Mastering Biology Assignment 5: Genetics Chapter PDF

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Mastering Biology Assignment on Genetics for ABIO 121. ...

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Mastering Biology Assignment 5 Item 1 A botanist has acquired a group of sweet pea plants. All of the plants have yellow pea pods (the recessive trait), except for one, which has green pea pods (the dominant trait). Pea pod color is a trait caused by a single gene. In this tutorial, you will determine how the botanist can identify the genotype of the green pea pod, and how this relates to Mendel’s laws and meiosis.

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How could the botanist best determine whether the genotype of the green-pod plant is homozygous or heterozygous? - Cross the green-pod plant with a yellow-pod plant. A cross between a plant of unknown genotype and one that is known to be homozygous recessive is called a test cross because the recessive homozygote tests whether there are any recessive alleles in the unknown. Because the recessive homozygote will contribute an allele for the recessive characteristic to each offspring, the second allele (from the unknown genotype) will determine the offspring’s phenotype.

Diagramming a cross using a Punnett square Punnett squares can be used to predict the two possible outcomes of the botanist’s test cross. The Punnett square on the left shows the predicted result if the unknown plant is homozygous (GG); the Punnett square on the right shows the predicted result if the unknown plant is heterozygous (Gg).

The genotypes in a Punnett square show all the possible combinations of alleles in offspring that could result from the particular cross. A Punnett square reveals the expected probabilities of each genotype among the offspring. For example, the Punnett square on the right reveals that there is a 50% chance that each offspring will have green pods and a 50% chance that each offspring will have yellow pods.

Relationship with Mendel’s findings

Suppose that the botanist carried out the test cross described in Parts A and B and determined that the original green-pod plant was heterozygous (Gg). Which of Mendel’s findings does her test cross illustrate? - Law of Segregation The law of segregation states that the two alleles for a gene separate during gamete formation, and end up in different gametes. In the case of the heterozygous green-pod plant (Gg), one gamete will receive the dominant allele (G), and the other gamete will receive the recessive allele ( g). The law of segregation accounts for the prediction that 50% of the offspring of the test cross will have green pods and 50% will have yellow pods.

Relationship of allele behavior to meiosis During which part of meiosis (meiosis I or meiosis II) do the two alleles of a gene separate? During which phase does the separation occur? - meiosis 1, anaphase Alleles separate from one another during anaphase of meiosis I, when the homologous pairs of chromosomes separate.

Item 2

A color-blind woman mates with a male with normal color vision. Which of these results would indicate that color blindness is caused by an X-linked recessive allele?

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All of the sons, and none of the daughters, are color-blind.

Since all of the sons have the same phenotype as the mother and all of the daughters have the same phenotype as the father, the color blindness is likely to be caused by an X-linked recessive allele. Color blindness is an X-linked recessive trait. A color-blind man has a daughter with normal color vision. What is the genotype of the daughter?

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XC Xc

The XC assures the daughter of normal color vision while she inherits a Xc from her father. Color blindness is an X-linked recessive trait. A color-blind man has a daughter with normal color vision. She mates with a male who has normal color vision. What is the expected phenotypic ratio of their offspring?

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2 normal vision females : 1 normal vision male : 1 color-blind male

The parents, XCXc x XCY, have four possible offspring: XCXC (homozygous normal vision female), XCXc (heterozygous normal vision female), XCY (normal vision male), XcY (color-blind male). Color blindness is an X-linked recessive trait. A woman who is homozygous for normal color vision mates with a color-blind male. What is the expected phenotypic ratio of their offspring?

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All offspring have normal color vision

The offspring of the parents, XCXC x XcY, are: XCXc (heterozygous normal vision female), XCXc (heterozygous normal vision female), XCY (normal vision male), XCY (normal vision male). Color blindness is an X-linked recessive trait. A color-blind woman mates with a male with normal color vision. What is the expected phenotypic ratio of their offspring?

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1 normal vision daughter : 1 color-blind son

The offspring of the parents, XcXc x XCY, are: XCXc (heterozygous normal vision female), XCXc (heterozygous normal vision female), XcY (color-blind male), XcY (color-blind male). Color blindness is an X-linked recessive trait. Under what conditions can an unaffected male have a colorblind daughter?

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He can't.

Unless there is a very unlikely mutation, an unaffected male cannot have a daughter who expresses an Xlinked recessive. Hypophosphatemia (vitamin D-resistant rickets) is inherited as an X-linked dominant. An unaffected woman mates with a male with hypophosphatemia. What is the expected phenotypic ratio of their offspring?

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1 daughter with hypophosphatemia : 1 normal son

The offspring of the parents, XhXh x XHY , are: XHXh (heterozygous female with hypophosphatemia), XHXh (heterozygous female with hypophosphatemia), XhY (normal male), XhY (normal male). Suppose that having three nostrils is a Y-linked character. A woman with two nostrils mates with a man with three nostrils. What is the expected phenotypic ratio of their offspring?

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1 daughter with two nostrils : 1 son with three nostrils

The offspring of the parents, XX x XYE, are: XX (female with two nostrils), XX (female with two nostrils), XYE (male with three nostrils), XYE (male with three nostrils). Item 3

Galactosemia is a human genetic disorder caused by a deficiency of the enzyme galactose-1phosphate uridylyltransferase (commonly called GALT). Children with galactosemia seem normal at birth, but begin to show symptoms as soon as they drink milk. This is because they cannot metabolize galactose, a simple sugar that results from the breakdown of lactose (the sugar found in milk). Early symptoms of galactosemia include vomiting and diarrhea. If untreated, affected babies will develop enlarged livers, cataracts, and mental retardation. Depending on the alleles the baby has inherited, early death may result. Construct a pedigree John and Jane Jones’ newborn baby girl, Leah, has just been diagnosed with galactosemia. Prior to their genetic counseling appointment, they submitted this family history:      

John and Jane’s older child, a son named Lee, does not have galactosemia. John is the only child of Hanna and Harry. Harry was an only child. Hanna has two older sisters, Hope (the oldest) and Holly. Both Harry and Hanna’s parents lived in good health into their 80s. Jane’s brother, Joe, is married to Jen. They have a son, Les, who is a healthy nine-yearold with myopia (nearsightedness) but no sign of galactosemia. Jane’s mother, Hilda, is alive and healthy; Jane’s father, Henry, passed away last year of a sudden heart attack. Hilda and Henry have no family history of galactosemia.

Use the information in the family history to construct a pedigree for this family.

Identify the mode of inheritance Look over the pedigree you constructed in Part A. Based on the inheritance pattern, which mode of inheritance must be the cause of galactosemia? - autosomal recessive Because neither Jane nor John has the same condition as their daughter, and there is no evidence of sexlinkage, galactosemia must be an autosomal recessive trait. Jane and John are considering having another child. Given the pedigree you constructed and the mode of inheritance for galactosemia, what is the risk that their next child will have the disorder? - ¼ (because they are both heterozygotes)

A Punnett square for two heterozygous (Gg) parents reveals that each child would have a 1/4 chance of receiving the gg allele combination, resulting in galactosemia.

Offspring genotype probabilities: 1/4 GG; 1/2 Gg; 1/4 gg Genetic testing If Jane and John want to have another child, they plan to see a genetic counselor to find out when it would be best to test for galactosemia. A newborn with galactosemia must be put on a lactoseand galactose-free diet as soon as possible after birth. Even on this diet, affected individuals may

still suffer from learning disabilities, ovarian failure (in young women), late-onset cataracts, and early death. Which of the following tests would be most useful for Jane and John to have? -

newborn screening (either assaying for the GALT enzyme or measuring excess galactose in the newborn’s blood)

Newborn screening by either of these methods would be effective in identifying the condition in a newborn. It would provide definitive information early enough to begin dietary restrictions, if needed. Fetal testing for this genetic mutation does not make sense because it carries greater risk than newborn screening, and there is no treatment prior to birth. Item 4

Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the relationship between the CB and CW alleles? - The relationship between the alleles is one of incomplete dominance The phenotype of the heterozygote falls between black and white. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the expected phenotypic ratio of a CBCW x CBCW cross?

- 1 black : 2 gray : 1 white In this instance the phenotypic and genotypic ratios are the same. Andalusian chickens with the genotype CBCB are black, those with the genotype CWCW are white, and those with the genotype CBCW are gray. What is the expected genotypic ratio of a CBCW x CBCW cross? - 1 CB CB : 2 CB CW : 1 CW CW In this instance the phenotypic and genotypic ratios are the same. Item 5...