genetics exam 2 quizzes & clicker questions

Réussis tes devoirs et examens dès maintenant avec Quizwiz!

Consider a recessive autosomal mutation a (wild-type is A) and a recessive X-linked mutation Xb (wild-type is X+). Consider the cross: AaX+Xb x AaX+Y What is the probability of a child having both mutant phenotypes? a. 1/16 b. 1/8 c. 1/4 d. 1/2 e. 0

a. 1/16

What is the probability that a child of III.1 and III.2 is affected? (use pedigree on chapter 7 quiz) a. 1/16 b. 1/2 c. 1/4 d. 1/64

a. 1/16

For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #2 will not be affected by the trait. A, B, and C

a. 100% b. 100%. If we assume that the trait is rare and that it is unlikely for the father of child 1 to be a carrier, then there will be a 100% chance that the child is unaffected. c. 100%.

how many pairs of autosomes do humans have? a. 22 b. 23 c. 46 d. 1

a. 22

In an experiment, you have a tall plant that produces round peas. Tall and round are dominant and the plant is heterozygous for both traits. The plant is then self-fertilized. What proportion of the offspring will be tall and have wrinkled peas? a. 3/16 b. 1/4 c. 1/16 d. 3/4

a. 3/16

For each of the following individuals, calculate the probability that their first son will be affected by the trait. a. III-3 b. III-5 c. III-9 d. III-10

a. 50% b. 25%. Her mother II-4 is heterozygous (and her father II-3 is unaffected), so she could have gotten either of her mother's X chromosomes. There is a 50% probability that she inherited the trait and a 50% probability that she will pass it on to her son. c. Assuming the trait is rare (so that II-6 is not a carrier) then the chance that III-9's son will be affected is 0% d. Again, assuming that the trait is rare, there is a 0% chance that III-10's son will be affected (remember that males pass on their Y-chromosome, not their X-chromosome to male offspring).

I just rolled an even number. What is the probability that it is a two?

-1/3 -(There are only 3 even numbers on the die. Since you already know it's an even number, the probability is out of 3 instead of 6.)

This pedigree shows the incidence of a rare genetic disorder in a family. Identify the mode of inheritance and assign genotypes as completely as possible to all individuals. (look at 2nd picture on camera role w/ white & black circles & squares)

-Autosomal dominant. -All individuals can be genotyped completely (no A_).

This question relates to Mendel's breeding experiments with peas. A true breeding plant producing smooth peas was crossed with a true-breeding plant producing wrinkled peas (cross 1). The offspring of cross 1 all produced smooth peas. The offspring of cross 1 were then crossed with the wrinkled-pea parent (cross 2). Which of the statements about the cross outcomes are true? Please select all that apply. -Half the offspring of cross 2 had smooth peas -All the offspring of cross 2 had smooth peas -All the offspring of cross 1 had the same genotype -The offspring of cross 2 had a 3:1 smooth:wrinkled ratio -The wrinkled allele is recessive

-Half the offspring of cross 2 had smooth peas -All the offspring of cross 1 had the same genotype -The wrinkled allele is recessive

A calico cat mates with a black male cat. What phenotypic ratios are expected among her offspring? (Both color and sex of the offspring are relevant phenotypes for this question.) (question 16)

-Males: 50% black, 50% orange -Females: 50% black, 50% calico

A male mouse with a pale yellow coat called cream is mated to a female mouse with a wild-type gray- brown coat. All of the F1 mice, both males and females have the wild-type gray-brown color. One of these F1 female mice is mated to cream-colored male a. What are the expected results if cream is X-linked and recessive?

-Males: 50% cream, 50% wild-type gray-brown -Females: 50% cream, 50% wild-type gray-brown

-Males: 50% missing cross veins -Females: 100% normal

A cell has the genotype AaBb. Each of the following is a list of the gentoypes of four cells that might arise from this cell undergoing meiosis. Which set of genotypes could only arise if crossing-over occurs (and is otherwise impossible)? a. AB, ab, AB, ab b. Ab, aB, Ab, aB c. AB, ab, Ab, aB

AB, ab, Ab, aB

Which of the following statements about meiosis and mitosis are true? Please select all that apply. -Meiosis involves two rounds of DNA replication -Mitosis involves one round of DNA replication -Meiosis involves four rounds of chromosome/chromatid separation -Mitosis involves one round of chromosome/chromatid separation

-Mitosis involves one round of DNA replication --Mitosis involves one round of chromosome/chromatid separation

Which of the statements about non-disjunction are true? Please select all that apply. -Non-disjunction at meiosis II leads to two normal gametes and two abnormal gametes. -Non-disjunction at meiosis II leads to three abnormal gametes and one normal gamete. -Non-disjunction at meiosis I leads to three normal gametes and one abnormal gamete. -Non-disjunction at meiosis I leads to four abnormal gametes. -Non-disjunction at meiosis II has a more severe impact than at meiosis I

-Non-disjunction at meiosis II leads to two normal gametes and two abnormal gametes. -Non-disjunction at meiosis I leads to four abnormal gametes.

2. Mitosis and meiosis are both processes of eukaryotic cell division, with the same names used to describe the different stages. 1. What are the significant cellular and chromosomal events that occur during each stage of mitosis? (prophase-->metaphase-->anaphase-->telophase/cytokinesis)

-Prophase: Chromosomes condense and spindles begin to form -Metaphase: Microtubules attached to the kinetochores begin to pull at the chromosomes creating tension that lines up the chromosomes along the center of the cell. -Anaphase: Cohesin releases and the sister chromatids are pulled towards opposite poles of the cell. -Telophase/Cytokinesis: The cell cleaves down the middle to form two new cells as the nuclear envelope starts to re-form in each cell and the chromosomes decondense.

While cleaning out the storage room in the basement of a Biology Department, you encounter a box of microscope slides. Most of the labels are illegible, but you can read that the slides are taken from ovaries some animal that has only two pairs of chromosomes (n=2). The slides appear to be stained with some fluorescent dye that allows you to see the chromosomes. Three of the slides are shown in Figure Q6-1. What stage of meiosis is shown in each slide and how can you tell?

-Slide A is the beginning of Meiosis II because there is only one chromosome of each type per cell, and not two like in Meiosis I. -Slide B is in the middle of Meiosis I, either at the end of Prophase I or the beginning of Metaphase I. The multiple copies of chromosomes indicates that this slide must be in Meiosis I, and the synapsing of the chromosomes means that this stage is before the chromosomes are pulled apart, but after the chromatin had condensed. -Slide C is close to the end of Meiosis II. This cell only has one chromatid of each type, indicating that this is after each chromosome has been split, but before the decondensation of the DNA.

Some organisms, e.g., birds, have a ZW/ZZ system of sex determination rather than the XY/XX system that is familiar from mammals. Which of the following characteristics are true of the ZW/ZZ system? Please select all that apply. -The sex-determining locus is on the W chromosome -Phenotypes resulting from Z-linked genes show up more frequently in males than in females -The Z chromosome in birds and the X chromosome in humans are genetically similar -The ZW/ZZ and XY/XX systems provide evidence of independent evolution of sex determination.

-The sex-determining locus is on the W chromosome -The ZW/ZZ and XY/XX systems provide evidence of independent evolution of sex determination. (The sex-determining locus is on the W chromosome. However, females are heterogametic (ZW), so Z-linked phenotypes show up more frequently in females than males. The Z chromosome most closely resembles (and shares an evolutionary history with) chromosome 9 in humans, providing evidence of independent evolution of these sex-determining systems.)

Individuals III.1 and III.2 are thinking about having children. A rare condition runs in both families and they want to know the chances of their child having it. 1. Identify the mode of inheritance: is the trait dominant or recessive? 2. Assign genotypes as completely as possible to all individuals using A for the dominant allele and a for the recessive allele. Use the underscore "_" to indicate an unknown allele. 3. Calculate the probability that their child will have the condition. Strategy: Work backwards. What is the series of unfortunate events that have to occur for the child to have the condition? What genotypes do III.1 and III.2 need to have for it to be possible? What has to happen for them to have those genotypes? Finally, multiply all probabilities. (look at picture on camera role w/ white & black circles & squares)

-This is a recessive trait. -For the child of III.1 and III.2 has the trait... --II.6 needs to be Aa. There is a 2/3 chance of this happening. --III.2 needs to be Aa. There is a ½ chance of this happening assuming II.6 is Aa. --The child needs to be aa. There is a ¼ chance of this happening assuming III.2 is Aa. -All three things need to happen for the child to have the trait. --2/3 x ½ x ¼ = 1/12

Consider the following cross: AABbCc x aaBbcc. What is the probability that an offspring will exhibit the dominant phenotype for all three genes?

-To exhibit the dominant phenotype for a gene, the genotype must be homozygous dominant or heterozygous. You can consider the probability for obtaining this result for each gene individually. --AA x aa -->A_ 100% of the time (4/4) --Bb x Bb -->B_ 75% of the time (3/4) --Cc x cc -->C_ 50% of the time (2/4) --4/4 x 3/4 x 2/4 = 3/8

non-disjunction during meiosis I is a failure of ________ ________ to separate evenly into the 2 daughter cells. non-disjunction during meiosis II is a failure of _______ _______ to separate evenly into the 2 daughter cells.

-homologous chromosomes -sister chromatids

__________ ensures that all daughter cells get a copy of each gene. __________ ___________ creates random combinations of alleles on non-homologous chromosomes. _________ ________ creates random combinations of alleles on homologous chromosomes.

-segregation -independent assortment -crossing over

if there is an inactivating mutation in the Xist gene what would happen?

-you would have gene dosage problems -there wouldn't be inactivation of the next chromosome

Roll of a six-sided die Probability of rolling a two Probability of rolling a four Probability of rolling a six

ALL 1/6

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation b. How many units of "square factor" activity would be encoded by the null mutation?

0 (zero)

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation c. How many units of "square factor" activity will mutant homozygotes have?

0 + 0=0 unites

Probability of the previous scenario NOT happening? (Probability of rolling an even number AND then rolling an even number (two rolls))

1 - (probability of it happening) = 1 - 1/4 = 3/4

Probability of rolling an even number AND then rolling an even number (two rolls)

1/2 x 1/2 = 1/4

Probability of rolling a two OR a four OR a six

1/6 + 1/6 + 1/6 = 3/6 = 1/2

Probability of rolling a six AND then rolling a six again (two rolls)

1/6 x 1/6 = 1/36

A gene known as lon-2 is X-linked in C. elegans; mutations in lon-2 result in worms that are unusually long. A lon-2 mutant male was mated to a wild-type hermaphrodite. b. What will be the cross-progeny of this mating if lon-2 is dominant?

100% of the females will be long and 100% of the males will be normal length.

A gene known as lon-2 is X-linked in C. elegans; mutations in lon-2 result in worms that are unusually long. A lon-2 mutant male was mated to a wild-type hermaphrodite. c. In fact, lon-2 is recessive. One of the F1 hermaphrodites is mated with wild-type males. What are the expected progeny among the offspring of this mating, and in what proportions will they be found.

100% of the females will be normal length and 50% of the males will be long.

A gene known as lon-2 is X-linked in C. elegans; mutations in lon-2 result in worms that are unusually long. A lon-2 mutant male was mated to a wild-type hermaphrodite. a. What will be the cross-progeny of this mating if lon-2 is recessive?

100% of the progeny (male and female) will be normal length. (check menly chapter 7 question 14)

Meiosis in a diploid cell produces ________ cells that are _________. -(I am talking about a cell that has gone through both meiosis I and meiosis II.)

4, haploid

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation e. How many units of "square factor" activity will wild-type homozygotes have?

40 + 40 = 80 units

Factor VIII and Factor IX are two X-linked genes in humans involved in blood clotting, and mutations in these genes result in Hemophilia A or Hemophilia B. Originally it was not known which form affected the royal families, but the inheritance pattern is the same in either case. Use Figure 7-6 to answer the following questions. c. Irene's sister Alice had a hemophiliac son, Viscount Tremation, as well as a daughter Mary, and a son who died as a newborn. What is the probability that this son was hemophiliac? (This is not known.)

50%

A male mouse with a pale yellow coat called cream is mated to a female mouse with a wild-type gray- brown coat. All of the F1 mice, both males and females have the wild-type gray-brown color. One of these F1 female mice is mated to cream-colored male b. What are the expected results if cream is autosomal and recessive?

50% cream, 50% wild-type gray-brown

A hermaphrodite produces gametes with X chromosomes in them, so when a hermaphrodite self-fertilizes, the offspring are all expected to have two X chromosomes. Therefore, none are expected to be male.

g. Synaptonemal complex

A ladder-like structure that forms between two homologues as they synapse and has an evolutionarily conserved width.

A mutation that is a complete loss-of-function; it makes no functional protein. Either no proteins is made at all (CRM or promoter mutation) OR a defect in the amino acid sequence makes the protein completely non-functional (coding sequence mutation).

(Looking back and ahead, and progressively challenging). In C. elegans, males are X0 while hermaphrodites are XX. Hermaphrodites have the somatic sex determination of females but they undergo both spermatogenesis and oogenesis, so they can use their own sperm to fertilize their ovum or they can mate with a male. Hermaphrodites cannot mate with other hermaphrodites. e. (Looking ahead). A wild-type male is mated to a hermaphrodite that is homozygous for a gene called dpy-11 on one of the autosomes, and heterozygous on the X chromosome for unc-3 and lon-2; that is, it had unc-3 on one X chromosome and lon-2 on the other X chromosome. The mutations in dpy-11, unc-3, and lon-2 are recessive. What are the expected cross-progeny offspring of this cross? Hint: This may help you answer Part D above.

All of the progeny will be wild-type for dpy-11 (they will be heterozygotes for this allele). This is the recessive allele for an autosomal gene that is used to distinguish self-progeny from cross-progeny. All of the females will also be wild-type for the Lon and Unc phenotypes (50% will inherit one copy of the mutant unc-3 allele and 50% will inherit one copy of the mutant lon-2 allele from the hermaphrodite, but all will inherit the wild-type X chromosome from the male). 50% of the males will inherit the mutant unc-3 allele from the hermaphrodite and be Unc. The other 50% will inherit the mutant lon-2 allele from the hermaphrodite and be Lon.

h. SPO11

An evolutionarily conserved enzyme among eukaryotes that creates double-stranded breaks for crossing over.

Shown below are three pedigrees for rare genetic traits in humans. Affected individuals are shown in black, while unaffected individuals are shown in white. Although each pedigree has too few members to be absolutely confident about how the trait is inherited in each case, certain modes of inheritance might be likely and some might be impossible. For each of the pedigrees below, which modes of inheritance—that is, X-linked recessive, X-linked dominant, Y-linked, autosomal recessive, and autosomal dominant—can be ruled out? Which is the most likely? -picture of pedigrees or on page 303 in textbook A.....

Autosomal dominant most likely. This trait is not X-linked, because a son inherits it from the father, and it is not Y-linked, because a daughter inherits it from the father. This trait could be autosomal recessive if the mother is a carrier; however, if we assume the trait is a rare trait, it is most likely that the mother is not a carrier and the trait is autosomal dominant.

Autosomal recessive most likely. The trait is not dominant, as there are affected children of unaffected parents. The trait cannot be Y-linked, because a daughter is affected. The trait cannot be X-linked because it is a recessive trait and an affected daughter must have a father with the trait.

f. Univalent and bivalent

Bivalent is a pair of homologous chromosomes or a tetrad. Univalent is an unpaired chromosome during meiosis.

A couple has a child with trisomy-21, that is, with three copies of chromosome 21 rather than two. Geneticists analyzed chromosome 21 of both parents and the child in order to understand the origin of this non-disjunction event. Some of the results are shown in Figure QB-4. b. The bottom part of the figure shows two different SNPs at different loci on chromosome 21 in each parent and the child. The line is a chromosome from each parent, with the dashed line indicating that the two loci are far apart on chromosome 21. The chromosomes are arranged in the same order as on the gel, so that the top chromosome shown on the gel (in the mother and the child) has the A at the first locus and the C at the second locus. What does this sequence information tell you about the "cause" of the non-disjunction event?

Crossing over did not occur between the mother's chromosomes.

Crossveinless is X-linked and recessive.

As discussed in the chapter, white eyes in Drosophila is an X-linked trait. Female flies which are heterozygous for white eyes have red eyes, and no heterozygotes have patches of red and white coloration. In fact, of the hundreds of X-linked recessive mutations in Drosophila, none of them shows a mosaic or chimeric phenotype in heterozygotes. Why is mosaicism or chimerism seen for X-linked traits in mammals but not in Drosophila?

Drosophila use a different dosage compensation system that does not involve inactivation of an X chromosome, thus there is no potential for a mosaic or chimeric phenotype. In Drosophila dosage compensation occurs through hyper transcription of x-linked genes in males.

Two people are both heterozygous (Aa) for a recessive trait. -If the child (F1) does NOT exhibit the trait, AND has a child (F2) with a homozygous recessive person (aa), what is the probability that their child (F2) will have the trait? a. 3/4 C. 2/3 D. 1/2 E. 1/3 F. 1/4

E. 1/3

Figure 6-24 is one of the most important unifying figures in the book. Explain this figure to a classmate.

Figure 6-24 shows how DNA is replicated through meiosis and inherited. Two homologous chromosomes with different SNPs (A/T vs G/C) are both replicated during meiosis, creating double copies of both chromosomes. The gametes created from these chromosomes will each carry one chromosome with one of the SNPs (two gametes will have the A/T SNP and two will have the G/C SNP). Simplified, half the gametes will each SNP. This shows the laws of mendelian inheritance and how genes are passed down from parent to child, through random assortment of chromosomes into gametes and then fertilization of a gamete from a gamete from the other parent.

(Challenging) As was shown with calico cats, female mammals that are heterozygous for an X-linked gene have some cells that express one allele and some that express the other allele. However, this difference between the two expression patterns does not make a difference in the phenotypes for most X-linked traits. In other words, calico cats have a very striking phenotype, but this pattern of cell differences is not seen for most other X-linked traits. What are some of the possible explanations for the lack of difference for most X-linked traits?

For some traits, the gene products are exported from the cell. For example, in a carrier for hemophilia, as long as the blood clotting protein is expressed in some cells, the organism will not show the disease. In addition, skin cells that produce color (the melanocytes) so not migrate or mix very much during development after they form so that the cells with the same inactivated X chromosome are clustered near each other in the skin. In some cases, the functions of genes on the X chromosome are duplicated in other genes on other chromosomes.

Individuals who are aneuploid for one of the sex chromosomes usually survive longer and have fewer health consequences than individuals who are aneuploid for one of the autosomes. Explain why this is true.

For species that have XX/XY sex determination systems, there is dosage compensation - otherwise the female would make too much product from genes carried on the X chromosome or the male would not make enough. In female mammals, one of the X chromosomes becomes inactivated and turns into a Barr body. This means that cells in the female are essentially operating with one copy of the X chromosome each, which is why individuals who are XO can survive well, although they are sterile. Individuals with too many X chromosomes (XXX or XXY) can inactivate as many as necessary to have one active X chromosome in each cell, so they also do not suffer from too much or too little gene product. Dosage compensation doesn't occur with autosomes. If there is only one of a particular autosome, then there will not be enough gene product for genes carried on that chromosome and there is no mechanism to inactivate one chromosome if there are three copies of an autosome. Thus aneuploidy in an autosome is likely to create health problems.

Question 7.26 on page 305 a. Why were the genes sequenced from the boy rather than one of his sisters?

Genes were sequenced from the boy because the boy was known to have hemophilia, an X-linked disease, so must have had a mutant form of one of the genes involved in blood clotting. The daughters have two X chromosomes, and may or may not have been carrying the mutant hemophilia allele. Therefore, to confirm the identity of the remains, only the boy's DNA needed to be analyzed.

(Looking back and ahead, and progressively challenging). In C. elegans, males are X0 while hermaphrodites are XX. Hermaphrodites have the somatic sex determination of females but they undergo both spermatogenesis and oogenesis, so they can use their own sperm to fertilize their ovum or they can mate with a male. Hermaphrodites cannot mate with other hermaphrodites. I. (Very challenging). In order to investigate which meiotic division was affected by each of the him genes, Hodgkin et al. (1979) did the following experiment. (This has been slightly modified.) A wild-type male was mated to a hermaphrodite that was homozygous for the him mutation, homozygous for dpy-11 on one of the autosomes, and heterozygous on the X chromosome for unc-3 and lon-2; that is, it had unc-3 on one X chromosome and lon-2 on the other X chromosome. In other words, it is the same cross as used in Part E, but the hermaphrodite is also homozygous for a him mutation. The offspring included a few hermaphrodites that were Unc but not Dpy or Lon, and Lon but not Dpy or Unc. Explain this result.

Half of the males' sperm have no X chromosome. If non-disjunction occurred in the mother during meiosis I, the disomic ovum would have two X chromosomes with the same allele—either both with lon-2 or both with unc-3. When these disomic ova are fertilized by a sperm with no X chromosome, the offspring will be hermaphrodites (since they have two X chromosomes) but Lon or Unc, having gotten both X chromosomes from the mother.

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation i. What is the term that can be applied to describe the genetic behavior of this system?

Haplo-insufficiency

Define the difference between heterogametic and homogametic sexes. Give an example of a species in which the male is heterogametic, and one in which the male is homogametic.

Heterogametic sexes have two different sex chromosomes while homogametic sexes have duplicates of the same sex chromosome. In humans males are heterogametic (XY) and females are homogametic (XX). In birds, males are homogametic (ZZ), while females are heterogametic (ZW).

(Looking back and ahead, and progressively challenging). In C. elegans, males are X0 while hermaphrodites are XX. Hermaphrodites have the somatic sex determination of females but they undergo both spermatogenesis and oogenesis, so they can use their own sperm to fertilize their ovum or they can mate with a male. Hermaphrodites cannot mate with other hermaphrodites. g. Mutations in genes that affect meiosis in C. elegans can be readily identified, and have a distinctive mutant phenotype. The mutant phenotype is called Him for "high incidence of male self-progeny", so the genes are named him-1, him-2, etc. Mutations in many him genes have greatly reduced fertility and lay many eggs that do not hatch. However, among the eggs that hatch, a relatively high proportion are X0 males. Explain this result, both the Him phenotype and the reduced fertility.

Him mutations could cause increased rates of non-disjunction in hermaphrodites. When non-disjunction occurs for the X chromosome, a male may be produced. Therefore, an overall increased rate of non-disjunction increases the likelihood that males will be produced. However, an increased rate of non-disjunction also increases the likelihood of non-disjunction in the autosomes, which results in gametes with too many or too few autosomes, which will not produce live offspring.

Why are the terms "X-linked" and "sex-linked" frequently used interchangeably? In at least one widely used domesticated species, the terms are not synonyms. What species is this and why are these terms not synonyms in that species?

In animals that use the XX/XY system of sex determination, sex-linked implies X-linked because the Y chromosome has almost no genes on it, while the X-chromosome carries many genes that are not related to sex determination. X-linked does not imply sex-linked in animals that use other systems of sex determination such as ZZ/ZW, found in chickens.

In summer squash, the fruit may be yellow or white and disk-shaped or sphere shaped. Disk-shaped is dominant over sphere-shaped, and white is dominant over yellow. A plant with white, disk-shaped fruit is crossed to a plant with white, sphere-shaped fruit. The cross yields 32 white/disk, 29 white/sphere, 9 yellow/disk, and 10 yellow/sphere plants. a. Give the genotypes for the P and F1 individuals. b. Determine the proportion (fraction) of each of the F1 phenotypes. In other words, what percent will be white/disk, etc.

I will assign the allele names thus: disk-shaped (M), sphere-shaped (m); white (H), yellow (h). -Parents (note we we're able to identify the exact genotype; no "_" necessary) --White/disk = HhMm --White/sphere = Hhmm -F1 --White/disk = H_Mm 3/8 of the progeny --White/sphere = H_mm 3/8 of the progeny --Yellow/dish = hhMm 1/8 of the progeny --Yellow/sphere = hhmm 1/8 of the progeny

Why is inheritance of X-linked recessive traits sometimes called crisscross inheritance?

If a female that expresses a an X-linked recessive trait is mated to a male that does not an express an X-linked recessive trait then the male offspring will express the trait and the females will not, as shown in this Punnett square. (picture of punnet square or question 3 for menley chapter 7)

Dosage compensation occurs by different methods in different organisms. Suppose that you found a mutation in which dosage compensation did not occur. What is the expected phenotype of a dosage compensation mutant a. in C. elegans?

In C. elegans, males are X0 and hermaphrodites are XX. Hermaphrodites carry out dosage compensation by making less of the product on each chromosome. A dosage compensation mutation would have no effect on a male, but a mutant female would make twice the amount of X chromosome product needed and could show a mutant phenotype. It was in fact the presence of such mutants that led to the recognition of the mechanism of dosage compensation in C. elegans.

In D. melanogaster, females are XX and males are XY. XY males compensate for the lack of a second X chromosome by upregulating transcription of genes on the X chromosomes. If mutant males couldn't upregulate the amount of X chromosome transcript they made, they would probably die. Mutant females would not be affected, because they do not perform dosage compensation. Thus, the mutants would be male-specific lethals. It was in fact the presence of such mutants that led to the recognition of the mechanism of dosage compensation in Drosophila.

In mice, females are XX and males are XY. Female mice perform dosage compensation via X inactivation. Therefore, males would be unaffected by a dosage compensation mutation, but females would make twice the necessary amount of product from the X chromosome, which could be harmful.

Why are male calico cats both rare and usually sterile?

In order for a cat to be calico it must be heterozygous for the B allele, which is on the X chromosome. Therefore, the cat must have two X chromosomes. In order for a male to be heterozygous for an X-linked trait, he has to have the genotype XXY. These males are not common because they are the result of non-disjunction of the X-chromosome. Males of this genotype will be sterile because they have 3 sex chromosomes.

Some parakeets and other birds of the parrot family have a phenotype known as lutino. Lutino is a much prized beautiful coloration pattern in which the feathers on the body and head are yellow, with silver patches on the cheeks, and light yellow tail feathers. (These birds are pretty enough that you might want to look at a picture of them on-line.) The birds also have red or pink irises in the eyes rather than the standard black irises. b. One of the male offspring in part A is mated to a lutino female. What are the expected outcomes of this mating?

In this cross the male was ZL/Zl and female was Zl/W. -In this cross, 50% of all offspring will be lutino and 50% will be green.

During the Korean War, soldiers (males) and nurses (females) were given antimalarial drugs such as chloroquine. Some soldiers developed a severe and life-threatening hemolytic anemia in which their red blood cells lysed; other soldiers were completely unaffected. Subsequent genetic analysis showed that the hemolytic response was due to a particular allele of the X-linked gene encoding the enzyme G6PD. a. Explain why the response of males to this drug was biphasic—some with a severe reaction and others with no reaction, with no intermediate phenotypes

Males are hemizygous for all alleles on the X chromosome. Therefore there they cannot have a heterozygous phenotype--they either have one or another allele for the enzyme G6PD. The males with one allele would have a strong reaction, while the males with the other allele would have no reaction.

Males could arise in self-progeny due to non-disjunction in the X chromosome. This creates gametes with two X chromosomes and gametes with no X chromosomes. When a normal gamete with one X chromosome combines with a gamete with no X chromosome, the offspring will be male.

What are the similarities and differences between the events that occur during each stage of mitosis and the corresponding stage of meiosis II?

Meiosis II is very similar to mitosis, except that the meiosis II occurs with half of the chromosomes, resulting in cells that are haploid instead of diploid.

Could this inheritance pattern occur if the trait were X-linked but dominant?

No. I-2, II-4, II-7, and III-2 are all carriers of the trait. If the trait were dominant then it would be expressed in all of those individuals.

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation h. Are functional wild-type alleles always dominant? Explain.

No. This is an example where the wild type phenotype appears only in the homozygous wild-type.

Non-disjunction in him-3 must occur for all chromosomes since the mutant is nearly sterile. If non-disjunction occurred from the autosome on which the dpy gene resides, a few of the offspring will be disomic for this autosome. In the actual experiment, he also used males that were mutant for him-3 so they were making sperm that were nullisomic for this autosome. Humans who have two copies of a chromosome but who are disomic for a maternal chromosome and nullisomic for a paternal chromosome (or vice versa) have rarely but occasionally been observed.

(Looking ahead) Caenorhabditis elegans has two sexes, males and hermaphrodites. Hermaphrodites are essentially females that make sperm for a few hours, and then shut off spermatogenesis for the remainder of their life. Thus, an hermaphrodite can either reproduce by self-fertilization of its own sperm with its own ova, or by cross-fertilization with a male. Hermaphrodites have five pairs of autosomes and a pair of X chromosomes (XX). As described in Chapter 7, males have five pairs of autosomes and a single X chromosome (X0); there is no Y chromosome in nematodes, so males have 11 chromosomes rather than 12. This mode of reproduction and sex determination has made meiotic mutants particularly easy to identify in C. elegans. a. Most of the offspring of self-fertilization by a hermaphrodite are also hermaphrodites. However, approximately one in 500 offspring of self-fertilization is a male. Explain these results.

Non-disjunction of the X chromosomes during meiosis caused one egg to not have an X chromosome. When this egg joins with a sperm that has an X chromosome, the fertilized egg will only have one X chromosome, making the fertilized egg male.

Explain why non-disjunction was considered proof of the chromosome theory of heredity.

Non-disjunction was considered the final proof of the chromosomal theory of heredity because non-disjunction during meiosis can be used to explain the unusual phenotypes that are observed as the result of having extra copies of sex chromosomes. This shows that the process of chromosome segregation, or failure thereof, during meiosis, can explain why traits are inherited in Mendelian patterns.

(Looking ahead) Caenorhabditis elegans has two sexes, males and hermaphrodites. Hermaphrodites are essentially females that make sperm for a few hours, and then shut off spermatogenesis for the remainder of their life. Thus, an hermaphrodite can either reproduce by self-fertilization of its own sperm with its own ova, or by cross-fertilization with a male. Hermaphrodites have five pairs of autosomes and a pair of X chromosomes (XX). As described in Chapter 7, males have five pairs of autosomes and a single X chromosome (X0); there is no Y chromosome in nematodes, so males have 11 chromosomes rather than 12. This mode of reproduction and sex determination has made meiotic mutants particularly easy to identify in C. elegans. b. Mutations that result in non-disjunction in the C. elegans hermaphrodite, are referred to as Him mutations. "Him" is the phenotypic designation for "High incidence of male progeny". Rather than having one male in 500 offspring, Him mutants have 3% or more male offspring, with the exact percentage depending on the gene and the mutant allele. Explain why mutations that affect meiosis have a Him phenotype.

Normal segregation of the chromosomes is the outcome after all of the steps of meiosis, so mutations that affect meiosis are likely to cause non-disjunction. If the chromosomes are not being segregated properly this is likely to create gametes that are missing an X chromosome, which will produce X0 male offspring if fertilized.

Factor VIII and Factor IX are two X-linked genes in humans involved in blood clotting, and mutations in these genes result in Hemophilia A or Hemophilia B. Originally it was not known which form affected the royal families, but the inheritance pattern is the same in either case. Use Figure 7-6 to answer the following questions. b. Note that Irene married her first cousin Henry of Prussia. Normally the marriage of first cousins results in a higher occurrence of recessive disorders. However, that is irrelevant to the explanation of the occurrence of hemophilia in this family. Why is this information on the family relationship not important?

Normally, marriage between cousins increases the chances of recessive disorders showing up in the offspring because being related increases the chance that both will be carriers for the same recessive genetic diseases. This increases the chance that their children will be homozygous for a recessive disease. However, men who have the X-linked gene for hemophilia will have the disease. This means that a healthy male does not have a recessive copy of the gene. Inheritance of the disorder is completely dependent on the mother. It doesn't matter how closely related the male is, just whether or not he has hemophilia.

(Looking ahead) Caenorhabditis elegans has two sexes, males and hermaphrodites. Hermaphrodites are essentially females that make sperm for a few hours, and then shut off spermatogenesis for the remainder of their life. Thus, an hermaphrodite can either reproduce by self-fertilization of its own sperm with its own ova, or by cross-fertilization with a male. Hermaphrodites have five pairs of autosomes and a pair of X chromosomes (XX). As described in Chapter 7, males have five pairs of autosomes and a single X chromosome (X0); there is no Y chromosome in nematodes, so males have 11 chromosomes rather than 12. This mode of reproduction and sex determination has made meiotic mutants particularly easy to identify in C. elegans. d. Although the Him mutants have a high incidence of male progeny, most of them also lay many eggs that do not hatch. Explain why most Him mutations have so many non-viable offspring.

Not only do these mutants have gametes that are missing an X chromosome, many of them will have non-disjunction of the other chromosomes as well. Monosomy for the X chromosome affects sex determination, while monosomy (and possibly trisomy) for the other chromosomes is likely to be lethal.

You identify a new species of fish, which clearly has two separate sexes, male and female. However, all of the chromosome pairs appear identical in the two sexes so it is not apparent if sex determination depends on an X/Y system, a Z/W system, or a single mating type locus. All of these systems of sex determination are known to occur in some species of fish, so any of them could be at work in this new species. You have found a locus that might be able to distinguish these possibilities. Using PCR, you amplify this sequence from a male and a female, and from two of their offspring, one male and one female. You separate the PCR products on an agarose gel with the results shown below. What is the most likely form of sex determination in this fish? Briefly explain your reasoning. -Page 304

Notice that this is a DNA sequence rather than a gene so it may or may not be coding for a protein. The male parent is heterozygous while the female parent is either homozygous (if she has two sex chromosomes) or hemizygous (if she has a single sex chromosome). The male offspring is not informative since no matter what system is involved, it will get one chromosome from the father and one from the mother, which is the result seen. The key is the female offspring, which has only one band at the same size as the one from her father; she has apparently not inherited the locus from her mother at all. These results are most easily explained if females are the heterogametic sex.

A Drosophila female that is missing cross veins on the wing is mated to a male with normal wings. The female F1 offspring have normal wings, but the male F1 offspring have crossveinless wings. Two F2 females are chosen, and each is put into a vial and mated with a male with crossveinless wings. One of these matings produces only wild-type flies. The other mating yields both crossveinless flies and normal flies. Explain these results. What are the expected proportions of crossveinless and normal flies in this vial?

One of the F2 females is 𝑋𝐴𝑋𝐴and the other F2 female is 𝑋𝐴𝑋𝑎.For the heterozygous female, 50% of the offspring (both the males and the females) will be missing cross veins.

10. C. elegans has six pairs of chromosomes. Mutants in the gene him-5 in C. elegans affect some process in pairing and synapsis prophase I, but precisely what process is affected is not clear. Different oocytes from a him-5 mutant hermaphrodite can have different karyotypes. How can the following karyotypes at the end of prophase I be explained? c. Five bivalents and two univalents

One set of homologous chromosomes was unable to synapse with each other during bouquet formation, while the other five paired and synapsed normally.

(Challenging) The genetic analysis of C. elegans as a model organism was begun in early 1960s. Mutations were isolated and numbered in the order in which they could be assigned to a location on a chromosome. Nearly all of the mutants with the lowest numbers are X-linked. Suggest an explanation for this.

Phenotypes resulting from mutations in genes found on the X chromosome would be visible in males, which are hemizygous, so would be the first mutations discovered and assigned numbers.

(Looking ahead) Caenorhabditis elegans has two sexes, males and hermaphrodites. Hermaphrodites are essentially females that make sperm for a few hours, and then shut off spermatogenesis for the remainder of their life. Thus, an hermaphrodite can either reproduce by self-fertilization of its own sperm with its own ova, or by cross-fertilization with a male. Hermaphrodites have five pairs of autosomes and a pair of X chromosomes (XX). As described in Chapter 7, males have five pairs of autosomes and a single X chromosome (X0); there is no Y chromosome in nematodes, so males have 11 chromosomes rather than 12. This mode of reproduction and sex determination has made meiotic mutants particularly easy to identify in C. elegans. c. What are the some of the biological processes that might be encoded by a him gene?

Possible answers could be synapsis, pairing, crossing over, structures of the synaptonemal complex, checkpoints, or any of the steps in meiosis.

C. elegans has six pairs of chromosomes. Mutants in the gene him-5 in C. elegans affect some process in pairing and synapsis prophase I, but precisely what process is affected is not clear. Different oocytes from a him-5 mutant hermaphrodite can have different karyotypes. How can the following karyotypes at the end of prophase I be explained? a. Six bivalents and no univalents.

Prophase I has proceeded correctly. All six chromosomes have been duplicated are paired with their homologous chromosome.

What are the similarities and differences between the events that occur during each stage of mitosis and the corresponding stage of meiosis I?

Prophase I includes bouquet formation and synapsis of homologues, unlike mitosis prophase. The synapsed chromosomes then cross over and recombine at the end of prophase I. During metaphase I, the synapsed chromosomes line up along the metaphase plate, with the tension from the spindle poles holding them in place. The main difference between metaphase in meiosis I and metaphase in mitosis is that homologous pairs line up along the center of the cell, rather than sister chromatids. In anaphase, the homologous chromosomes split apart from one another, instead of the sister chromatids splitting apart in Mitosis. The separate cells that are formed during telophase I will be haploid, but will have two sister chromatids for each homologue--unlike a cell resulting from mitosis, which will be diploid and have one copy of each chromatid.

Every eukaryote that has been studied has an ortholog of SPO11, but it is very difficult to obtain and maintain a mutation that knocks out the function of SPO11. What do you predict is the phenotype of a spo11 mutant that will make it so hard to maintain, and why?

SPO11 is the enzyme that creates double stranded breaks in the homologous chromosomes during prophase I of meiosis to facilitate crossing over. The homologous chromosomes for an SPO11 mutant would therefore be unable to cross over, so non-disjunction would occur for every chromosome. SPO11 mutants would therefore be unable to produce offspring. It would be difficult to maintain a strain of SPO11 mutants since they will be unable to produce offspring.

Factor VIII and Factor IX are two X-linked genes in humans involved in blood clotting, and mutations in these genes result in Hemophilia A or Hemophilia B. Originally it was not known which form affected the royal families, but the inheritance pattern is the same in either case. Use Figure 7-6 to answer the following questions. a. Queen Victoria's daughter Alice had two daughters, Irene and Alexandra. Both of these daughters were carriers for hemophilia, as seen from their children. What is the probability that the two daughters born to Alice would both be carriers of the disease?

Since Alice is a carrier, there is a 50% chance that her daughter will inherit the mutant allele for hemophilia. Since this is true of each daughter, the probability of both daughters being carriers is (0.5*0.5) = 0.25 or 25%.

A pea plant with smooth green peas is crossed with a different plant that also has smooth green peas. The following F1 phenotypes result: 55 smooth/green, 17 smooth/yellow, 18 wrinkled/green, and 5 wrinkled/yellow. Assign genotypes to the F1 and parent plants.

Smooth (S) is dominant to wrinkled (s). Green (C) is dominant to yellow (c). -Parents (both): SsCc -Smooth/green progeny: S_C_ -Smooth/yellow progeny: S_cc -Wrinkled/green progeny: ssC_ -Wrinkled/yellow progeny: sscc

What are two important differences between the processes of spermatogenesis and oogenesis in animals?

Spermatogenesis results in four viable sperm being generated from one primary spermocyte. Oogenesis produces cells of different sizes: three polar bodies and one (much larger egg) as the cytoplasm is not divided evenly. All oocytes are made during gestation and then pause during Prophase I in meiotic (dictyate) arrest until puberty. When ovulation occurs during reproductive years, the primary oocytes complete meiosis I and then arrest again at metaphase of meiosis II, which is only completed if the egg is fertilized.

Starting with a hermaphrodite that is homozygous for an X-linked trait and males lacking this allele who also have dominant mutations on an autosome will help to distinguish self and cross-progeny. The most common method uses a recessive allele for an unrelated autosomal gene in hermaphrodites, and the wild-type dominant allele in the male. Another widely used method is to count only the males since they arise from cross-fertilization.

e. Synapsis

The "zipping up" of two chromosomes that have been paired together.

The F1 hermaphrodites will all be wild type because they will inherit the unc-3 allele from the hermaphrodite parent and the wild-type allele from the male parent. All of the F1 males will be Unc because they will inherit the unc-3 allele from the hermaphrodite and no allele from the male.

Define the following terms as they are used in this chapter. a. START

The G1 checkpoint at which time the cell commits to starting the cycle of cell division.

Joe has a white cat named Sam. When Joe crosses Sam with a black cat, he obtains 3 white kittens and 3 black kittens. When the black kittens are interbred, all the kittens that they produce are black. On the basis of these results, would you conclude that white or black coat color in cats is a recessive trait? Explain your reasoning.

The black coat color is likely recessive. When Sam was crossed with a black cat, one-half the offspring were white and one-half were black. This ratio potentially indicates that one of the parental cats is heterozygous dominant while the other parental cat is homozygous recessive—a testcross. The interbreeding of the black kittens produced only black kittens, indicating that the black kittens are likely to be homozygous, and thus the black coat color is the recessive trait. If the black allele was dominant, we would have expected the black kittens to be heterozygous, containing a black coat color allele and a white coat color allele. Under this condition, we would expect one-fourth of the progeny from the interbred black kittens to have white coats. Because this did not happen, we can conclude that the black coat color is recessive.

Triploid plants are typically seedless while tetraploid plants often produce seeds. Explain why this is true.

The chromosomes of triploid plants will not pair correctly during meiosis because there are three copies of each chromosome, which results in unequal and partial pairing. Unequal numbers of chromosomes will result in gametes the will not produce live offspring. Tetraploids are more frequently fertile because the even numbers of chromosomes in a tetraploid can be pair and split evenly, producing viable gametes.

i. Non-disjunction

The failure of chromosomes to segregate properly during meiosis.

(Looking back) The woman II-2 shown in the pedigree in Figure Q6-5 Panel A is pregnant with her first child. Both her family and her husband have a sister who is affected by an autosomal recessive trait as shown. b. A geneticist uses RFLP analysis (described in Tool Box 3-1) in combination with a polar body assay to determine if their child is likely to be affected by this trait. In this analysis, the size of the band on the gel represents a chromosome that has either the "normal" allele or the "disease" allele. The gel is shown in Figure Q6-5 Panel B, with each lane of the gel showing the results for one member of her family. The lane labeled "PB" is the analysis of the first polar body. Based on this analysis, what is the probability that their child will not be affected by this trait?

The first polar body shows the "disease" allele, just like II-1, who is recessive for the trait. This means that the child has no probability of having the disease, because at best the child can be heterozygous for the trait, because the secondary oocyte only has "normal" copies of the allele and therefore the egg that will be created will also have a "normal" copy of the allele.

The homologous chromosomes were unable to initiate or maintain synapsis so no chromosome is paired with its homologue.

Compare the orientation of the kinetochore during mitosis with the orientations during meiosis I and meiosis II. How does this change in the orientation affect these divisions?

The kinetochore has an inner and outer face and the microtubules of the spindle will only attach to the outer face of the kinetochore. The inner side of the kinetochore is associated with the DNA of the centromere. In meiosis I, the kinetochores of the sister chromatids are facing the same direction, and the kinetochores of homologous chromosomes are facing opposite directions. This means that the sister chromatids get pulled together, but the homologous chromosomes get separated. In meiosis II, the kinetochores of the sister chromatids are oriented in opposite directions so that microtubules can attach on either side and pull them apart towards the poles.

A couple has a child with trisomy-21, that is, with three copies of chromosome 21 rather than two. Geneticists analyzed chromosome 21 of both parents and the child in order to understand the origin of this non-disjunction event. Some of the results are shown in Figure QB-4. a. The gel at the top of the figure shows a polymorphic locus on chromosome 21 referred to as a copy number variation (CNV). At this locus, different individuals have a different number of copies of a repeated sequence; a higher copy number results in a larger band (closer to the top of the gel) than a low copy number. The locus itself has nothing to do with meiosis; it is simply a marker to track the parents' and child's chromosomes. Based on this gel, which parent experienced the non-disjunction, and how can you tell?

The mother of this child experienced non-disjunction during meiosis because two of the three bands on the child's gel match up with the mother, and only one matches with the father.

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation g. Given this information, which allele is dominant: the wild-type or null mutant? Explain your reasoning.

The mutant null allele is the dominant. In heterozygotes 40 + 0 units = 40 units of square factor are produced. But you are told the threshold amount to be square is 50 units. This does not meet the threshold amount, therefore, the heterozygote is the mutant phenotype. The mutant allele is dominant.

The probability is 8/9.It is simpler to work out the probability that the child will be affected. That can only occur if both parents are heterozygotes. There is a 2 in 3 chance that either parent is heterozygous for the trait as both parents have parents that are heterozygous for the trait but are not themselves affected. The chance that both parents are heterozygous for the trait is 4 in 9 (⅔ x ⅔= 4/9). If they are both heterozygous, there is a 1 in 4 chance that their child will homozygous recessive, which makes the probability that both parents are heterozygous and their child is homozygous recessive 4/36 or 1/9 (4/9 x ¼ = 4/36 or 1/9). Therefore the chance that that the child does not have the trait is 8/9.

One of the female skeletons was tentatively identified as Anastasia, based on the age of the girl and the presence of some jewelry. Anastasia died before she could have children. Based on the evidence above, what is the probability that her first child would be a boy with hemophilia?

The probability that Anastasia's first child would have been a boy is 50%. Since Anastasia was a carrier, the probability that her first son would have had hemophilia is 50%. The probability that Anastasia's first child would have been a boy with hemophilia is therefore (0.5*0.5) = 0.25 or 25%.

d. Cohesion

The process by which sister chromatids stay together after replication but before metaphase of mitosis or meiosis II.

c. Centromere and kinetochore

The protein structure that holds together two sister chromatids (the centromere) and the proteins around this structure that provide a place for microtubules to attach (the kinetochore).

b. Centrosome

The structures in an animal cell from which microtubules are grown and organized.

It was long thought that the appearance of tufts of hair on the outer ear was inherited as a Y-linked trait in humans. What distinctive inheritance pattern would have led to this proposal? Why might have been evidence that this is not a Y-linked trait?

The trait could appear to be Y-linked if is only expressed in males. A female expressing this trait would provide evidence that the trait is not Y-linked.

Some parakeets and other birds of the parrot family have a phenotype known as lutino. Lutino is a much prized beautiful coloration pattern in which the feathers on the body and head are yellow, with silver patches on the cheeks, and light yellow tail feathers. (These birds are pretty enough that you might want to look at a picture of them on-line.) The birds also have red or pink irises in the eyes rather than the standard black irises. a. A lutino male parakeet is mated with a normal green female. The female offspring are all lutino, while the male offspring are all green. Explain this result.

The trait lutino is Z-linked recessive, while green is dominant. (Males are ZZ and females are ZW in birds.) Male parents were Zl/Zl and female parents were ZL/W. (question 11)

The original white-eyed fly that the Morgans found was a male. Based on what you know about X-linked inheritance and sex determination in flies, why was it much more likely that the first mutant would be a white-eyed male rather than a white-eyed female?

The white-eyed mutation is recessive and X-linked. Because a female has two X-chromosomes, she would have to have to be homozygous for the recessive allele to be white-eyed. Because males are hemizygous for all X-linked genes, a male fly would only need one mutant gene to have white eyes, and there is a greater likelihood of a male with one mutant allele than a female with two.

(Looking back and ahead, and progressively challenging). In C. elegans, males are X0 while hermaphrodites are XX. Hermaphrodites have the somatic sex determination of females but they undergo both spermatogenesis and oogenesis, so they can use their own sperm to fertilize their ovum or they can mate with a male. Hermaphrodites cannot mate with other hermaphrodites. h. A hermaphrodite that is homozygous for one of the him mutations and homozygous for unc-3 was mated to a wild-type male. In addition to the results that you deduced for part C above, some of the offspring are wild-type males. Explain this result. Hint: The offspring also include unc-3 hermaphrodites, but unless one of the methods from part D is used, these cannot be distinguished from self-progeny.

The wild-type males must be the results of non-disjunction for the X chromosome in the hermaphrodite, such that the hermaphrodite produced eggs without a X chromosome that were fertilized by male sperm containing wild-type chromosomes. These are called patroclinous males—their X chromosome came from their father's side. The Unc-3 hermaphrodites can be explained as the product of self-fertilization.

A male mouse with a pale yellow coat called cream is mated to a female mouse with a wild-type gray- brown coat. All of the F1 mice, both males and females have the wild-type gray-brown color. One of these F1 female mice is mated to cream-colored male c. In fact, cream is X-linked. When the F1 female mouse is examined more closely, it is apparent that she has patches of fur that are cream colored rather than gray-brown. Explain this result.

This can be explained by X inactivation. Different X chromosomes form a Barr body in different cells of the mouse. In some cells, the chromosome containing the dominant allele for gray-brown fur was inactivated, resulting in the recessive allele for cream fur being expressed, resulting in cream colored fur.

The pedigree below is an example of an X-linked recessive trait. (picture or question 11 on meenly chapter 7) a. Could this inheritance pattern occur if the trait were autosomal? If so, what additional assumptions would need to be made to explain this inheritance pattern?

This pattern could occur if the trait was autosomal recessive but the trait would have to be assumed to be a common one, since II-1, II-3, and II-7, all of whom married into the family, would have to be carriers. Therefore, X-linked recessive inheritance is a much better explanation of this pedigree.

The single base change shown above appears to have changed the splicing of the exons in the gene encoding factor IX. How would this mutation alter the function of factor IX and lead to hemophilia?

This single base change appears to cause an alteration in splicing that increases the length of exon 4 by a single base. This results in a frameshift in Exon 4 that eventually codes for an early stop codon. This shortening of the protein may cause it to lose its function and lead to reduced blood clotting.

The dimensions of the synaptonemal complex are similar in different organisms, but the amino acid sequences of proteins that make up the central element (the "rungs of the ladder") are not similar. What does this suggest has been the evolutionarily important feature of the synaptonemal complex?

This suggests that while the amino acid composition of the ladder is not evolutionarily conserved, the width of the ladder is, suggesting that the spacing between the chromosomes is the evolutionarily important feature of the synaptonemal complex and not the protein sequence.

The pedigree below shows the inheritance of a rare trait in humans. What is the probability that the individual shown by the ? will not be affected by this trait? -page 304

This trait is not X-linked, because the affected female is the daughter of an unaffected father, and it is not dominant, because two unaffected parents produced an affected child. The trait must be autosomal recessive. The probability that the father of the child (?) is a carrier is ⅔, since we know that he is unaffected. The probability that the mother of the child is a carrier is 1, since she must have inherited a recessive allele from her affected father. If both parents are carriers, there is a ¼ chance that they will have an affected offspring. Therefore, the probability of the child inheriting the trait is ⅔ x 1 x ¼ = ⅙.

Three pairs of homologous chromosomes did not maintain synapsis while the other three pairs synapsed.

Using OMIM or another source for human genetics, what are examples of X-linked recessive traits in humans?

Traits that are X-linked in humans include hemophilia, color-blindness, Duchenne muscular dystrophy, and hundreds more.

What is the most common underlying cause of trisomy 21 in humans?

Trisomy 21 is usually caused by a failure to cross over in the mother. This means that the homologous chromosome 21 pair can't be held at the metaphase plate, resulting in non-disjunction, and an egg cell with two copies of this chromosome instead of one. The disomic egg is fertilized by a sperm with one copy of chromosome 21 to produce an embryo with three copies of chromosome 21.

Hairlessness in American rat terriers is recessive to the presence of hair. Suppose that you have a rat terrier with hair. How can you determine whether this dog is homozygous or heterozygous for the hairy trait?

We will use h for the hairless allele and H for the dominant. Because H is dominant to h, a rat terrier with hair could be either homozygous (HH) or heterozygous (Hh). To determine which genotype is present in the rat terrier with hair, cross this dog with a hairless rat terrier (hh). If the terrier with hair is homozygous (HH), then no hairless offspring will be produced. However, if the terrier is heterozygous (Hh) then we would expect one-half of the offspring to be hairless.

During the Korean War, soldiers (males) and nurses (females) were given antimalarial drugs such as chloroquine. Some soldiers developed a severe and life-threatening hemolytic anemia in which their red blood cells lysed; other soldiers were completely unaffected. Subsequent genetic analysis showed that the hemolytic response was due to a particular allele of the X-linked gene encoding the enzyme G6PD. b. (Challenging) Surprisingly, some female nurses also showed this severe hemolytic response, although the allele is rare enough that no nurse was expected to be homozygous. Postulate an explanation for this response.

When X-inactivation occurs in the embryo, either X chromosome can be inactivated. Perhaps the hematopoietic system (which gives rise to the red blood cells) is derived from only a few cells at the time when inactivation occurred. By chance then, some females will have the same X chromosome inactivated in all of their hematopoietic cells. By knowing either when X-inactivation occurs or how many cells give rise to the hematopoietic system in the embryo at this time, it is possible to estimate the other parameter from the fraction of women who show the hemolytic response. If the hematopoietic system is derived from only a single cell when X-inactivation occurs, then half of women should show this hemolytic anemia; if it is derived from two cells that inactivate their X chromosomes independently, then about 25% of women should show the hemolytic responses, and so on.

When a hermaphrodite mates with a male, 50% of the offspring are expected to be male, since each egg has a 50% chance of being fertilized with a sperm that contains no X chromosome. In reality, because the hermaphrodite also carries out self-fertilization, somewhat less than 50% of the offspring are males but half of the offspring from cross-fertilization are expected to be males.

In examining a large pedigree for a human trait, what would be an indication that the trait is X-linked? What would provide the clearest evidence that the trait could not be X-linked?

X-linked genes tend to be more frequent in males than females while autosomal genes are inherited at equal frequencies in males and females. The clearest evidence that a trait is not X-linked would come from an affected father having an affected son, since fathers do not pass X chromosomes to their sons.

Y-linked trait most likely. This trait cannot be X-linked because it is passed from a father to his sons. It is possible for the trait to be autosomal dominant or autosomal recessive, but the fact that it is passed only from the father to the sons suggests that it is likely to be a Y-linked trait.

For the most likely mode of inheritance in each pedigree, calculate the likelihood that child #1 will be affected by the trait. for A, B, and C

a. 50%. Since the trait is autosomal dominant, there is a 50% chance the child 1 will inherit the dominant allele and be affected. b. 0%. If the trait is autosomal recessive, then there is a 66% chance that the father of child 1 is a carrier. However, if we assume that the trait is rare and that it is unlikely for the mother of child 1 to be a carrier, then there will be a 0% chance that the child is affected. c. Assuming the trait is Y-linked, there is 100% chance if child 1 being affected if he is a boy, and a 0% chance child 1 being affected if she is a girl. Since there is a 50% chance of the child being a boy, there is a 50% chance of the child being affected.

What is the genotype of individual I.1? a. A_ b. Aa c. aa

a. A_

Xist is an example of a(n)... a. RNA that isn't translated b. oncogene c. transcription factor

a. RNA that isn't translated

When analyzing a pedigree showing the inheritance of a rare condition, what should we initially assume about individuals from outside the family? Group of answer choices a. They are not carriers of the trait b. They are carriers of the trait c. We should never make an initial assumption

a. They are not carriers of the trait

A certain gene in a species of jellyfish produces a fluorescent protein that allows the jellyfish to "glow in the dark." A mutant allele arises in this species that produces only a quarter of the amount of fluorescent protein as the wild-type allele. Heterozygous individuals still glow but homozygous mutant individuals do not. Which term best describes the wild-type allele in this context? a. haplo-sufficient b. haplo-insufficient

a. haplo-sufficient (heterozygote displays the wildtype phenotype rather than the mutant phenotype)

One form of color-blindness is a X-linked recessive. Which term would most correctly describe a color-blind male (XcY)? a. hemizygous b. homozygous c. heterozygous

a. hemizygous

An individual with the genotype Aa is... a. heterozygous b. homozygous for the dominant allele c. homozygous for the recessive allele

a. heterozygous

We can use the phenotypic ratios from controlled crosses of individuals to determine the genotypes of the parents. This, however, only works... a. if there are many offspring b. with complex traits c. with plants

a. if there are many offspring

Imagine that scientists developed an inhibitor of the SRY gene. What effect would this inhibitor have if given to a young, prepubescent mouse with an XY genotype? a. nothing b. the mouse would develop as female c. the mouse would develop as male d. the testes would disapear

a. nothing -its given to a mouse who has already developed testes & when they are already there then you don't really need the SRY -the mouse would develop as male -tetes already developed when born (if in the womb maybe would develop as female)

Mendel's Law of Segregation refers to the fact that the two alleles for a gene in a diploid organism separate during gamete formation and each gamete receives only one of these alleles. a. true b. false

a. true

Sex is not always inherited. a. true b. false

a. true

Two people are both heterozygous (Aa) for a recessive trait. -If they have two children, what is the probability that neither of them exhibit the trait? a. ¾ x ¾ b. ¾ + ¾ c. ¾

a. ¾ x ¾

What is the genotype of individual I.1? (allele super-scripts: "+" is wild-type, "m" is mutant, and "_" is unknown) (use pedigree on chapter 7 quiz) a.X+Y b. X+X_ c. XmY d. X_Y e. X+Ym

a.X+Y

Two people are both heterozygous (Aa) for a recessive trait. -If they have two children, what is the probability that at least one of them exhibits the trait? a. ¾ x ¾ b. 1- (¾ x ¾) c. ¼ x ¼ d. ¼ + ¼

b. 1- (¾ x ¾)

Two people are both heterozygous (Aa) for a recessive trait. -If they have a child, what is the probability that the child does not the trait? a. 1 (100%) b. 3/4 c. 2/3 d. 1/2 e. 1/3 f. 1/4

b. 3/4

An individual expresses the dominant phenotype associated with a gene, but it is unknown whether they are a carrier. What would be the most accurate way to write their genotype? a. AA b. A_ c. Aa

b. A_

Which part of meiosis explains Mendel's Law of Segregation at the chromosome level? a. Separation of homologues at meiosis II b. Separation of homologues at meiosis I c. Separation of sister chromatids at meiosis I d. Separation of sister chromatids at meiosis II

b. Separation of homologues at meiosis I

Use this scenario for the following 3 questions: Guinea pig coat coloration can be banded or solid. A female with banded coloration was mated to a male with solid coloration. The cross produced 3 solid females and 3 banded males. -Is inheritance of this coloration trait most likely autosomal or X-linked? a. autosomal b. X-linked

b. X-linked

What is the mode of inheritance for this trait? a. recessive b. dominant (look at pedigree in camera roll or on chapter 5 quiz under question 5)

b. dominant

A heterozygous individual expresses the phenotype associated with the... a. recessive allele b. dominant allele

b. dominant allele

Use this scenario for the following 3 questions: Guinea pig coat coloration can be banded or solid. A female with banded coloration was mated to a male with solid coloration. The cross produced 3 solid females and 3 banded males. -Are the female progeny homozygous or heterozygous? a. homozygous b. heterozygous

b. heterozygous

A certain gene in a species of jellyfish produces a fluorescent protein that allows the jellyfish to "glow in the dark." A mutant allele arises in this species that produces only a quarter of the amount of fluorescent protein as the wild-type allele. Heterozygous individuals still glow but homozygous mutant individuals do not. which term best describes this mutation? a. null b. leaky c. neutral d. GOF

b. leaky

b. mutant

Is dosage compensation needed in reptiles like sea turtles? a. yes b. no

b. no

The X chromosome selected for inactivation in cats is... a. the same for all cells b. occurs randomly among cells

b. occurs randomly among cells

Dosage compensation is achieved in humans through... a. inactivation of sex-related genes b. shutting down the entire X chromosome c. increasing expression of X-linked genes in males

b. shutting down the entire X chromosome

b. solid

A cross between tall plants (Tt genotype) and short plants (tt genotype) produced 6 offspring. What is the probability that all 6 offspring are tall? a. 1/2 b. 1/16 c. 1/64 d. 1/128

c. 1/64

Two people are both heterozygous (Aa) for a recessive trait. -If the child does NOT exhibit the trait, what is the probability that the child is a carrier? a. 1 (100%) b. 3/4 c. 2/3 d. 1/2 e. 1/3 f. 1/4

c. 2/3

What is the probability that offspring "?" is affected? a. 33% b. 25% c. 50% d. 75%

c. 50%

Which phase of the cell cycle has two copies of the genome in the cell for the entire phase? a. G1 b. S c. G2 d. M

c. G2

When does crossing-over (recombination) occur? a. Prophase of meiosis II b. Prophase of mitosis c. Prophase of meiosis I

c. Prophase of meiosis I

The genotype of an organism with a dominant phenotype but unknown genotype can be inferred using a test cross. What genotype should a plant A_ be crossed to in a test cross? a. Aa b. A_ c. aa d. AA

c. aa

An individual with the genotype w+/w+ is... a. homozygous for the mutant allele b. heterozygous c. homozygous for the wild-type allele

c. homozygous for the wild-type allele

This cell is part of a diploid organism and is in anaphase. What type of cell division is the cell undergoing? Do not consider the number of cells shown (i.e. that there is 1 vs. 2 or 4) when making your decision. Remember that non-homologous chromosomes are shown as different sizes. (check picture from question 7 on chapter 6 quiz) a. mitosis b. meiosis 1 c. meiosis 2

c. meiosis 2

Two people are both heterozygous (Aa) for a recessive trait. -Ok -back to just one child.What is the probability that the child is a carrier (heterozygote)? a. 1 (100%) b. 3/4 c. 2/3 d. 1/2 e. 1/3 f. 1/4

d. 1/2

Crossing over occurs in which phase of cell division? a. Prophase II (meiosis) b. Crossing over occurs during all of these phases c. prophase (mitosis) d. Prophase I (meiosis) e. Both Prophase (mitosis) and Prophase I (meiosis)

d. Prophase I (meiosis)

What is the mode of inheritance of the trait shown in the pedigree, below? (use pedigree on chapter 7 quiz) a. X-linked recessive b. X-linked dominant c. autosomal recessive d. autosomal dominant

d. autosomal dominant

Two people are both heterozygous (Aa) for a recessive trait. -If they have a child, what is the probability that the child exhibits the trait? a. 1 (100%) b. 3/4 c. 2/3 d. 1/2 e. 1/3 f. 1/4

f. 1/4

evaluate the following statements based on your understanding of how genes are often named: the wild-type BRCA1 (breast cancer 1) gene causes cancer. true or false

false

Place the following stages of mitosis in chronological order from first (1) to last (4). -metaphase -prophase -telophase -anaphase

prophase-->metaphase-->anaphase-->telophase

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation d. What will be the phenotype of mutant homozygotes?

round

In a hypothetical diploid organism, squareness of cells is due to a threshold effect: more than 50 units of "square factor" per cell will produce a square phenotype, and fewer than 50 units will produce a round phenotype. The wild-type allele for the square factor (Sf) gene causes the synthesis of the functional square factor. Each wild-type allele contributes 40 units of square factor. A mutant allele in the Sf gene arises; it is a null mutation f. What will be the phenotype of wild-type homozygotes?

square

The X chromosome in humans contains genes unrelated to sex determination. true or false

true

in males, the X and Y chromosomes pair during metaphase of meiosis I true or false

true

Example problem: For Mendel's pea plants, "tall" stem length is the dominant phenotype and "short" stem length is recessive. You have a pea plant with an unknown genotype and you can't tell if it has a tall or short phenotype because you forgot to water it for a week and now it's looking pretty sad. To identify its genotype, you cross it with a short plant. Of the 100 resulting F1 plants, 52 are tall and 48 are short (you remembered to water them this time). What is the genotype of your unknown plant?

•P: __x short plant (P_ is unknown & we cross it w/ a short plant) •Short has to be homozygous recessive (tt) •Tall has to be heterozygous (Tt) •F1--> 52 tall & 48 short •Tt x tt--> do a punnet square and see that all boxes have at least one lowercase t (half the boxes have to be short: tt & half are tall: Tt)

genetics exam 2 quizzes & clicker questions

Ensembles d'études connexes

Marketing exam 2

Chapter 1

chapter 8&9

English Lesson 5 Vocabulary Words and Roots

Bilirubin and Liver

HIST 101 Ch 7

ITE exam 2

History Topic9

substance abuse test 3

SIE Missed exam questions :)

us government - unit 1: the history of civics and government

Chapter 62 - Pituitary and Adrenal Gland issues

Chapter 11: Maternal Adaptation During Pregnancy

Patho Chapt 17

2210 Test 2 Chapters 65-68

study

cs: ch 12-14

TEXTBOOK: Ch. 5: Analyzing the Industry and Market

Vocabulary Workshop Level E Unit 8 - Synonyms and antonyms

Vocabulary Workshop Level C Unit 6 sentences