genotype cannot be determined QUESTION 2 Suppose that extra fingers and toes are caused by a recessive trait but only appears in 25% of the homozygous recessive individuals. If two heterozygotes conceive a child, what is the probability that this child will have extra fingers? O a. 1/4 O b. 1/2 O c. 1/16 O d. 1/3 O e. 1/8 QUESTION 3 o Mendel's nea plante mund shane (R) is dominant to wrinkled shane (r) and vallow color Vis dominant to green color (u) The alleles for coad chane and coad.c Save and Submit to save and submit. Click Save All Answers to save all answers. 20 ODO MacBook Air

genotype cannot be determined QUESTION 2 Suppose that extra fingers and toes are caused by a recessive trait but only appears in 25% of the homozygous recessive individuals. If two heterozygotes conceive a child, what is the probability that this child will have extra fingers? O a. 1/4 O b. 1/2 O c. 1/16 O d. 1/3 O e. 1/8 QUESTION 3 o Mendel's nea plante mund shane (R) is dominant to wrinkled shane (r) and vallow color Vis dominant to green color (u) The alleles for coad chane and coad.c Save and Submit to save and submit. Click Save All Answers to save all answers. 20 ODO MacBook Air

Name: genotype cannot be determinedQUESTION 2Suppose that extra fingers and
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Description: genotype cannot be determinedQUESTION 2Suppose that extra fingers and toes are caused by a recessive trait but only appears in 25% of the homozygous recessive individuals. If two heterozygotes conceive achild, what is the probability that this child

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 14QP: Analysis of Autosomal Recessive and Dominant Traits Huntington disease is a rare, fatal disease that...

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A woman who is heterozygous, Bb, has brown eyes; B (brown) isthe dominant allele, and b (blue) is recessive. One of her eyes,however, has a patch of blue color. Give three different explanationsfor how this might have occurred?
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Fruit flies with one allele for curly wings (Cy) andone allele for normal wings (Cy+) have curly wings.When two curly-winged flies were crossed, 203curly-winged and 98 normal-winged flies wereobtained. In fact, all crosses between curly-wingedflies produce nearly the same curly : normal ratioamong the progeny.a. What is the approximate phenotypic ratio in theseoffspring?b. Suggest an explanation for these data.c. If a curly-winged fly was mated to a normalwinged fly, how many flies of each type would youexpect among 180 total offspring?
Homozygous dominant parent (PP), and a Homozygous recessive or just simply say recessive parent (pp): a. Fill in the Punnett square. Each box represents a genotype possibility for an offspring. b. Place the allele donated by each parent in the corresponding box. Now list the possible genotypes and their corresponding phenotype. c. If an individual's genotype is heterozygous, the dominant trait will be expressed in the phenotype. Give the percent possible for the phenotypes. P P p p Genotype: ______________ Phenotype: ______________ Phenotype % probable: __________ 2. Cross between a Homozygous dominant parent (PP) and a Heterozygous parent (Pp). Fill in as in step one. 3. Cross between a Heterozygous parent (Pp), and another Heterozygous parent (Pp). Fill-in as before. 4. Test cross: A test cross is between a recessive parent (pp), and a Heterozygous parent…
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Pedigree attached shows an autosomal recessive genetic disease. G is the normal allele and g is the disease-causing allele. Individual 1’s father is heterozygous (*) and his mother is homozygous dominant. Other individuals in the pedigree may be carriers, but are not marked. The question mark (?) indicates that you do not yet know anything about this individual’s phenotype with regard to the disease. part a) What is the probability that individuals 1 and 2 will have a child (5) who is a boy with the disease (the child is unborn and the sex is not yet known)? a)1/8 b)1/4 c)0 d)1/16 part b) What is the probability that the daughter (6) that individual 3 and 4 just had will have the disease? a)1/8 b)1/6 c)1/4 d)1/12