenotypes based on them. Predict the results of this cross: In case both mutations are recessive; In case both mutations are dominant; In case vestigial wings are recessive and the sepia eyes are dominant.

Name: We have crossed true-breeding flies today. Wild type to mutant. Their
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Description: We have crossed true-breeding flies today. Wild type to mutant. Their phenotypes were wt and vg se. (Let’s not worry about males and females for now, just assume both genes are autosomal, which is normal in a Mendelian experiment).The results of a cr

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 31QP

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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?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. Should the parents be concerned about the heterozygous condition as well as the homozygous mutant condition?
A couple was referred for genetic counseling because they wanted to know the chances of having a child with dwarfism. Both the man and the woman had achondroplasia (MIM 100800), the most common form of short-limbed dwarfism. The couple knew that this condition is inherited as an autosomal dominant trait, but they were unsure what kind of physical manifestations a child would have if it inherited both mutant alleles. They were each heterozygous for the FGFR3 (MIM 134934) allele that causes achondroplasia. Normally, the protein encoded by this gene interacts with growth factors outside the cell and receives signals that control growth and development. In achrodroplasia, a mutation alters the activity of the receptor, resulting in a characteristic form of dwarfism. Because both the normal and mutant forms of the FGFR3 protein act before birth, no treatment for achrondroplasia is available. The parents each carry one normal allele and one mutant allele of FGRF3, and they wanted information on their chances of having a homozygous child. The counsellor briefly reviewed the phenotypic features of individuals with achondroplasia. These include facial features (large head with prominent forehead; small, flat nasal bridge; and prominent jaw), very short stature, and shortening of the arms and legs. Physical examination and skeletal X-ray films are used to diagnose this condition. Final adult height is approximately 4 feet. Because achondroplasia is an autosomal dominant condition, a heterozygote has a 1-in-2, or 50%, chance of passing this trait to his or her offspring. However, about 75% of those with achondroplasia have parents of average size who do not carry the mutant allele. In these cases, achondroplasia is due to a new mutation. In the couple being counseled, each individual is heterozygous, and they are at risk for having a homozygous child with two copies of the mutated gene. Infants with homozygous achondroplasia are either stillborn or die shortly after birth. The counselor recommended prenatal diagnosis via ultrasounds at various stages of development. In addition, a DNA test is available to detect the homozygous condition prenatally. What is the chance that this couple will have a child with two copies of the dominant mutant gene? What is the chance that the child will have normal height?
Mice of the genotypes A/A ; B/B ; C/C ; D/D ; S/S anda/a ; b/b ; c/c ; d/d ; s/s are crossed. The progeny are intercrossed. What phenotypes will be produced in the F2 andin what proportions? [The allele symbols stand for thefollowing: A = agouti, a = solid (nonagouti); B = blackpigment, b = brown; C = pigmented, c = albino; D =nondilution, d = dilution (milky color); S = unspotted,s = pigmented spots on white background.]
This question deals with Mendelian Genetics. Read each part carefully before you answer it. In the following questions, assume green pea pods are dominant over yellow pea pods. a) If a pea plant that is Heterozygous for pod color is crossed with a pea plant with yellow pods, what do you expect to see in the offspring (the f1 generation)? Specifically, what genotypes, what phenotypes, and in what ratios? (Use G for the green pod gene and g for the yellow pod gene.) (It is usually helpful to write out all the possible genotypes and give the phenotype for each genotype before answering the specific question.)
type P = n! (p)* (q)"* х! (n - х)! Practice Problem: You cross a true-breeding pea plant with red flowers to a true-breeding pea plant with white flowers. All of your offspring have red flowers. Which gene is dominant? Why? What is the genotype of your offspring? You then cross the offspring to each other. What ratio do you expect? Why? You count 1000 plants and look at their flowers. Your results are as follows: 740 red 260 white Does this follow a simple Mendelian inheritance pattern? Why or why not? DADT 2 MEA SUDI ND D LUT IONS
The non-wild-type alleles are k (clipped wings), l (long tail), and m (magical powers). The parental stocks are homozygous doubly recessive flies of genotype k+/k+ · l/l · m/m and homozygous singly recessive flies of genotype k/k · l+/l+ · m+/m+. From this cross, triply heterozygous progeny of genotype k/k+ · l/l+ · m/m+ are obtained, and females of this genotype are testcrossed to triple recessives of genotype k/k · l/l · m/m. The genotypes determining the eight progeny types from this testcross (CROSS C) are shown here with their numbers, out of a total sample of 1,572 flies. What is the correct order of the genes? What number of progeny for each genotype would you predict in the case of independent assortment? What is a non-recombinant genotype? What is a genotype of a single crossover? What is a genotype of a double crossover? What is the longest distance between any of the two dragon alleles above? What is the shortest distance between any of the two dragon alleles above?…
The following diagram shows a hypothetical diploid cell. The recessive allele for albinism is represented by a, and d represents the recessive allele for deafness. The normal alleles for these conditions are represented by A and D, respectively. Please use the symbols A, a, D and d as appropriate to help me understand your answer. a а D 1. According to Mendel's principal of segregation, what is segregating in this cell? 2. According to Mendel's principle of independent assortment, what is independently assorting in this cell? 3. What is the phenotype of this individual? © Cengage Learning
Two fruit flies (Drosophila melanogaster) were crossed. The cross was between a homozygous red-eyed, wingless female and a white-eyed male heterozygous for wings. Recall that red eyes (R) is dominant over white eyes (r) and is inherited on the X chromosome (sex-linked) and as such, eye colour alleles should be written as superscript. In addition, the autosomal trait wings (W) is dominant over wingless (w). 1. List the female genotype 2. List the male genotype 3. Construct a Punnett Square for this dihybrid cross on a piece of paper, to determine what the offspring of such a cross would be with respect to sex, eye colour & wings. 4. Using the information from your Punnett Square answer the following: a. What genotypes did you get (list all genotype combinations; if you have more than one of the same kind, you do NOT have to re-list it.) b. What are the phenotypic ratios of the offspring with respect to sex, eye colour and wings?
Using the pedigree below for an autosomal dominant disorder to determine the phase (which ones are recombinants and which are non-recombinants? Identify them by pedigree position (11, 112, etc.) (SLO4) 1 11 III 2/2 1/2 1/1 -O 2/12 2/2 2/2 2/2 2/2 1/2 2/2 1/2 2/2 1/2 2/2 a. 0.01 b. 0.1 c. 0.2 Calculate the LOD score for each theta below for the pedigree above. Note: keep lots of places behind the decimal until the very end for accuracy. (SLO4)
You believe that the gene you're investigating, eyes absent, is an autosomal recessive mutant allele F7 ge in Drosophila. You set up a cross between a true-breeding mutant male and a true-breeding wildtype female to obtain the F1 generation. You then cross the F1 flies. In the F2 offspring, you observe the following: 237 wild-type males 180 wild-type females 70 mutant males 83 mutant females Based on these findings, is the mutant allele autosomal recessive? How do you know?
In the common daisy, genes A and B control flower color. Both genes have a dominant allele (A or B) and a recessive allele (a or b). At least one copy of each dominant allele is required for flowers to be colorful instead of white. (Explain and Justify your answers) 21.1) Predict the genotypes and phenotypes of the F1 progeny of a cross between two white-flowered plants, one homozygous AA and the other homozygous BB. A) AA bb, white B) aa BB, white C) Aa Bb, colorful D) Aa Bb, white E) aa bb, colorful 21.2) Predict the phenotypic ratio of the F2 progeny of a cross between two white-flowered plants, one homozygous AA and the other homozygous BB. A) 3 colorful : 1 white B) 9 colorful : 7 white C) 9 white : 7 colorful D) 15 white : 1 colorful E) 15 colorful : 1 white 21.3) The inheritance pattern of daisy flower color provides an example of what type of gene interaction? A) additivity…