Biology Chapter 14

An individual heterozygous for eye color, skin color, and number of eyes mates with an individual who is homozygous recessive for all three characters; what would be the expected phenotypic ratio of their offspring? [Hint: B = black eyes, b = orange eyes; G = green skin, g = white skin; C = two eyes, c = one eye]

1 black eyes, green skin, two eyes : 1 black eyes, green skin, one eye : 1 black eyes, white skin, two eyes : 1 black eyes, white skin, one eye : 1 orange eyes, green skin, two eyes : 1 orange eyes, green skin, one eye : 1 orange eyes, white skin, two eyes : 1 orange eyes, white skin, one eye This is a tough problem; you had to expand your Punnett square to accommodate another character. However, 1:1:1:1:1:1:1:1 ratio is the expected outcome of a BbGgCc x bbggcc cross.

Law of segregation and law of independent assortment

1.Law of segregation: The two copies of a gene are separated during meiosis and end up in different gametes 2. When gametes form, the alleles of a particular gene segregate during meiosis independently of the alleles of other genes.

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?

1/4 (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.

Suppose you mate two mice with the genotypes AYaBb x AYaBb . Considering only the live-born offspring, what would be the expected frequency of mice with yellow fur? (For help getting started, see Hint 1.) Express your answer as a fraction using the slash symbol and no spaces (for example, 1/16).

2/3. Because the presence of the AY allele is epistatic to (masks expression of) the B/b gene, the B/b gene does not need to be taken into consideration in this problem. For the AYa x AYa cross, 1/4 of the offspring would have the AYAY genotype, which is lethal before birth. For the live-born offspring, 2/3 would be AYa, and thus have yellow fur.

Two mice are heterozygous for albinism (Aa) . The dominant allele (A) codes for normal pigmentation, and the recessive allele (a) codes for no pigmentation. What percentage of their offspring would have an albino phenotype?

25% The offspring would be in a 3:1 ratio of normally pigmented mice to albino mice.

What would be the expected frequency of agouti brown offspring in the litter?

3/16. Because the two traits are determined by unlinked genes, they assort independently. As a result, you need to use the multiplication rule to calculate the probability of agouti brown offspring (A_ bb) from AaBb parents. The probability of A_offspring is 3/43/4, and the probability of bb offspring is 1/4 1/4. The combined probability is therefore 3/4 x 1/4 = 3/16.

What number and types of chromosomes are found in a somatic cell in an animal with a diploid number of 48?

48 chromosomes (24 from each parent, including one sex chromosome from each) Review: All somatic cells in animals have two sets of chromosomes, one haploid set (n) from each parent, and each set includes one sex chromosome. For an animal with a diploid number of 48 (2n = 48), the number of chromosomes in the somatic cells is 48 and the number of chromosomes in each haploid set is 24 (n = 24). Humans have a diploid number of 46 (2n = 46), and the number of chromosomes in each set is 23 (n = 23), including one sex chromosome.

Drag the correct symbols to their appropriate locations on the pedigree below. Not all symbols will be used. See the Hints for help getting started.

A pedigree provides a pictorial representation of the genetic relationships in a family and shows each person's sex and phenotype with respect to the trait in question. In this family, Leah is the only family member (on both sides) that has galactosemia.

Which of these is a testcross

A? x aa. A testcross is used to determine whether an individual expressing the dominant phenotype is homozygous dominant or heterozygous.

What is the difference between heterozygous and homozygous individuals?

All of the gametes from a homozygote carry the same version of the gene while those of a heterozygote will differ. Since homozygotes carry two identical copies of a gene, all of the gametes will carry the same version. Heterozygotes have two different versions, so there will be two different types of gametes.

In the following cross the genotype of the female parent is BbGg. What is the genotype of the male parent? [Hint: B = black eyes, b = orange eyes, G = green skin, g = white skin]

BBGg. All of the offspring have black eyes, and there is a 3:1 ratio of skin color.

The result of the following cross indicates that the genotype of the male parent is _____.

Because wild type head top is dominant to flat head top, the only genotype that produces the flat head top phenotype is homozygous recessive.

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? State your answer as meiosis I or meiosis II followed by a comma and the name of the phase (for example, if your answer is meiosis II and metaphase, enter meiosis II, metaphase).

Meiosis I, anaphase Alleles separate from one another during anaphase of meiosis I, when the homologous pairs of chromosomes separate.

That each gamete contains a single allele of the eye color gene is an illustration of _____.

Mendel's law of segregation. Mendel's law of segregation is based on gametes getting only one copy of each allele of each gene.

The result of the following cross indicates that the genotype of the female parent is _____.

RR. Because wild type head top is dominant to flat head top and because all of the F1 generation are wild type head top, the genotype of the female parent must be homozygous dominant.

You cross a true-breeding red-flowered snapdragon with a true-breeding white-flowered one. All of the F1 are pink. What does this say about the parental traits? See Concept 14.3 (Page)

Red shows incomplete dominance over white, and the F1 is therefore pink.

In the video, the paternal parent carries the allele for purple flowers, and the maternal parent carries the allele for white flowers. What would happen if the maternal parent carried the purple allele, and the paternal parent carried the white allele?

The F1 hybrid pea plants would still produce purple flowers. In the case of F1 hybrid pea plants, it does not matter which parent carries the allele for purple flowers. It only requires one copy of the purple allele to result in purple flowers.

You continue your analysis by crossing the forked and twist lines. Your results are as follows: Which of the following statements best explains the outcome of this cross?

The forked mutation and the twist mutation are codominant alleles of the same locus. The absence of wild-type progeny in the F2 of a cross indicates a monohybrid cross: The two lines crossed are mutant at the same locus. Any other explanation that involves two loci (for example, recombination) fails because any cross with two loci will produce at least some F2 individuals with a wild-type genotype for both loci and therefore a wild-type phenotype. In this case, the F1 expresses both mutant phenotypes, indicating that the forked and twist alleles are codominant.

True or false? The same phenotype can be produced by more than one genotype.

True. Since there exist dominant and recessive versions of many genes, a phenotype that is based upon the dominant version will be expressed in both homozygous (AA) and heterozygous (Aa) genotypes.

A cross between two individuals that are heterozygous for eye and skin color would be an example of a _____ cross.

dihybrid. This is a cross of heterozygotes for two characters.

Consider pea plants with the genotypes GgTt and ggtt . These plants can each produce how many type(s) of gametes? See Concept 14.2 (Page)

four ... one GgTt individuals can produce the following gametes: GT, Gt, gT, and gt. A ggtt plant can produce only gt.

The information contained in DNA is used to make which of the following product(s)?

proteins, mRNA, and tRNA. Review: The information that programs all of a cell's activities is encoded in the sequence of DNA nucleotides. DNA directs RNA synthesis and, through mRNA, directs the order of the amino acids during protein synthesis; this entire process is called gene expression. Two types of RNA required for protein synthesis are mRNA and tRNA.

A plant grown from a [round, yellow] seed is crossed with a plant grown from a [wrinkled, yellow] seed. This cross produces four progeny types in the F1:[round, yellow], [wrinkled, yellow], [round, green], and [wrinkled, green].Use this information to deduce the genotypes of the parent plants. Indicate the genotypes by dragging the correct label to the appropriate location.

yellow round = RrYy yellow wrinkled= rrYy Answering this question requires two logical steps: First, eliminate those genotypes that are inconsistent with the phenotypes of the parents. Then examine the remaining possibilities: Which are consistent with the phenotypes of the progeny? Because the cross produces green progeny (and both parents are yellow), both parents must be Yy: If either parent were YY, only yellow progeny would result.The cross also produces wrinkled progeny, so the [round, yellow] parent must be heterozygous (Rr): If it were RR, all of the progeny would be Rr and have a round phenotype.

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.

If B represents the allele for black eyes (dominant) and b represents the allele for orange eyes (recessive), what would be the genotypic ratio of a cross between a heterozygous black-eyed MendAlien and an orange-eyed MendAlien?

0 homozygous black (BB): 1 heterozygote (black) (Bb): 1 homozygous orange (bb)

In a situation in which genes assort independently, what is the ratio of the gametes produced by an AaBB individual?

1 AB: 1 aB

If B represents the allele for black eyes (dominant) and b represents the allele for orange eyes (recessive), what would be the phenotypic ratio of a cross between a heterozygous black-eyed MendAlien and an orange-eyed MendAlien?

1 black : 1 orange The heterozygous black-eyed MendAlien has the genotype Bb. The orange-eyed MendAlien has the genotype bb. The result of this cross is two Bb offspring and two bb offspring. This means that two of the offspring have black eyes and two of the offspring have orange eyes. This 2:2 ratio then reduces to 1:1.

Black eyes are dominant to orange eyes, and green skin is dominant to white skin. A male MendAlien with black eyes and green skin, has a parent with orange eyes and white skin. A female MendAlien with orange eyes and white skin. If these male and female MendAliens were to mate, the predicted phenotypic ratio of their offspring would be _____.

1 black eyes, green skin : 1 black eyes, white skin : 1 orange eyes, green skin : 1 orange eyes, white skin Sam's genotype is BbGg, and Carole's genotype is bbgg.

Two organisms with genotype AaBbCcDdEE mate. These loci are all independent. What fraction of the offspring will have the same genotype as the parents? See Concept 14.2 (Page)

1/16 Breaking the question down into individual loci makes it simpler. The offspring of Aa ×Aa will be Aa 50% (one-half) of the time. The same is true for Bb, Cc, and Dd. Two EE individuals can only have EE offspring, so that probability is 1. The chance of an offspring being identical to the parent is therefore (1/2)^4, or 1/16.

One character in peas that Mendel studied was yellow versus green seeds. A cross between a homozygous yellow line (YY) and a homozygous green line (yy) will result in F1 plants that are heterozygous (Yy) for this trait and produce yellow seeds. When an F1 plant undergoes meiosis, what gamete types will it produce, and in what proportions?

1/2 Y 1/2 y Mendel's law of segregation states that allele pairs segregate equally into gametes during meiosis. This means that a gamete will have only one allele of any given gene, and that the probability of a gamete having one allele or the other is equal (and therefore ½, or 50%, for either allele).

Punnett squares are convenient ways to represent the types and frequencies of gametes and progeny in experimental crosses.This Punnett square shows the results of a Yy x Yy cross to form F2 progeny.Use your understanding of Mendel's law of segregation and the rules of probability to complete the Punnett square for this cross. First identify the gametes. Use pink labels to identify the male and female gamete types and white labels to identify the gamete frequencies. Then identify the F2 progeny. Use pink labels to identify the progeny genotypes and white labels to identify the progeny frequencies.

A Punnett square is a convenient method for representing Mendel's law of segregation in a visual form. Using a Punnett square allows one to easily see gamete types and frequencies, as well as the genotypes and frequencies of progeny formed by random gamete fusion. The genotype frequencies inside a Punnett square are calculated using the multiplication rule: The probability of two independent events occurring simultaneously is the product of their individual probabilities. In this example, the genotype frequencies inside the square (¼) are the product of the gamete frequencies that led to their formation (½ x ½ = ¼).

You continue your genetic analysis by crossing the forked and pale mutant lines with each other. The leaves of the F1 are light green (intermediate between pale and wild-type leaves) and forked. The F2 has six phenotypic classes, as shown below. You designate the forked mutant allele as F (wild type = f+ ) and the pale mutant allele as p (wild type = P). Consider the alleles for leaf color first. Drag the labels to the targets in Group 1 to identify the genotype of each F2 class. Remember that p (the pale mutant allele) and P (the wild-type allele) are incompletely dominant to each other. Consider the alleles for leaf shape next. Drag the labels to the targets in Group 2 to identify the genotype of each F2 class. Remember that F (the forked mutant allele) is dominant to f + (the wild-type allele). Labels may be used once, more than once, or not at all. For help getting started, see the hints.

Alleles P and p are incompletely dominant to each other. Therefore, each genotype has a distinct phenotype. That is why you are able to assign definite genotypes for leaf color to each F2 plant. Allele F is dominant to f +. Therefore, you cannot be certain whether forked leaves are homozygous dominant or heterozygous. That is why you must assign the genotype F_ to the F2 plants in the top row. For the unforked leaves in the bottom row, you know they must be homozygous for the wild-type allele.

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.

Around the mid-1850s, Mendel crossed true-breeding purple-flowered pea plants with true-breeding white-flowered pea plants. The results of his research provided us with the basic principles of heredity. Drag the terms on the left to the appropriate blanks on the right to complete the sentences. Not all terms will be used.

Every gene is a sequence of DNA nucleotides at a specific position along a chromosome called a locus. Diploid cells have two sets of chromosomes, one set inherited from each parent, that form homologous pairs. The homologs of a chromosome pair contain the same genetic loci. Therefore, each genetic locus is represented twice in a diploid cell. Variations in inherited characteristics is due to the presence of , which are alternative versions of genes. A gene is a sequence of DNA nucleotides that contains hereditary information. There is a specific location on a chromosome where a gene can be found, which is known as the locus. In a diploid cell, there are two copies of each chromosome (one from each parent), and thus two copies of each genetic locus. Alternative versions of genes, with different sequences of nucleotides, are called alleles.

True or false? In diploid organisms, a dominant phenotype will only be expressed if the individual is homozygous dominant for that trait.

False A dominant phenotype is indeed expressed if the individual is homozygous dominant for that trait, but the dominant phenotype is also expressed if the individual is heterozygous for the trait. In fact, heterozygous expression is the definition of dominant.

When constructing a Punnett square, the symbols on the outside of the boxes represent _______, while those inside the boxes represent _______.

Gametes, progeny The Punnet square is representing all of the possible combinations of the gametes from each parent, with the progeny represented in the interior of each box.

Which of the following do you expect if an individual is heterozygous for the sickle-cell trait? See Concept 14.3 (Page)

He or she will show some symptoms of the disease. A heterozygote will produce both normal and abnormal hemoglobin.

Which of the following is true about a plant with the genotype AABbcc? See Concept 14.1 (Page)

It is homozygous at two loci. An organism having a pair of identical alleles at a given locus is said to be homozygous for that locus. Here, the plant is homozygous at loci A and C.

Folk singer Woody Guthrie died of Huntington's disease, an autosomal dominant disorder. Which statement below must be true? See Concept 14.4 (Page)

It is very likely that at least one of Woody Guthrie's parents also had the allele for Huntington's disease. Unless the disease is caused by a new mutation, which is quite rare, individuals with a dominant condition must have inherited the dominant allele from one of their parents. As it happens, Guthrie's mother also died of Huntington's disease.

You decide to designate the twist allele as FT to distinguish it from the forked allele F. Using the following allele symbols, identify the genotypes of the three F2 classes in Part C by dragging one label to each class. Labels may be used once, more than once, or not at all.

Like incomplete dominance, codominance produces a distinct phenotype for each of the three genotypes in a monohybrid cross. Unlike incomplete dominance, however, codominance results in both alleles fully expressing their phenotype in the heterozygote.

A BbGg x bbgg cross yields a phenotypic ratio of approximately 5 black eyes, green skin : 5 orange eyes, white skin : 1 black eyes, white skin : 1 orange eyes, green skin. Which of the following best explains these results?

Mendel's law of independent assortment is being violated. If the genes for eye color and skin color assorted independently, then the outcome of this cross would have been a 1:1:1:1 ratio.

The observed distribution of alleles into gametes is an illustration of _____.

Mendel's laws of segregation and independent assortment. The events seen here illustrate both the law of segregation and the law of independent assortment.

If an organism with the genotype AaBb produces gametes, what proportion of the gametes would be Bb?

None. Alleles of the same gene must separate during gamete formation; thus, the two B alleles would be distributed to different gametes.

In his breeding experiments, Mendel first crossed true-breeding plants to produce a second generation, which were then allowed to self-pollinate to generate the offspring. How do we name these three generations? See Concept 14.1 (Page)

P ... F1 ... F2 Mendel started with true-breeding P generation plants, and in a typical experiment generated two subsequent generations, called F1 and F2.

Mendel studied pea plants dihybrid for seed shape (round versus wrinkled) and seed color (yellow versus green). Recall that the round allele (R) is dominant to the wrinkled allele (r) and the yellow allele (Y) is dominant to the green allele (y).The table below shows the F1 progeny that result from selfing four different parent pea plants.Use the phenotypes of the F1 progeny to deduce the genotype and phenotype of each parent plant. Complete the table by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

Plant 1: green round, Rryy, green round, green wrinkledPlant 2: yellow round, RrYy, yellow round, yellow wrinkled, green round, gren wrinkledPlant 3: yellow round, RRYy, yellow round, green roundPlant 4: green wrinkled, rryy, green wrinkled The ability to deduce an organism's genotype from the phenotype(s) of its progeny is an important skill in solving genetics problems. In this example, the logic was simplified because the parent plants were selfed, and therefore only one parental genotype was involved.

Use the completed Punnett square in Part B to answer the questions below about the F2 generation. Drag the probabilities on the left to the blanks on the right to answer the questions. Terms can be used once, more than once, or not at all.

Question 1 simply asks about the probability of forming a yellow seed in the F2, which is 3/4. Question 2 introduces the concept of conditional probability. Although there are two genotypes in the F2 that will breed true (YY and yy), only one of them is yellow. Thus, the question adds a new condition: Among yellow seeds only, 1/3 are true-breeding. Questions 3 and 4 require a full evaluation of all eight possible groupings of three randomly selected F2 seeds. Each grouping has its own probability, shown in the table below. For question 3, there are seven possible groupings that have at least one yellow seed (shown in blue in the table); the sum of their individual probabilities yields the final answer (63/64). Note that you could have also calculated this answer by subtracting the probability of the one category that doesn't fit the criterion (Green, Green, Green) from the total probability (1 - 1/64 = 63/64). For question 4, there are three possible groupings that have one green and two yellow seeds (shown in bold in the table); the sum of their individual probabilities yields the final answer (27/64). Possible grouping Probability 1. Green, Green, Green 1/4×1/4×1/4=1/641/4×1/4×1/4= 1/642. Yellow, Green, Green 3/4×1/4×1/4=3/643/4×1/4×1/4= 3/643. Green, Yellow, Green 1/4×3/4×1/4=3/641/4×3/4×1/4= 3/644. Green, Green, Yellow 1/4×1/4×3/4=3/641/4×1/4×3/4= 3/645. Yellow, Yellow, Green 3/4×3/4×1/4=9/643/4×3/4×1/4= 9/646. Yellow, Green, Yellow 3/4×1/4×3/4=9/643/4×1/4×3/4= 9/647. Green, Yellow, Yellow 1/4×3/4×3/4=9/641/4×3/4×3/4= 9/648. Yellow, Yellow, Yellow 3/4×3/4×3/4=27/643/4×3/4×3/4= 27/64

For the cross in Part B, predict the frequencies of each of the phenotypes in the F1 progeny, and determine the genotype(s) present in each phenotypic class. Complete the diagram by dragging the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

The frequencies of the four phenotypic classes are determined by applying the multiplication rule.The green progeny, for example, make up ¼ of the total progeny because the probability of their formation depends on receiving a y allele from both parents (probability = ½ x ½ = ¼).Similarly, the wrinkled progeny make up ½ of the total progeny because the probability of their formation depends on receiving an r allele from both parents (probability = 1 x ½ = ½).Because the [wrinkled, green] progeny are simultaneously green and wrinkled, and these events are independent, apply the multiplication rule to obtain their frequency: ¼ x ½ = 1/8.

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). Drag the labels to the correct locations on the Punnett squares. (G is the symbol for the green-pod allele and g is the symbol for the yellow-pod allele.) You can use a label once, more than once, or not at all.

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.

In pea plants, the purple allele is sufficient for making purple flowers, even if one of the homologous chromosomes carries the white allele. Which of the following statements are true in this case?

The purple allele is the dominant allele The white allele is the recessive allele One purple allele results in enough purple pigment for the flowers to be purple Explanation: In flowering pea plants, one or two purple alleles will result in plants with purple flowers. This is because one purple allele leads to the synthesis of enough pigment to produce all purple flowers. Therefore, the purple allele is dominant, and the white allele is recessive. White flowers are only produced in a plant with two white alleles.

The result of the following cross indicates the orange eyes are _____ black eyes.

The result of this cross indicates that both parents are heterozygous for eye color.

You know that alleles are alternative versions of a gene. What makes alleles different from each other?

They have different sequences of DNA nucleotides. Variation in inherited characteristics occurs because alleles have different sequences of DNA nucleotides, resulting in different phenotypes, such as different colored flowers. The alleles at a single locus in a pair of homologous chromosomes may be the same, or they may be different.

Pedigree 2 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal dominant condition. Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.

You can deduce the genotype of an individual in a pedigree based on two types of information: the individual's phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children. For autosomal dominant conditions: 1.Unaffected individuals are homozygous for the recessive, wild-type allele. 2.Affected individuals with only one affected parent are heterozygous. 3.Affected individuals with any unaffected children are heterozygous. 4.Affected individuals with two affected parents may be homozygous dominant or heterozygous.

Pedigree 3 from Part A is shown below. Recall that this pedigree shows the inheritance of a rare, autosomal recessive condition.Note that individual II-3 has no family history of this rare condition. Fill in the genotypes for the indicated individuals in the pedigree by dragging the best label to the appropriate location. Labels can be used once, more than once, or not at all.

You can deduce the genotype of an individual in a pedigree based on two types of information: the individual's phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children. For autosomal recessive conditions: 1.Affected individuals are always homozygous recessive. 2.Unaffected children of an affected parent are always carriers (heterozygous). 3.Both parents of affected individuals must have at least one recessive allele. 4.If both parents are carriers, their unaffected children may be carriers or homozygous for the dominant, wild-type allele. 5.For rare conditions, you can assume that individuals marrying into a family do not carry the recessive allele if there is no evidence that they are carriers.

Height in humans generally shows a normal (bell-shaped) distribution. What type of inheritance most likely determines height? See Concept 14.3 (Page)

a combination of polygenic inheritance and environmental factors Several genes (polygenic inheritance) control height in humans, giving an overall normal distribution. Environmental factors such as nutrition smooth out the curve.

Normal hemoglobin is a tetramer, consisting of two molecules of B-globin and two molecules of a-globin; normal hemoglobin molecules do not associate with each other. In sickle-cell disease, the change in a single amino acid results in mutant hemoglobin tetramers, which associate with each other and assemble into large fibers. Based on this information alone, what can we conclude about the changes in the structure of sickle cell hemoglobin as compared to normal hemoglobin?

altered primary and quaternary structure Review: The primary structure of a protein is its sequence of amino acids, secondary structure is the result of hydrogen bonds between the repeating constituents of the polypeptide backbone, tertiary structure is the 3-d shape stabilized by interactions between side chains, and quaternary structure results when two or more polypeptides associate with each other. Normal hemoglobin tetramers do not associate with each other. The single amino acid change that causes sickle-cell disease alters the primary and quaternary structure of the protein, and as a result, the abnormal hemoglobin tetramers aggregate into chains. Not enough info is given to determine whether secondary and tertiary interactions are affected.

What is an allele? See Concept 14.1 (Page)

an alternative version of a gene A diploid organism carries two alleles for each autosomal gene. The two alleles are found at comparable locations (loci) on homologous chromosomes. The alleles may be identical or slightly different, but they affect the same genetic character.

The result of the following cross indicates that genotypically the offspring _____.

are all Rr. This cross indicates that the wild type head top allele is dominant to the flat head top allele, so all offspring are indeed Rr.

The pedigrees below show the inheritance of three separate, rare autosomal conditions in different families. For each pedigree, decide if the condition is better explained as recessive or dominant. Drag the correct label to the appropriate location. Labels can be used once, more than once, or not at all.

autosomal dominant condition, dominant condition, recessive condition. If an individual has a genetic condition that neither parent has, then that condition must be recessive. Dominant conditions require that every affected individual have at least one affected parent. In situations where the inheritance mode of a rare condition cannot be definitely determined, the most likely mode is the one that requires the fewest unrelated individuals to have the condition-causing allele.

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 sex-linkage, galactosemia must be an autosomal recessive trait.

In order to determine the genotype of a MendAlien with black eyes and green skin, you would cross this individual with a(n) _____ individual.

bbgg. This is an extension of a testcross for a single character.

What is the genotype of the parent with orange eyes and white skin? (Note: orange eyes are recessive.)

bbgg. This result of the cross indicates that both orange eyes and white skin are recessive.

A phenotypic ratio of 9:3:3:1 in the offspring of a cross indicates that _____.

both parents are heterozygous for both genes. Such a result indicates that the genes assort independently and that, for each gene, the alleles exhibit a dominant/recessive relationship.

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 lactose- and 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.

A tall, purple-flowered pea plant (TtPp) is allowed to self-pollinate. (The recessive alleles code for short plants and white flowers.) The phenotypic ratio of the resulting offspring is 9:3:3:1. What is the genotype of the plant whose phenotype appeared once out of every 16 offspring (the "1" in the 9:3:3:1 ratio)?

ttpp The smallest phenotypic group consists of the homozygous recessive plants, which in this case are short and white flowered. In the case of this problem-- 9 tall purple flowers TTPP 3 tall heterozygote white TTpp (Ttpp, TTpp, tTpp) 3 short heterozygote purple ttPP (ttPP, ttPp, ttpP) 1 short white flower ttpp

You decide to conduct a genetic analysis of these mutant lines by crossing each with a pure wild-type line. The numbers in the F2 indicate the number of progeny in each phenotypic class.

twist- The mutant allele is dominant to its corresponding wild-type allele forked- the mutant allele is dominant to its corresponding wild-type allele pale- The mutant allele is neither dominant nor completely recessive to its corresponding wild-type allele An allele is never intrinsically "dominant" or "recessive." Instead, these terms describe a relationship between two alleles. This relationship is evaluated by examining a heterozygote: The allele that determines the phenotype of the heterozygote is dominant to the other (recessive) allele. Some heterozygotes have a phenotype that is intermediate between the phenotypes of the two homozygotes. This situation is the result of incomplete dominance: Neither allele is completely dominant to the other.