Reciprocal Crosses for Autosomal Recessive Inheritance mutant female X wild-type male wild-type female X mutant male X AA (genotype) P: AA X a a (genotype) (genotype) (genotype) Genotypes & Ratios: Phenotypes & Ratios: Genotypes & Ratios: Phenotypes & Ratios: F,: F, intercross: F, intercross: X (genotype) (genotype) (genotype) (genotype) Genotypes & Ratios: Phenotypes & Ratios: Genotypes & Ratios: Phenotypes & Ratios:

Reciprocal Crosses for Autosomal Recessive Inheritance mutant female X wild-type male wild-type female X mutant male X AA (genotype) P: AA X a a (genotype) (genotype) (genotype) Genotypes & Ratios: Phenotypes & Ratios: Genotypes & Ratios: Phenotypes & Ratios: F,: F, intercross: F, intercross: X (genotype) (genotype) (genotype) (genotype) Genotypes & Ratios: Phenotypes & Ratios: Genotypes & Ratios: Phenotypes & Ratios:

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 30QP: Variations on a Theme by Mendel Pea plants usually have white or red flowers. A strange pea-plant...

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Incomplete dominance Codominance Dosage Dominant negative LOF ||| ||| An allele (or gene) that produces a new functional product. A mutation whose gene product produces a misshaped protein in the recessive condition and whose recessive allele will also cause the heterozygous protein product to misfold. Wild type or the phenotype of the typical form of a species as it occurs in nature. A mutation that results in a LOF. An allele (or gene) that no longer produces ✓ Open functional product.
CO-DOMINANCE/I... % Ø Oll ^ 6- 3. A homozygous red gummy bear is crossed with a homozygous yellow gummy bear (red and yellow mixed together make orange). You predicted that all the offspring will be orange gummy bears. Create a Punnett Square, then explain if your prediction was correct. 13 r y 2 / 3 | - h ? & 71 u * 82 87% i + | © k ( 93 O D § ) 014 Р 11 -½ { [ [ ^ 11 = ¾ 11 8 '{' \
X-linked Recessive Inheritance A gene is described as X-linked when it occurs on the X chromosome and not the Y. Our convention is to indicate X-linkage by attaching the appropriate gene symbol as a superscript on the letter X. Commonly, the wild-type (+) allele is indicated with only a "+" to avoid having to type a superscript on a superscript. For example, a female that is heterozygous and carrying a recessive mutant allele is indicated as X+Xm. Note the convenience of the shorthand + for m+ in this situation. A mutant male has the genotype XmY. When working with X-linked inheritance, always include the X and Y chromosomes in the descriptions of genotypes, and include the sex (male or female) in the descriptions of the phenotypes (e.g., mutant male, wild-type female, etc.). Here are the genotypes and associated phenotypes for X-linked recessive inheritance: X+X+ Wild-type female X+Xm Wild-type female xmxm Mutant female X+Y xmy Wild-type male Mutant male
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Genetic cross with dependent assortment (linked traits). Repeat the previous problem but assume that the genes are located very close to each other on the same chromosome, and are therefore always linked (passed on together). Assume that in these individuals, each chromosome contains the dominant allele for one gene and the recessive allele for the other gene. Note: this is similar to the dependent assortment example in your notes, but the linkage is different. Construct a 2x2 Punnett square, and answer the questions below: GgBb x GgBb GB Gb gB gb GB GGBB GGBb GgBB GgBb Gb GGBb GGbb GgBb Ggbb gB GgBB GgBb ggBB ggBb gb GgBb Ggbb ggBb ggbb a) From your Punnett square, how many individuals are dihybrids? b) How many individuals are both homozygous recessive for coat color and heterozygous for eye color (ggBb)? c) Fill in the blanks to state the expected phenotypic ratio of the offspring:…
pls don’t forget to answer the last bottom question as well. Thank you!
Please answer. Thank you and Godbless.
. Chromosome 3 of corn carries three loci (b for plant-color booster, v for virescent, and lg for liguleless). A testcross of triple recessives with F1 plants heterozygous forthe three genes yields progeny having the followinggenotypes: 305 + v lg, 275 b + +, 128 b + lg, 112 + v +,74 + + lg, 66 b v +, 22 + + +, and 18 b v lg. Give the genesequence on the chromosome, the map distances between genes, and the coefficient of coincidence.
Analysis of X-Linked Dominant and Recessive Traits Suppose a couple, both phenotypically normal, have two children: one unaffected daughter and one son affected with a genetic disorder. The phenotype ratio is 1:1, making it difficult to determine whether the trait is autosomal or X-linked. With your knowledge of genetics, what are the genotypes of the parents and children in the autosomal case? In the X-linked case?
Variations on a Theme by Mendel Pea plants usually have white or red flowers. A strange pea-plant variant is found that has pink flowers. A selfcross of this plant yields the following phenotypes: 30 red flowers 62 pink flowers 33 white flowers What are the genotypes of the parents? What is the genotype of the progeny with red flowers?
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?
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?