Concept explainers
The accompanying pedigree shows a family in which one child (II-1) has an autosomal recessive condition. On the basis of this fact alone, provide the following information.
a. Using A for the dominant allele and a for the recessive allele, give the genotypes for I-1, I-2, and II-1.
b. Using the same alleles, give the possible genotypes for II-2, II-3, and II-4.
c. What are the probabilities for each of the possible genotypes for II-2, II-3, and II-4?
d. What is the probability that all three of the children in generation II who have the dominant
e. What is the chance that among the three children in generation II who have the dominant phenotype, one of them is AA and two of them are Aa? (Hint: Consider all possible orders of genotypes.)
Trending nowThis is a popular solution!
Chapter 2 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- The condition phenylketonuria is caused by a recessive allele. There are two carriers who have progeny.a. Give the gene notation. b. Give the expected genotypic and phenotypic ratios. c. What is the probability that their child will be heterozygous if they have a normal child?d. What is the probability of having two affected children and one normal child if they have three children?arrow_forwardCystic fibrosis (CF) is an autosomal recessive trait. A three-generation pedigree is shown below for a family that carries the mutant allele for cystic fibrosis. Note that carriers are not colored in to allow you to figure out their genotypes. Normal allele = F CF mutant allele = f What is the genotype of individual #13? A) ff B) FF C) Ff D) it is impossible to tellarrow_forwardPlease consider the pedigree below. There are no cases of false paternity. I II III IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the AB0 and H loci? Please label your answers a and b, Il-2: and Ill-2:.arrow_forward
- ii) State the genotypes of individuals # 1-5 in the following table using the letter "A". Use the uppercase letter to represent the dominant allele and lowercase letter to represent the recessive allele. Genotype Individual #1 # 3 #4 #5 If individuals # 2 and 3 have another son what are the chances that this son will be affected? b) i) What is the most likely mode of inheritance for this pedigree? ii) State the genotypes of individuals # 6-8 in the following table using the letter "B". Use the uppercase letter to represent the dominant allele and lowercase letter to represent the recessive allele. Individual # 6 7 # 7 # 8 Genotypearrow_forwardPlease consider the pedigree below. There are no cases of false paternity. I B II A 2 3 III AB (A IV в 1 a. Which individual/s definitely has/have Bombay phenotype in the descendants of I-1 and I-2? b. What are the genotypes of individuals II-2 and III-2 at the ABO and H loci?arrow_forwardThis pedigree traces the inheritance of a rare disease in humans. a. Based on this pedigree, is the allele for this disease dominant or recessive? Explain. b. What genotypes are possible for the individuals labeled 1, 2, and 3?arrow_forward
- The pedigree shows a family in which several members have suffered from one and the same disease (look at the picture to be able to answer) a) Is it a dominant or recessive allele that causes the disease? Motivate your answer. b) Is allele autosomal or sex-linked? Motivate your answer. c) What is the probability that III-3 and III-4 will have a healthy child? Motivate your answer.arrow_forwardIn goats, a beard is produced by an autosomal allele that is dominant in males and recessive in females. We’ll use the symbol b for the beard allele and + for the beardless allele. Another independently assorting autosomal allele that produces a black coat (W) is dominant over the allele for white coat (w). Part A and D have been completed for you and is shown in the image. Give the phenotypes and their expected proportions for the crosses B and C:b) +bWw male x +bww femalec) ++Ww male x bbWw femalearrow_forwardTay–Sachs disease is caused by recessive alleles on anautosome. In which case(s) could two parents with anormal phenotype have a child with Tay–Sachs?a. Both parents are homozygous for a Tay–Sachs allele.b. Both parents are heterozygous for a Tay–Sachsallele.c. One parent is homozygous for a Tay–Sachs allele,and the other is heterozygous.arrow_forward
- Take the example of B-thalassemia, an autosomal recessive genetic disease that particularly affects people from around the Mediterranean. This disease is associated with an anomaly of hemoglobin, a protein essential for the transport of oxygen, which is composed of four chains: two alpha (a) and two beta (B). In case of B-thalassemia, the ẞ chains are produced in insufficient or no quantity in an individual homozygous recessive resulting in insufficient production of overall hemoglobin leading to anemia and other physiological challenges. The gene that controls the synthesis of the ẞ chains is located on chromosome 11. Here is part of the coding portion of this gene (which controls a total of 146 amino acids and of which you only see the portion 36 to 41) and one of the targeted mutations: 1. Give the sequence of amino acids from the template and mutated strands. 2. What type of point mutation is it? 3. Using the principles of the theory of evolution, explain briefly and generally why…arrow_forwardCystic Fibrosis (CF) is an autosomal recessive condition. Therefore, heterozygous (Cc) carriers do not display symptoms. Two parents who are carriers plan to start a family and you are a genetic counselor helping to advise them about their chances of having children affected by CF. a) Suppose the couple has 4 children, each one year apart. What is the probability that all 4 children will inherit CF? b) What is the probability that any 3 of their 4 children will not inherit CF, but 1 will be affected? c) What is the probability that their first child will not inherit CF, but the younger 3 children will inherit CF?arrow_forwardThe following pedigree shows the pattern of inheritance for an uncommon human disease. Filled symbols indicate individuals with the disease: open symbols indicate normal individuals. Using the symbols A and a, answer the following questions about this pedigree. 1. What are the most probable genotypes of individuals II-9 and II-10 2. If individuals II-1 and II-2 have another child, what is the probability that it will have the disease?arrow_forward
- Human Heredity: Principles and Issues (MindTap Co...BiologyISBN:9781305251052Author:Michael CummingsPublisher:Cengage Learning