Albinism is homozygous recessive (aa). A sister with normal coloration has a sister that has albinism and both her parents have normal coloration. What is the probability that the sister with normal coloration is a carrier of the albinism allele? a) 1/8 b) 1/4 c) 1/2 d) 1/16 e) 2/3
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- As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women. During his tenure as guard, he had numerous clandestine affairs. In each case, children resulted. Subsequently, some of the childrenwho had no way of knowing that they were relatedmarried and had children of their own. Assume that two pairs of genes determine height. The genotype of the 7-foot-tall Potsdam Guard was A9A9B9B9, and the genotype of all of his 5-foot clandestine lovers was AABB. An A9 or B9 allele in the offspring each adds 6 inches to the base height of 5 feet conferred by the AABB genotype. a. What were the genotypes and phenotypes of all the F1 children? b. Diagram the cross between the F1 offspring, and give all possible genotypes and phenotypes of the F2 progenyWhat could be the genotypic and phenotypic ratios for a typical mendelian trait showed in the picture?In humans, brown eyes are dominant to blue eyes, and the ability to roll your tongue is dominant to not being able to roll your tongue. If a mother who was heterozygous for eye colour and homozygous recessive for tongue rolling had a baby with a father that was heterozygous for both traits, what is the probability that their child would have blue eyes and could roll their tongue?
- Farmer Brown had two hogs, Olga and Hank. Olga and Hank were normal, healthy hogs, but whenever Farmer Brown crossed them, a few of the piglets had cleft palates. Assuming a simple autosomal basis for cleft palate, what are the genotypes of Olga and Hank, and about what proportion of the piglets are born with a cleft palate?Gregor Mendel discovered the basis of heredity with his sweet pea plant experiments. In his studies, he determined that certain traits, such as pod color and pea shape, express complete dominance. A cross occurs between a plant with heterozygous yellow pods and wrinkled peas and a plant with green pods and heterozygous round peas. What is the probability that the offspring will exhibit recessive genes for both traits? A - 1/2B - 1/4C - 1/8D - 1/16A woman with fair skin, blond hair, and blue eyes gives birth to fraternal twins; the father has dark brown skin, dark hair, and brown eyes. One twin has blond hair, brown eyes, and light skin, and the other has dark hair, brown eyes, and dark skin. What Mendelian law does this real-life case illustrate and explain what this means in terms of the inherited alleles for these genes?
- In humans, the genetic disease cystic fibrosis is caused by a recessive allele (a). The normal (healthy) allele is dominant (A). What is the genotype of someone who has cystic fibrosis? What are the two different genotypes that a healthy person could have? If two people were both heterozygous for the cystic fibrosis gene, what fraction of their children would be likely to have this disease? Hint: Draw a Punnett square to figure it out.Albinism is inherited as a recessive trait in humans. If the parents are both heterozygous for the trait, what is the probability of obtaining the following: a. 3 albino children? b. 2 normal girls? c. 4 albino boys? d. first 4 children will be normal and the last 2 children will be albino? e. 4 normal children and 2 albinos in any order?Ben, Marie, and Korra are siblings. Ben is blood type O, Ana is blood type B, and Korra is type A. What could be the possible genotypes of their parents?
- In classical Mendelian genetics, how can one check the genotype of a parent (A) expressing the characters of a dominant allele? Select one: a. By performing a back cross with a recessive homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character. If the parent A was, instead, a heterozygote, then 50% of the F1 progeny will express the recessive character (homozygote recessive) and 50% the dominant one (heterozygotes). b. It is impossible to check such genotype without using specific molecular assays. c. By performing a back cross with a dominant homozygote parent (B). If the A parent is homozygote for the dominant allele, then all the individuals from the F1 will display the dominant character.The following cross is set: DDEeFFggHh X DdEEFfGGhh a. What is the maximum number of additive alleles possible in their offspring? b. What is the maximum number of non-additive alleles possible in their offspring?In terms of a chi-square analysis, what ratio would be tested for the offspring of a monohybrid cross involving a pure-breeding dominant condition mated with a pure-breeding recessive condition?