Section I: Monohybrid Simple Mendelian GeneticsThe following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit thedominant trait, W. Affected individuals are indicated by a dark square or circle.IIIIIV1ww2 3 45B. Ww2WwB. 100%62 31. What is the sex of the individual in II-3 AND does the person have the W trait?A. Female....yesC. Male.....yesB. Female.....noD. Male....no82. What is the genotype of individual II-5A. ww or WwC. wwD. WW3. What is the likelihood the progeny of IV-3 and IV-4 will have the trait?A. 0%C. 75%D. 50%

The genotype is the combination of alleles for a trait. The allele for a trait exists in two forms…

Section I: Monohybrid Simple Mendelian Genetics The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. I III IV 1 ww 2 3 4 5 B. Ww 2 Ww B. 100% 6 1. What is the sex of the individual in II-3 AND does the person have the W trait? A. Female....yes C. Male.....yes B. Female.....no D. Male....no 2 3 2. What is the genotype of individual II-5 A. ww or Ww C. ww D. WW 3. What is the likelihood the progeny of IV-3 and IV-4 will have the trait? A. 0% C. 75% D. 50%

Section I: Monohybrid Simple Mendelian Genetics The following question refer to the pedigree chart in the figure below for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle. I III IV 1 ww 2 3 4 5 B. Ww 2 Ww B. 100% 6 1. What is the sex of the individual in II-3 AND does the person have the W trait? A. Female....yes C. Male.....yes B. Female.....no D. Male....no 2 3 2. What is the genotype of individual II-5 A. ww or Ww C. ww D. WW 3. What is the likelihood the progeny of IV-3 and IV-4 will have the trait? A. 0% C. 75% D. 50%

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter12: Mendel, Genes, And Inheritance

Section: Chapter Questions

Problem 13TYK: Some recessive alleles have such a detrimental effect that they are lethal when present in both...

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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 if the couple wanted prenatal testing so that a normal fetus could be aborted?
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?
Some recessive alleles have such a detrimental effect that they are lethal when present in both chromosomes of a pair. Homozygous recessives cannot survive and die at some point during embryonic development. Suppose that the allele r is lethal in the homozygous rr condition. What genotypic ratios would you expect among the living offspring of the following crosses? a. RRRr b. RrRr
t 2. ng. Questions and Problems Given the pedigree below, answer the following questions separately for each pedigree. 1. What is the mode of inheritance? 2 Write the genotype of the individuals in the pedigree based on the mode of inheritance given. date distance Pedigree 1 gous Pedigree 2 Recombination zzel 9rd To ya hisiga of elds ed bittorie 10 Tube heterozygpitanjonosenis the fasiria cell while the tester, taken as the raingoz, mix on the allele of the nie of Deteraniol are now adva Hol Tidmoos DNG sevel. This cross will produce filial generation (2) walthen purple ves f 11020 12 13 encont 000 sg nomi 2 is importan gies to do this, scientists preously 4 méte based monso nomiskt sild devon S01 000 05 1uodo 910 919/3 2m al m 29098 3291T 20 comondo of pes. The value is, the OTO the value is the closer 1 2 nesten che 3 4 5 6 I snag- niboo to 19dmun srid wordt oo Jadigrappen va bo trog do disp ambe 000 61 251110010 aliso ist was by The song Shear Besa STS GRAD color. In his experimenti…
age 105 of 114 3 5 € BUD > D Q Autosomal Genes 1. Parental Cross: It is possible that your mutation is autosomal (dominant or recessive). In this case, you only need one Punnett square to make predictions for the F1 generation that will result from the parental cross, as it does not matter if the female or male in the parental generation is the true breeding mutant. a. Provide a Punnett square to predict the genotype of the F1 generation. b. Q Search If the mutant allele is dominant, predict the phenotype of the F1 generation. Make sure to also summarize the prediction as a ratio, comparing Wild-type : Mutant. c. If your mutant allele is recessive, predict the phenotype of the F1 generation. Make sure to also summarize the prediction as a ratio, comparing Wild-type : Mutant. 2. F1 Self-Cross: Take the F1 results from question #1 to create a Punnett square predicting the F2 generation genotypes. 104 a. What are the predicted F2 phenotypes if the mutant allele is dominant? Make sure to…
Part 1: Mendelian Genetics 1. For each of the genotypes Purple Flowers are Dominant to white flowers PP - Purple Pp - Purple pp-White Straight hair is dominant to curly hair below, list all possible phenotypes. Brown eyes are dominant to blue eyes Straight - HH, Hh Curly - hh BB- Bb bb- 2. For each of the phenotypes below, list all possible genotypes. (Remember: use only 1 letter to designate a trait, big and small per scenario) Pointed heads are dominant to round heads Black fur is dominant to white fur Pointed - Round - FF - Ff- ff - Jagged leaves are dominant to smooth leaves Jagged - Smooth -
PEDIGREES: Problem (continued) This pedigree shows the inheritance of cystic fibrosis in this family. I • QUESTIONS ••. 5. What is the genotype of individual II-3? Use the letter "f" to 1 2 represent the disease allele. II 1 2 3 6. Individuals II-I and II-2 are sisters. Explain how it is possible for one sister to have cystic fibrosis but NOT the other. III 1 2 3
A. HUMAN PEDIGREE CASE ANALYSIS1. One couple has three children with the following sexes and ages: one son (40 y.o.) and two daughters (35 y.o. and 33 y.o.), all of them have normal pigmentation. Another couple has a son (35 y.o) and a daughter (20 y.o.) and all of them also have normal pigmentation. Both couples have normal pigmentation. The younger daughter from the first couple married the son of the second couple and they had three children. Their eldest daughter (5 y.o.) has normal pigmentation while their only son (3 y.o.) and one daughter (1 y.o.) have albinism. a. Draw the pedigree of this family. Follow protocols in making a pedigree. Provide the genotype of all individuals in the pedigree. Please provide also the gene notation. b. What is the mode of inheritance of this trait? c. Justify your answer in letter (b).d. For their normal daughter, what is the probability that she is a carrier? Show solution. e. If they will have a fourth child, what is the probability that the…
BB AB AB A. None B.3 с.1 D.2 ВВ OE. The father AB In the pedigree presented above, an autosomal dominant disease which causes significant visual loss and eventual blindness, is segregating in the family. The disease gene causing this sight loss looks as if it may be linked to a marker locus. The alleles of this marker locus that are present in this family are allele A and allele B. Are there any recombinant individuals in this pedigree? AB ВВ AB ВВ AB ВВ AB AB ВВ
97. This family is (picture above) affected with blindness. Individual (IV.1) is clinically unaffected. What is the chance that he is homozygous for the normal allele? A-½ B-% C- almost 0 D-1/3 E- almost 100% ANS: IV TO O 560-556
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.