a. What specific mode of inheritance is exhibited? b. Using the letters A and B, assign alleles to the traits c. Based on the 9:3:3:1 ratio, what are the genotypes of the following:  7 children with black hair ________________  2 children with red hair ___________________  1 child with blond hair ____________________

a. What specific mode of inheritance is exhibited? b. Using the letters A and B, assign alleles to the traits c. Based on the 9:3:3:1 ratio, what are the genotypes of the following:  7 children with black hair ______  2 children with red hair _  1 child with blond hair ____________

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter4: Pedigree Analysis In Human Genetics

Section: Chapter Questions

Problem 1CS: Pedigree analysis is a fundamental tool for investigating whether or not a trait is following a...

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Alkaptonuria is a metabolic disorder in which affected people produce black urine. Alkaptonuria results from an allele (a) that is recessive to the allele for normal metabolism (A). Sally has normal metabolism, buther brother has alkaptonuria. Sally’s father has alkaptonuria, and her mother has normal metabolism. Q. Give the genotypes of Sally, her mother, her father, and her brother.
In humans, a dimple in the chin is a dominant characteristic controlled by a single gene.a. A man who does not have a chin dimple haschildren with a woman with a chin dimple whose mother lacked the dimple. What proportion of theirchildren would be expected to have a chin dimple?b. A man with a chin dimple and a woman who lacksthe dimple produce a child who lacks a dimple.What is the man’s genotype?c. A man with a chin dimple and a nondimpledwoman produce eight children, all having the chindimple. Can you be certain of the man’s genotype?Why or why not? What genotype is more likely,and why?
In a breed of mice fur coat color is determined by two genes. For coat color BB results in black, BT results in black with tan spots and TT results in tan. However, to have coat color there must be at least one dominant allele of gene A where aa is white (no color). (a)For each trait, give the type of inheritance demonstrated. Color of fur and if there is color. A (1) tan mouse heterozygous for coat color is crossed with a (2) Black and tan spotted mouse that is homozygous for color (b) What is the genotype of the two parents involved in the cross? (c) When crossed, what is the chance of getting a tan mouse? Show your work (you may either use a Punnett square or Fork method).
raccoons may have wide, medium-sized, or narrow bands around their tails. They may also havethe habit of washing all, or some of their food, or do not wash their food at all. a) assign genotypes to the phenotypes mentioned (see attached table) b. What mode of inheritance would most likely be exhibited by these traits if crosses were made? c. If two raccoons with medium-sized tail bands and have the habit of washing some of theirfoods will be crossed, what is the probability of having F1 raccoons with: c.1 wide tail bands that won’t wash any of their food? c.2 the same genotype as the parent raccoons? d. If a raccoon with a wide tail band that washes only some of its food is crossed with a raccoonwith a narrow tail band that doesn’t wash any food, what percentage of their offspring wouldbe medium-tailed and washes all its food? Show COMPLETE cross.
A. Identify the most likely inheritance pattern in the pedigree below. B. What are the genotypes of individuals II-3 and Il-4? Use any letter for your example. 1 2 II 1 2 3 4 5 II 1 2 3 4 5 6
Skin color is one of the traits in a human which is determined by polygenic inheritance system because it is possibly involving as many as 9 genes. To make this simple, let us consider the influence of 2 genes: A and B., where dark skin color is dominant. Suppose a man who is AABb marries a woman who is Aabb, what would be the genotypes of their children. Prove your answer by using a Punnet Square.
Wavy hair is intermediate to straight and curly hair. Similarly, medium-sized nose may be inherited if one parent has a large nose and the other parent has a small nose. Wendel has wavy hair, large nose, and type A blood. His father is type O. What is the complete genotype of Wendel? Krisha has straight hair, medium-sized nose, and type AB blood. 2. What is the complete genotype of Krisha? Assuming that Krisha and Wendel would marry, what is the probability that they will have a: 3. Son with the same traits as Wendel? 4. Daughter with the same traits as Krisha? 5. Child with curly hair, small nose, and type O blood?
In humans, there are three alleles for blood type: A, B, and O. A and B are codominant over O. A male with type AB blood and a female with type A blood have a child. The male’s parents both had type AB blood. The female’s mother had type A blood and her father had type B blood. What is the potential phenotype of the child? A. type A (25%), type B (25%), type AB (25%), type O (25%) B. type A (25%), type B (50%), type AB (25%) C. type A (25%), type B (25%), type AB (50%) D. type A (50%), type B (25%), type AB (25%)
The B allele confers black coat color and the b allele brown coat color in Labrador dogs. When another gene, E, is inherited, the coat is golden no matter what the B phenotype is. A dog of genotype ee expresses the black phenotype. What unusual phenomenon is at work in this disorder?
In Labrador retriever dogs, two alleles (B and b) determine whether coat colour will be black (B)or brown (b). Black coat colour is dominant. A second pair of alleles, E and e, are on a separatechromosome from B and b. The homozygous recessive condition, ee, prevents the expressionof either allele B or b, and produces a dog with a yellow-coloured coat. Some examples ofgenotypes and phenotypes for Labrador retrievers are shown below.Genotype PhenotypeBBEe blackbbEe brownBbee yellow Two dogs, each with the genotype, BbEe, were crossed. What is the percentageprobability that their offspring will have yellow coat colour? (Record your answer as awhole number percentage)
The term epistasis is used to refer to a situation in which the expression of a gene is influenced by another independently inherited gene. In labs, the gene B determines how much pigment is made. The dominant allele B results in black fur and the recessive allele b results in brown (chocolate) fur. A separate gene, E, codes for a protein that determines whether the pigment is deposited in the hair. (E = pigment deposited; e = pigment not deposited). What genotypes are possible for a black lab? What genotypes are possible for a chocolate lab? What genotypes are possible for a yellow Lab? In a cross between a black lab that is homozygous for both alleles and a yellow lab that is homozygous for both alleles, what would the genotype and phenotype of the offspring be? Imagine you cross two heterozygous parents. What are the genotypic and phenotypic ratios of the offspring?
Match the pattern of inheritance to the appropriate term. A. Heterozygotes with different alleles of the DTL1 gene survive better than homozygotes B. The disease is usually passed from a mother to all children C. Pure-breeding pumpkin plants grown on sandy soil have bigger seeds than plants of the same pure-breeding lineage grown on peaty soil. D. A cross between a true-breeding plant with serrated leaf edges and a true- breeding plant with smooth leaf edges produces an F1 generation with 88% plants with serrated leaves. E. A cross between two heterozygotes, DdFf, produces offspring in three phenotypic groups; Long wings in offspring with genotype D_F_, short wings in offspring with D_ff, and no wings in offspring with the genotypes ddF_ and ddff. F. Individuals homozygous for a mutation in SME have a heightened sense of smell and large earlobes. G. Grey chickens bred together have offspring that are black, grey and white in a 1:2:1 ratio H.Alleles at locus S…