# Solving a Pedigree Between Heterozygous Half-Cousins

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A man who is a known heterozygous carrier of oculocutaneous albinism marries his half-cousin (they share one common grandparent) as shown in the pedigree below. This trait is transmitted as a fully penetrant autosomal recessive. What is the probability that this couple will produce a child with this disorder?

``A. 1/2 B. 1/4 C. 1/8 D. 1/16 E. 1/64``

My calculation was 1/2 (for II-2 as he must be carrier) * 1/2 (for II-3) * 1/2 (for III-2) * 1/4 = 1/32. Please help me out by proper explanation to your steps and point out my mistakes. The correct answer given is 1/64. Pedigree shown in the image below.

From what I can tell, you are not including the probability of the grandfather being a carrier (it could be the grandmother who is the carrier). Calculation should go like:

P(Child is born with disorder) = P(II-2 is carrier) * P(I-2 is carrier) * P(II-3 is carrier) * P(III-2 is carrier) * P(Child is homozygous recessive)

P(Child is born with disorder) = (1/2) * (1/2) * (1/2) * (1/2) * (1/4)

P(Child is born with disorder) = 1/64

## Bookshelf

NCBI Bookshelf. A service of the National Library of Medicine, National Institutes of Health.

Griffiths AJF, Miller JH, Suzuki DT, et al. An Introduction to Genetic Analysis. 7th edition. New York: W. H. Freeman 2000.

• By agreement with the publisher, this book is accessible by the search feature, but cannot be browsed.

## INTRODUCTION

Problem solving has been defined in the literature as engaging in a decision-making process leading to a goal, in which the course of thought needed to solve the problem is not certain (Novick and Bassok, 2005 Bassok and Novick, 2012 National Research Council, 2012 Prevost and Lemons, 2016). Ample research shows that students have difficulty learning how to solve complex problems in many disciplines. For example, in biology and chemistry, students often omit critical information or recall information incorrectly and/or apply information incorrectly to a problem (Smith and Good, 1984 Smith, 1988 Prevost and Lemons, 2016). Furthermore, across many disciplines, researchers have found that experts use different procedural processes than nonexperts when solving problems (Chi et al., 1981 Smith and Good, 1984 Smith et al., 2013). While students often identify problems based on superficial features, such as the type of organism discussed in a problem, experts identify primary concepts and then link the concept with strategies on how to solve such a problem (Chi et al., 1981 Smith and Good, 1984 Smith et al., 2013). Experts also often check their work and problem solutions more frequently than nonexperts (Smith and Good, 1984 Smith, 1988). Given the difficulties students have in problem solving and the value of such skills to their future careers, there is clearly a need for undergraduate educators to assist students in developing problem-solving skills (American Association for the Advancement of Science, 2011 National Research Council, 2012).

Two kinds of knowledge have been described in the literature as important for solving problems: domain specific and domain general. Domain-specific knowledge is knowledge about a specific field, including the content (declarative knowledge), the procedural processes used to solve problems (procedural knowledge), and how to apply content and process when solving problems (conditional knowledge Alexander and Judy, 1988). Domain-general knowledge is knowledge that can be used across many contexts (Alexander and Judy, 1988 Prevost and Lemons, 2016). A third category, strategic knowledge, is defined as knowledge about problem-solving strategies that can be domain specific or domain general (Chi, 1981 Alexander and Judy, 1988). Research suggests that domain-specific knowledge is needed, but may not be sufficient, for applying strategic knowledge to solve problems (Alexander and Judy, 1988 Alexander et al., 1989). Thus, helping students learn to solve problems likely requires teaching them how to activate their content knowledge, apply their knowledge to a problem, and logically think through the problem-solving procedure.

Previous research suggests that receiving help in a variety of forms, including procedure-based prompts (Mevarech and Amrany, 2008), a combination of multiple content- and procedure-based prompts (Pol et al., 2008), and models (Stull et al., 2012), can be beneficial to learning. Not surprisingly, accessing relevant prior knowledge has been shown to positively influence performance (Dooling and Lachman, 1971 Bransford and Johnson, 1972 Gick and Holyoak, 1980). For example, in genetics, successful problem solvers often identify similarities between problems, whereas unsuccessful problem solvers do not (Smith, 1988). Previous research also suggests that receiving procedural guidance can be beneficial to learning. In a study that asked students to examine different problems with related solutions, prompting students to consider previously reviewed problems helped most students subsequently solve a challenging problem (Gick and Holyoak, 1980). In another study, when students received guidance that included identifying similarities to other problems as well as other procedural skills, such as planning and checking their work, they were better able to solve subsequent problems than in the absence of such guidance (Mevarech and Amrany, 2008). However, although accessing prior knowledge is important, it is also important that students understand how to apply their prior knowledge to a given problem (Bransford and Johnson, 1972). Thus, while students may realize they need additional information to solve a problem, if they cannot make sense of this information in the context of a given problem, the information is unlikely to be useful.

In addition to knowledge, students need practice. Within the field of psychology, many studies have examined the association between practice and performance. Completing a practice test leads to better performance on a subsequent final test compared with other conditions in which students do not test themselves, such as studying or completing an unrelated or no activity (e.g., Roediger and Karpicke, 2006 Adesope et al., 2017). In a meta-analysis, this effect, termed the “testing effect,” was found to occur regardless of whether feedback was given and regardless of the time between the practice test and the final test (Adesope et al., 2017). The benefits of practice testing on later performance can occur not only when using the same questions (retention) but also when students are asked to transfer information to nonidentical questions, including questions that require application of concepts. In one of the few studies on the testing effect using transfer questions, students who took practice tests performed better on transfer questions on a final test for both factual (i.e., a single fact in a sentence) and conceptual (i.e., a cohesive idea across multiple sentences) questions than those who studied but did not take practice tests (Butler, 2010). This study also found that those who performed well on their practice tests were more likely to do well than those who performed poorly on their practice tests 1 week after practice on a subsequent final test, which included conceptual questions that required application (Butler, 2010).

In the current study, we focused on whether students who are incorrectly solving a problem can apply content knowledge given to them as a prompt to correctly solve subsequent genetics problems. We address the following questions: 1) Does providing a single content-focused prompt help students answer similar questions during subsequent practice, and does this practice help on later exams? 2) When unable to apply content prompts, what content errors and omissions do students make that lead them to continue to answer incorrectly?

## Genetics multiple choice questions with answer key in mendelian genetics, molecular genetics, linkage, crossing over, cytogenetics, modified mendelian ratios.

Punnet squares are used to predict and compare the genetic variations that will result from a cross. This worksheet will take about 20 minutes for most students, i usually give it to them after a short lecture on solving genetics problems. Mendel in his charts showed the probability of dominant and recessive genes being passed on to offspring. In genetics, for example, probability is used to estimate the likelihood of gene distribution from one generation to the next. The sum rule and product rule. If the alleles are not identical, then the individual is heterozygous. Probability is used in genetics to determine the possibilities of offspring having a particular trait. An explanation of how this solution is derived. pdf simple genetics practice problems key simple genetics practice problems key this worksheet will take about 20 minutes for most problems worksheet with answer keys photos collection can be a guidance for you, deliver you more references and of course make you have a. Basic notation applying basic probability to mendelian genetics conditional probability probability in statistical analysis the the aim of this tutorial is to guide you through the basics of probability. Rare alleles have a higher probability of being lost during a population bottleneck event. Probability tutorial for biology 231. Genetics multiple choice questions with answer key in mendelian genetics, molecular genetics, linkage, crossing over, cytogenetics, modified mendelian ratios. There are 3 parts to this exam. The passing of traits from parents to offspring. Who is known for being the 'father of genetics'? Some of the worksheets displayed are work mendel and genetic crosses mendelian genetics work mendels pea plants work gregor mendel answer key meiosis and mendels. Evolutionary biology exam #1 fall 2017. Skin cells have a short life span—skin is easily damaged and must often be 27. Where do you find the genetic material that is passed from parent to offspring 7. Christopherso answer key applied probability models with optimization applications by sheldon m. Individuals b, d, i, and n are colorblind. The seed color controlled by the dominant allele has a 50% probability of being. Key concepts and learning objectives. (aabbccddee) x (aabbccddee) predict the probability of recovering offspring of each of the following genotypes aabbccddee aabbccddee aabbccddee aabbccddee 1/32 1/128 1/32 0 answer key mendelian genetics problem set 2: Suppose a mother and father is both heterozygous for the characteristics of brown eyes and brown hair, ie they have brown eyes and. For both dice to turn up a three, the probability is determined by multiplying the probability of each event happening independently, or 1/6 x 1/6 =1/36. It allows geneticists to emphasize desirable traits in food, plants, and animals. Reebop genetics answer key search » all » science. What is the probability of their next child having a widows peak? Reconstructing the last universal common ancestor (luca) is challenging.

## Difficult biology (genetics) question with pedigree

٩(๏̯͡๏)۶This is a very difficult question I’ve come across in my studies I would sincerely appreciate anyone’s assistance. This is coming from an 11th – 12th grade exam not in any particular textbook. (THE QUESTION/PROBLEM*** Human beings are very complex and so is human genetics. There are some traits in humans determined by a single gene such as tongue rolling. Some people are able to roll their tongue up at the edges others are not. Is this a dominant or recessive trait? Study this pedigree which shows the appearance of this trait in three generations of a family. http://i234.photobucket.com/albums/ee300/tundrawindimvu/activity.png From this chart would you conclude the trait to be dominant or recessive? Why? From the pedigree would you conclude the male in the first generation to be heterozygous or homozygous? Why? (MY CONCLUSION *** I have come to the conclusion that it is impossible for the of offspring of the first generation to have a ratio of 3 : 4 tongue rollers if the dominant trait always shows over the phenotype of a recessive trait. Also my problem solving processes of making many punnett squares leads me to believe if the male has a dominant tongue rolling trait then the offspring have either a 100% or 50% chance of being able to roll their tongue. If the gene is recessive then the female must also have the trait BUT the pedigree provided shows parental female as an empty circle which means she does not have the trait…. )

THANK YOU EVERYONE FOR YOUR HELP!

• 감사함니다 • I chose to write: A) The trait, tounge rolling is dominant because the first generation male must be [Tt] heterozygous and female mate [tt] homozygous in order for the rest of the pedigree to result or make sense. For example, one daughter of the first generation does not have a penotype for tounge rolling.I assume she has a genotype of [tt] which means the father must have a ressesive trait to combine with the mothers. B)The male in the first generation is heterozygous because the key given indicates his mate cannot carry the trait. Also if he is dominant homozygous then his daughter in the second generation most likely could not have the same phenotype as her mother. Note to everyone, the key given says that the empty circle a.k.a female mate of the first generation, is a female without the trait.Sooooo I had to chose dominant over ressesive even though I know all and any is possible ..

Just waiting to graduate with a Biology BSc, here's what I think. (Disclaimer: I will explain everything as clearly as I can, please don't take offense if you already know it)

If a trait is dominant, a person can be hetero- or homozygous and show the phenotype (tongue rolling in this case)

If a trait is recessive, a person has to be homozygous for it to show the phenotype but can be a heterozygous carrier.

Assume recessive

The male is generation 1 would have to be homozygous for the recessive allele and the female would have to be heterozygous for the female in the 2nd gen. to not show the phenotype. The probabilities from the punnet squares make it unlikely that so many individuals would be affected if the trait is recessive.

Assume dominant

The male in generation 1 would have to be heterozygous and the female homozygous non-rolling. This would allow the female in 2nd gen. to be non-tongue rolling (homozygous). Her partner could be homozygous dominant which would mean all offspring are heterozygous and therefore show the tongue rolling.

My conclusion In terms of probabilities with the offspring, it's most likely that the trait is dominant because it would be more likely to produce this family tree according the Mendelian genetics. The male in the 1st gen. would have to be heterozygous to permit the rest of the tree.

I hope this helps and if you find out the correct answer Iɽ be interested to hear! Also, if you can any questions I'm more than willing to answer :)

Just finished me MSc in biology here. I concur with confused_penguin. P1 (parental generation) Male Tt female tt (large T dominant for tongue rolling)

F1 (child generation) All black boxes Tt with the white box being tt

F2 on the left could be mated to another heterozygous or homozygous, can't tell. On the right mated to a homozygous is likely though still could be heterozygous (as its a 50% 50% chance each child).

Male in P1 is definitely heterozygous because there is a recessive phenotype in the offspring. Edit: and his mate is homozygous recessive

Hope this answers what I'm assuming is a take home exam/hw question for you.

It is dominant. One thing that is VERY important to remember is that statistics (aka the probabilities you get out of your punnet squares) are not at all exact. That is merely what would be expected to happen given a large sample size. Even if there is a 50/50 chance to have/not have the phenotype, that DOES NOT mean that every set of 4 offspring will have 2 tongue rollers and 2 non tongue rollers it just means that that is the most likely outcome. In most cases, you want to go with the answer that most closely fits with your punnet square probability projections. If the trait is dominant, the 1st gen male has to be heterozygous (to produce any non tongue rolling offspring) and the predicted phenotypes of the offspring will be 50/50. If the trait was recessive, the prediction for gen 2 would again be 50/50 so generation 2 tells you nothing. The right side of generation 3, however, tells you all you need to know. 100% tongue rollers would only be predicted if the trait was dominant and the male that bred with the non-tongue roller female is homozygous dominant. If it is a recessive trait, he would be homo recessive, she would be a het, and the predicted ratio would be 50/50. As you can see, the offspring are 100% tongue rollers. Because statistics are NEVER 100% true predictions (merely the best guess based on the data available), you can't 100% say which of these cases is true, but you can say that one fits better than the other. The explanation that fits the data better is that the trait is dominant, and the principle of parsimony says go with the simplest answer so there you have it. Sorry if that seems like a soft answer, but the reality of statistics is that nothing is certain. Even with an infinite sample size, the data just keeps getting closer and closer to the predicted ratios, but almost never fits perfectly with those predictions.

OK so first things first.
I am fairly certain that this is a trick question.

In F1 You have to have a het crossed with a homo. If they were both homo you would get them all the same offspring (100% het). If you had a het X het cross you might be tempted to say you could get the offspring based on a 25%/50%/25% product. however if this was the case both parents would be expected to have the same phenotype here which they dont. So guaranteed you have a het X homo cross in F1

Now so you KNOW that there is a het X homo cross in F1. So if you pick either recessive or dominate for the trait based on a homo X Het mating you dont get what you are seeing perfectly. (r/r)X(R/r)=50%(R/r) and 50% (r/r). So the recesive vs dominate is equally probably from the F1 cross

Now to look at the F2 crosses. Lets start with the left. If the trait is dominate then the male is R/r (we know this because it is a homo x het) crossed with a R/?. If the female is homozygous for R then you get 100% offspring with the trait. If the female is het then you get 25% R/R 50% R/r and 25% r/r or 75% with the trait. the offspring are all expressing the trait so if it is dominate then you dont get much information from this cross because you dont know the female and the F3 population is low.

So now if the trait is recessive the left mating will be a r/r X r/r which matches what you see cause 100% of the offspring have the trait. So thats nice.

Now to the right F2 mating. If it is dominate then the female must be r/r. Now the male could be R/r or R/R. so if you have a R/R X r/r you get 100% R/r so thats possible. a R/r x r/r gives you 50% R/r and 50% r/r which doesnt fit but is still possible due to low population of F3 generation.

So if you look at the right mating and assume recessive the female is R/r and the other is r/r. This would be a 50% R/r and 50% r/r and does not match the data perfectly.

So basically what you have here is an example where the trait could be either recessive or dominate and either the left or the right F2 mating is weird based on what you pick. Alternativly you could pick dominate and say both of the outside mating individuals are homozygous for the trait and then everything works out smashing. I would say that the correct answer to this question is one that is internally consistent. So if you pick recessive you need to point out which one is wierd and which one fits and reverse it if you pick dominate. Also you need to talk about what those outside mating couples mean. If it is dominate then you would expect that both of those outside mating partners are homozygous (R/R) to get what you see. However the probability of that is going to be either low or high based upon the propensity of the trait to show up in the population.

Another possibility that I wont get into here is that the trait is X linked. If that was the case you can get a lot more complicated but I worked out the math and it doesnt work perfectly either so that is just adding in complexity that doesnt really help you understand the problem.

Anyways good luck. I typed this up between experiments so I kinda rushed. If you dont understand something just ask and I will try to explain it better for you. Also if you see a problem with my logic say something and I will look at it again to make sure I am right.

## Solving a Pedigree Between Heterozygous Half-Cousins - Biology

*Differentiate between quantitative observations, qualitative observations and inferences.

*Biogenesis -- Spontaneous Generation Debate over Time: Experiments by Redi & Pasteur

*Demonstrate an understanding of the parts, functions, and proper usages of compound microscopes

*What microscopes do to images. Example: the letter “e”.

Characteristics of living things

*define, describe and give an examples of each.

* Determine if something is alive, defend your position with scientific knowledge.

*Parts of an atom: protons, neutrons, electrons

*Interpret how atoms interact to form bonds, which which are used to form bonds in complex chemical compounds.

*Valence electrons are used to bond. (Two types of bonds: ionic & covalent.)

*pH scale range from 0-14. Acids = 0—7 and base range above 7—14

*Relate the structure of a water molecule to its unique properties including polarity, adhesion, cohesion, and capillary action.

*universal solvent vs. solute

*Explore the unique properties of the carbon atom and its role in organic compounds.

*Investigate the roles of the four basic biomolecules and their roles in living organisms:

carbohydrates, proteins, lipids, and nucleic acids

*Making polymers from monomers and then breaking them down (dehydration synthesis vs. hydrolysis)

Reactions and Enzymes

*What a reaction is- rearranging bonds

*Parts of a reaction- reactants, products, catalysts

*Activation energy, what catalysts do to activation energy, and graph

*Lock and Key model (enzyme-substrate comlex)

*Effect of temperature & pH on enzyme activity (denaturing)

*Enzyme lab: identifying the substrate and the products

*Categorize organisms based on their cell structure (plant, animal, prokaryote, eukaryote).

*Identify and relate the function of the organelles to their structures. The types of organelles present or absent in different types of organisms (prokaryotes/eukaryotes, plants vs. animals) *What the organelles do & their location/appearance(cell wall, cytoplasm, cell membrane, ribosome, nucleus, nucleolus, nuclear membrane(envelope), lysosome, smooth and rough endoplasmic reticulum(ER), chloroplast, mitochondria, centrioles, golgi, vacuole)

Cell membrane and transport through the cell membrane

*Phospholipid bilayer structure of the cell membrane, cell membrane is semi-permeable

*Predict the movement of particles based on concentration and explain which method (simple diffusion, osmosis, active transport (needs ATP), facilitated diffusions) would be used to cross a semi-permeable membrane in different situations to reach equilibrium

ATP & Photosynthesis & the leaf & Cellular Respiration

*parts of ATP molecule, how energy is stored/released from ATP molecule

*The equations for photosynthesis and cellular respiration including identifying reactants and products

*The role of chloroplasts/chlorophyll in capturing energy for the formation of ATP in autotrophic organism and the role of the mitochondria in cellular respiration

*Label the mesophyll(spongy layer/palisade layer), cuticle, epidermis, stoma, guard cells, xylem, phloem and chloroplasts in a cross section of a leaf and describe their functions.

*compare/contrast aerobic vs. anaerobic respiration(fermentation) in terms of energy production

*two types of fermentation: lactic acid and alcoholic

DNA, RNA, & protein synthesis

*Watson & Crick double helix DNA model which utilized Franklin’s X-ray diffraction photo

*Identify parts of DNA & RNA nucleotides(sugar, phosphate, base)

*Base pairing rules, purines with pyrmidines. Hydrogen bonds “hold” bases together

*Comparing/contrasting DNA and RNA

*Protein synthesis including the processes of transcription(writing the mRNA codons from DNA) and translation(using the chart to look up codons and determine amino acids)

Cell cycle, Mitosis, Meiosis

*Phases of Cell Cycle: Interphase (G1, S, G2), Mitosis (PMAT)

*Phases of Mitosis: (PMAT): What happens in each? What do they look like?

*Cancer: Causes, What makes it so dangerous?

*Mitosis Products: Makes two identical, diploid body cells (2n).

*Meiosis Products: Makes four genetically different haploid (n), sex cells (gametes)

*Homologus chromosomes: What makes them homologous? Crossing Over: Prophase I: tetrad

Mendel & Complete Dominance (normal punnett squares)

*Mendel: Who was he? What did he study? What were his conclusions? (Mendel’s 3 Laws)

*Phenotype vs. Genotype (Homozygous, Heterozygous, hybrid, pure-bred)

*Solving Regular Punnett Squares: What do their parts represent? Importance of probability and ratios.

Incomplete dominance and co dominance

*Incomplete Dominance: “In-between” phenotype for heterozygous

*Codominance: Two dominant traits both in phenotype for heterozygous

*Solving Problems: Punnett Squares, Ratios

*Multiple alleles: A-dominant, B-dominant, O-recessive

*Possible Genotypes for each blood type

*Blood Transfusions: Antigens/Antibodies, Which types work with which? Why?

*Identify and know the function of male and female reproductive parts of a flower.

*Explain what Mendel did, in terms of flower parts F1 generation, Parent generation

*Sex Chromosomes: Male (XY) and Female (XX), What do X and Y chromosomes look like?

*X-linked Recessive Disorders: Hemophilia, Colorblindness, Muscular Dystrophy

*What sex is more likely to get them? Why? (Punnet Square)

*Possible Genotypes: Males and Females: Which is called a carrier?

Human Genetics: Common Disorders and Testing

*Using diagnostic tools such as amniocentesis followed by karyotyping, pedigrees, genetic counseling to identify various type of common genetic disorders

*define “nondisjunction” and discuss the effects on offspring- trisomy, monosomy, polyploidy

*genetic disorders caused by nondisjunction: Down syndrome, Turner syndrome, Klinefelters syndrome

*follow the inheritance of a trait through a family pedigree recognizing symbols used for male/female and identifying affected people by shading

*determining genotypes for individuals on a pedigree

Genetic engineering

*define and give examples for each of the following topics related to genetics: polygenic traits, selective breeding, hybridization, polyploidy/induced mutation, and inbreeding

*define, identify final results of and give examples for the following types of genetic engineering: transgenic organisms(both plant and animal), cloning, transformation, stem cells, and recombinant DNA

*recognize and interpret diagrams illustrating genetic engineering processes listed above such as cloning, recombinant DNA, and transformation

*Given images of mutated genes or chromosomes identify the type of mutation that has happened

*mutagens- define and give examples

*distinguish between different types of mutations: for example Point vs frame shift

*Gene mutations - Insertion, deletion, substitution

*Chromosomal mutation- duplication, deletion, inversion, translocation

*nondisjunction during meiosis resulting in too many or too few chromosomes in gametes

## NES Profile: Biology (305)

Which of the following is the most likely cause of a base pair mutation?

1. Sister chromatids fail to separate during meiosis.
2. A small portion of one chromosome is added to another chromosome.
3. A segment of a chromosome breaks, flips, and reinserts itself.
4. The DNA is not accurately duplicated during replication.

D. This question requires the examinee to analyze the types and causes of chromosomal and gene mutations, the consequences of these genetic changes, and the genetic basis of common disorders and diseases. Mutations that involve nucleotide base pairs typically result from the substitution, insertion, or deletion of one or more incorrect bases. These most commonly occur during the replication of DNA when the new complementary strands are being produced.

### Descriptive Statements:

• Analyze meiosis and fertilization and their roles in sexual life cycles.
• Analyze patterns of inheritance and the relationship between genotypic and phenotypic frequencies.
• Demonstrate knowledge of the chromosomal basis of inheritance and its relationship to observed inheritance patterns and of the characteristics of extranuclear inheritance in plants and animals.
• Solve genetics problems.

### Sample Item:

The key provides symbols for Affected Male, Affected Female, Unaffected Male, and Unaffected Female. In the first generation, an unaffected female crosses with an affected male. This pair produces 4 offspring, which from left to right are an unaffected male who crosses with an unaffected female, an affected female who crosses with an unaffected male, an unaffected male who crosses with an unaffected female, and an affected female who crosses with an unaffected male. The first pair from the second generation produces 2 unaffected females. The second pair produces an unaffected male, an affected female, an affected male, and an unaffected female. The third pair produces an unaffected female, an unaffected male, and an unaffected female. The fourth pair produces an affected male and an unaffected female.

The pedigree above shows the sex-linked inheritance pattern of a gene. This pattern of inheritance suggests the allele is:

A. This question requires the examinee to analyze patterns of inheritance and the relationship between genotypic and phenotypic frequencies. The allele cannot be inherited as Y-linked, since females in the first generation of offspring express the trait. If the trait is X-linked recessive, the affected females of the first generation must be homozygous recessive and their offspring should show sex linkage, with the males affected and the females unaffected. This is not the case. The pattern in the pedigree is consistent with an X-linked dominant, however. The affected females of the first generation must be heterozygous for the trait so the expression of the trait in their offspring would not show sex linkage. This is the pattern in the pedigree.

### Descriptive Statements:

• Demonstrate knowledge of population genetics (e.g., Hardy-Weinberg), the mechanisms of natural and artificial selection, and the sources and significance of variation in populations.
• Analyze evolutionary patterns and the mechanisms of speciation.

### Sample Item:

The occurrence of similar features in organisms from different phylogenetic lineages can be attributed to:

2. stabilizing selection.
3. convergent evolution.
4. uniformitarianism.

C. This question requires the examinee to analyze evolutionary patterns and the mechanisms of speciation. Organisms that are only distantly related may adapt by developing similar characteristics in response to similar selective pressures. This is known as convergent evolution.

### Descriptive Statements:

• Demonstrate knowledge of the geologic history of Earth, current scientific theories on the origin of life, biologically significant events in Earth's history, and the fossil record.
• Demonstrate knowledge of the principles of biological classification, phylogenetic trees and their cladistic basis, evolutionary relationships of major groups of organisms, and evolution as a unifying principle in biology.
• Analyze different kinds of scientific evidence for evolution.

### Sample Item:

Which of the following would provide the strongest evidence of a relatively close evolutionary relationship between two species?

1. Both species occupy similar niches in their respective ecological communities.
2. Fossils of ancestors of both species are found in the same geographic area.
3. The embryos of both species look almost identical until late stages of development.
4. Both species use the same 20 amino acids to synthesize peptides and proteins.

C. This question requires the examinee to analyze different kinds of scientific evidence for evolution. Homologous patterns of development that appeared early are shared by all members of an evolutionary lineage. Homologies shared by some but not all members of a lineage suggest that the species sharing these homologies have a more recent common ancestor than other members of the lineage. Thus species whose embryos differ from each other only in the very late stages of development are likely to be more closely related to each other than species that diverged at earlier stages of development.

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## Multiple Choice Questions

What can be the correct conclusion for the following family?

1. Mother is heterozygous.
2. Parents could not have a normal daughter for this character
3. The trait under study could not be color-blindness
4. Father is homozygous dominant

Ans: (1) Mother is heterozygous.

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## NCERT Solutions For Class 12 Biology Chapter 5 Principles of Inheritance and Variation

Get NCERT Solutions for Class 12 Biology Principles of Inheritance and Variation in this step by step solution guide. In a number of State Boards and CBSE schools, students are taught through NCERT books. As the chapter comes to an end, students are asked few questions in an exercise to evaluate their understanding of the chapter. Students often need guidance dealing with these NCERT Solutions. It&rsquos only natural to get stuck in the exercises while solving them so to help students score higher marks, we have provided step by step NCERT solutions for all exercises of Class 12 Biology Principles of Inheritance and Variation so that you can seek help from them. Students should solve these exercises carefully as questions in the final exams are asked from these so these exercises directly have an impact on students&rsquo final score. Find all NCERT Solutions for Class 12 Biology Principles of Inheritance and Variation below and prepare for your exams easily.

### Mention any two autosomal genetic disorders with their symptoms.

1. Sickle cell anaemia: This autosomal recessive genetic disorder is caused by the substitution of glutamic acid by valine at the sixth position of the beta-globin chain of the haemoglobin molecule. This point mutation causes a physiological change in the haemoglobin, which in turn changes the shape of the RBC from biconcave disc to sickle-like structure.

Symptoms: Shortness of breath, dizziness, headaches, cold hands and feet and pale jaundiced skin.

1. Phenylketonurea: This autosomal recessive genetic disorder leads to the absence of an enzyme that converts the amino acid phenylalanine to tyrosine. This causes the accumulation of phenylalanine, which in turn leads to the associated symptoms.

Symptoms: Mental retardation, seizures, delayed development, behavioural problems, and psychiatric disorders.

### Who had proposed the chromosomal theory of the inheritance?

The chromosomal theory of inheritance was proposed by Walter Sutton and Theodore Bovery in 1902. The theory linked the inheritance of traits described in Mendelian Laws to the inheritance of chromosomes.

Explanation: The theory explains that chromosomes carried the factors described in Mendelian inheritance. It also stated that the chromosomes are linear structures with genes located at specific sites called loci along their length.

### What is point mutation? Give one example.

A gene mutation involving the substitution, addition, or deletion of a single nucleotide base is called a point mutation.

Sickle-cell anaemia is caused by the substitution of glutamic acid by valine at the sixth position of the beta-globin chain of the haemoglobin molecule. This substitution is caused due to a point mutation that changes the nucleotide A to T in the coding strand of the DNA.

Sequence for Normal Haemoglobin (HbA gene)

Sequence for Sickle Cell Haemoglobin (HbS gene)

Point mutations can have one of three outcomes.

Altered codon corresponds to the same amino acid

Altered codon corresponds to a different amino acid

Altered codon corresponds to a stop signal

Point mutations may arise from spontaneous mutations that occur during DNA replication. The rate of mutation may be increased by mutagens which are physical, chemical or biological agents that increase the frequency of mutations above the normal level. Mutagens associated with cancers are often studied to learn about cancer and its prevention.

Sickle-cell anaemia is an autosome linked recessive trait that is transmitted from parents to offspring when both parents are carriers of the point mutated gene. The disease is controlled by a single pair of allele, HbA and HbS. Double recessive individuals of genotype HbSHbS show the disease phenotype. The heterozygous individuals are only carriers of the disease and are unaffected.

### Explain the following terms with example (a) Co-dominance (b) Incomplete dominance

(a) Co-dominance is a genetic phenomenon where the F1 generation resembles both parents i.e. the two alleles of a gene are equally dominant and both are expressed in a heterozygous condition. ABO blood group is an example of co-dominance. The ABO blood group is controlled by gene I which has three alleles IA, IB and i. IA and IB produce two different types of antigens on the RBC while the I allele doesn’t produce any antigen. In an individual, with the genotype IAIB both the A and B antigens are expressed on the RBC because of co-dominance.

(b) Incomplete dominance refers to a phenomenon whereby one allele does not completely dominate another allele, and therefore the progeny resembles neither of the parents resulting in a new phenotype which is a mixture of parental phenotypes. For example, a cross between the purebred red (RR) and white (rr) flowered plants of snapdragon (Antirrhinum sp.). The F1 generation (Rr) of such a cross yields 100% pink flowers as opposed to the expected red colour. This is because the RR genotype is partially dominant in the recessive trait of white flowers. Therefore, the white pigment of the flowers is also expressed resulting in pink colour flowers.

Parents: Red (RR) X White (rr)

Phenotype – 100% flowers Pink

Genotype – 100% Rr (Red trait shows incomplete dominance)

Parents: Pink (Rr) X Pink (Rr)

### A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offspring.

The child of a set of parents that have blood group A and B can only be of blood group O if the genotype of the parents is heterozygous having one allele as the gene “i” that expresses neither antigen A or B. Therefore the genotypes of the parents and the offspring will be:

The following Punnet cross explains how the various genotypes and phenotypes are derived:

Parent: Father (I A i) X Mother (I B i)

Progeny Genotype – I A I B : I A i: I B i:ii::1:1:1:1

Progeny Phenotype – AB:A:B:O:: 1:1:1:1

Human blood type is determined by co-dominant alleles. It means that in a heterozygous situation, both alleles are expressed. In the case of blood type, there are three different alleles that result in variable expression of antigens on the RBCs.

No Antigen (neither A nor B)

Each of us has two ABO alleles, one from each parent. A pair of allele is called the genotype for that trait. Since there are three alleles 6 different genotypes are possible at the ABO genetic locus.

*Two recombinations result in AB genotype resulting in 6 different genotypes.

### How is sex determined in human beings?

Sex determination in humans is associated with sex chromosomes that are different between male and female individuals. Normal human females have two sex chromosomes - XX. The normal human male has 2 sex chromosomes - XY. The males produce two different types of gametes one having the X chromosome and one with the Y chromosome, while all-female gametes have an X chromosome. The sex of the baby is determined by whether the fertilising sperm contains the X or the Y chromosome. There is an equal probability of having a girl or a boy for each fertilisation event.

Parent: Male (XY) X Female (XX)

### What is pedigree analysis? Suggest how such an analysis can be useful.

Pedigree analysis is a type of genetic analysis that analyses the pattern of inheritance of a particular trait, disease or abnormality. This type of analysis is made in several generations of a family using special symbols and lines.

Uses of pedigree analysis:

1. Can help to determine if the trait is dominant or recessive
2. Whether the trait is linked to the sex chromosome

Predict and trace the pattern of inheritance of single mutation disorder like Haemophilia, sickle cell anaemia etc so that parents who are carrying the mutated genes or belong to families afflicted by these disorders can make an informed decision about their progeny.

### Briefly mention the contribution of T.H. Morgan in genetics.

Morgan formulated the theory of inheritance of chromosomes. He performed several dihybrid crosses in Drosophila to demonstrate linkage and recombination (the two terms coined by him) of genes on the X chromosome. He saw that some pair of contrasting characters did not segregate in the ratio of 9:3:3:1 (the expected outcome when the two genes are independent). Morgan attributed this to the genes being physically linked in such a way that they didn’t segregate independently of each other. Recombination was the event that was used to describe the generation of non-parental gene combinations. He demonstrated that tightly linked genes showed very low recombination while others that were loosely linked showed a higher rate of recombination. Morgan won the Nobel Prize in Physiology or Medicine in 1933 for his contribution to genetics.

### Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dibybrid cross?

When two genes are linked they do not follow the expected ratio for a dihybrid cross between heterozygous parents as seen in Mendel’s crosses (9:3:3:1). Instead, the phenotype ratio will be like that of a monohybrid cross if the two genes are very tightly linked because they will be inherited together. Recombinant phenotypes may also appear in low numbers or varying numbers depending on the distance/extent of linkage between the two loci.

### When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be (a) Tall and green (b) Dwarf and green

A dihybrid cross between two parents differing in two pairs of contrasting traits: Plant height and seed colour was made using a Punnet square.

Number of offspring with phenotype

Explanation: This phenomenon is based on Law of Independent assortment coined by Mendel in which he states that when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

### Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Let us assume that the man is heterozygous for the blood group antigen AB and the female is homozygous BB. The first filial generation of the two parents will have the following distribution:

Parents: Female (BB) X Male (AB)

Therefore in this situation, the phenotypic and genotypic ratio of the progeny will be the same. Also, the genotype of the female and male will be equally expressed.

### Define and design a test-cross.

A test cross is when an organism exhibiting the dominant phenotype is crossed with the homozygous recessive parent to determine the genotype of the former by analyzing proportions of offspring displaying the recessive phenotype. If all offspring display the dominant phenotype, the individual in question is homozygous dominant if the offspring displays both dominant and recessive phenotypes, then the individual is heterozygous.

Plants with violet flowers (dominant trait unknown genotype WW/Ww) were crossed with a purebred line yielding white flowers. The outcome of the experiment determines the genotype of the parent with the dominant trait. If all offspring yield violet flowers the parent with the dominant trait would be homozygous (WW) whereas, if the offspring yield both violet and white flowers 50% of the times the parent with the dominant trait would be heterozygous (Ww).

Parents: Violet (unknown parent could be WW/Ww) X White (ww)

Phenotype – 100% flowers Purple

Inferred parental genotype - WW

Parents: Violet (unknown parent could be WW/Ww) X White (ww)

Phenotype – 50% flowers purple, 50% flowers white

Inferred parental genotype - Ww

Offspring Genotype – 50% Ww, 50% ww

### Explain the Law of Dominance using a monohybrid cross.

According to Mendel’s Law of Dominance, a physical trait is controlled by a pair of factors (also called alleles). When the factors are dissimilar one factor dominates (dominant factor) the other (recessive factor). But the recessive factor is not lost and appears in the second generation.

This law is explained by a monohybrid cross which is a cross of two parents that are pure for one contrasting trait, for example, tall (TT) and dwarf (tt). The F1 generation yields all the progeny with a heterozygous genotype Tt that express the dominant trait – Tall. The F2 generation which is a self cross (TtxTt) yields progeny with a phenotypic ratio of 3:1 . tall:short and a genotypic ratio of TT:Tt:tt::1:2:1. The recessive trait (dwarf plant) appears in the F2 generation which was masked by the dominant trait (tall plant) in the F1 generation.

Parents: TT (tall) X tt (dwarf)

Parents: Tt (tall) X Tt (tall)

### A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Loci means the position on the chromosome at which a gene lies. If an organism is heterozygous for 4 loci it means that for four genes the organism has dissimilar alleles. During meiosis the alleles will segregate in 16 possible ways to form 16 different types of gametes given that the genes are not linked. If the genes are linked then the outcome could vary as two or more alleles could co-segregate.

Explanation: To calculate the number of possible gametes when an organism is heterozygous for a locus the formula is as follows:

Number of heterozygous genes in the organism

Number of types of gametes

*The above holds true for genes that are not linked and segregate independently.

### Differentiate between the following – (a) Dominance and Recessive (b) Homozygous and Heterozygous (c) Monohybrid and Dihybrid

(a) A dominant factor or allele expresses itself in the presence or absence of the recessive allele. For example in a pea plant tall, violet flowers, green pod etc. are dominant factors. A recessive factor or allele cannot express itself in the presence of the dominant character. For example in a pea plant dwarf, white flowers, yellow pods are recessive factors.

(b) A homozygous genotype is the one in which both the alleles for a trait are similar. The genotype can be homozygous for a dominant (TT) or a recessive (tt) trait. All gametes produced by the homozygous organism will carry the same allele. A heterozygous genotype is one in which the two alleles for a trait are different. This type of genotype will have both dominant and recessive allele for example Tt. The gametes produced will be of two kinds one will only carry the dominant allele and one will carry the recessive allele.

(c) A monohybrid cross is a cross between two pure breed parents that differ in only one pair of contrasting character. For example a cross between tall (TT) and dwarf (tt) pea plants. A dihybrid cross is a cross between two parents that differ in two pairs of contrasting characters. For example a cross between pea plants having round yellow (RRYY) and wrinkled green (rryy) seeds.

### Mention the advantages of selecting pea plant for experiment by Mendel.

Mendel chose the pea plants to carry out his genetics experiment because of the following features:

1. Pea plants have naturally occurring visible contrasting traits such as purple/white flowers, tall/dwarf plants, yellow/green pods etc.
2. Peas are self-pollinating plants i.e. pollen from a flower fertilises the eggs of the same flower giving rise to purebred lines having the same trait generations after generations.
3. Pea plants can be easily cross-pollinated by emasculation whereby, the stamen of the plant is removed and the pistil is dusted with the stamen of the desired parent.
4. The life cycle of pea plants is short and they produce many seeds in one generation enabling the statistical analysis of the result.

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## Solving a Pedigree Between Heterozygous Half-Cousins - Biology

Note: Click on the figures to obtain a clearer image.

1) 12 pts . Consider the following pedigrees, all involving a single trait. Briefly explain for each case whether any of the following modes of inheritance can be excluded: X-linked dominant, X-linked recessive, autosomal dominant, autosomal recessive.

X-Dom: excluded, no affected parents.

X-rec: possible, mother could be heterozygous

A-Dom: excluded, no affected parents

A-rec: possible, both parents heterozygous.

X-Dom: excluded, no affected parents.

X-rec: excluded, father would have had to be affected.

A-Dom: excluded, no affected parents

A-rec: possible, both parents heterozygous.

X-Dom: excluded, all daughters should have expressed.

X-rec: possible, mother could be heterozygous

A-Dom: possible, male parent heterozygous.

A-rec: possible, female parent heterozygous.

2) 8 pts. Among native Americans, two types of cerumen (ear wax) are seen, dry and sticky. A lucky geneticist assigned to study the inheritance of this trait observed the offspring produced by different kinds of matings. The results were:

(a) How is earwax type inherited? Briefly explain.

Earwax seems to be inherited as a simple dominant or recessive Mendelian trait. Sticky appears to be dominant and Dry is recessive. If so, crosses between "Dry" individuals should produce only dry offspring while crosses between sticky and dry and sticky and sticky would be expected to give rise to both types.

(b) Why are there no 3:1 or 1:1 ratios in the data?

In order to get Mendelian ratios you have to examine true breeding lines or individuals with known genotypes. In this case the Sticky individuals were unknown combinations of homozygous DD and heterozygous Dd individuals. Only the "Dry" d/d individuals can be expected to be true breeding.

3) 8 pts . A married couple went to see a genetic counselor because each had a sibling with cystic fibrosis. CF is a recessive disease and neither member of the couple nor any of their four parents are affected.

(a) What are the chances that their child will be affected with CF?

The change that either parent is a carrier is 2/3. Thus the chances of having a child with genotype cf/cf is 2/3 X 2/3 X 1/4 = 4/36 = 1/9.

(b) What is the probability that their child will be a carrier of the CF mutation?

The child would be a carrier if its genotype were Cf/cf (or cf/+).

This could occur if both parents were heterozygotes:

Or if one parent is heterozygous:

Thus the probability is 2/9 + 2/9 = 4/9.

4) 18 pts. Short Answer section : Answer the following questions with a short (one or two sentence) explanation or by the appropriate calculations.

(a) The red fox has 17 pairs of large long chromosomes. The arctic fox has 26 pairs of smaller shorter chromosomes.

(i) What do you expect to be the chromosome number in the hybrid somatic tissues?

Each parent would contribute its haploid complement of 17 or 26 chromosomes giving a total of 43 chromosomes per somatic cell.

(ii) What problems would the hybrid face in terms of mitosis and meiosis? Would you expect the animals to be viable and/or fertile? Briefly explain.

Because the homologs do not pair in mitosis, there should be no problem with segregation and the progeny should develop into adults.

Because many chromosomes would be expected to be unpaired they will not align in meiosis. In meiosis I the sister chromatids will both go to one pole leaving the other cell deficient for this information. With so many chromosomes it's vanishingly unlikely that any gamete will receive a balanced euploid set so the hybrids would be sterile.

(b) What is the standard used to establish linkage in LOD score analysis? What is the meaning of a negative (-) value for a LOD score?

A LOD score of 3.0 is taken as certain evidence of linkage. This indicates that the probability of a certain degree of linkage is at least 1000 X greater than the probability that the genes are unlinked. A negative LOD score means the assortment of the markers in the pedigree is better explained by independent assortment than by linkage.

(c) Recall the dominant piebald spotting trait in cats.

(i) When comparing the phenotype of one animal to another, what genetic term is used to explain the Piebald phenotype?

Piebald exhibits variable expressivity. Different S/s animals express different degrees of the white spotting phenotype, but all have some white spots.

(ii) From the perspective of comparing the individual hair-producing cells in an Ss cat, what genetic term would you use to describe the Piebald phenotype?

Individual S/s hair cells will be either white or colored. This is a qualitative difference and each cell has a variable penetrance of the phenotype. Some S/s cells express white others do not. For part (i) all cats express some white, but the degree of expression varies.

5) 12 pt . (a) Why do two different calico cats have their patches of orange and black fur in different places? Explain whether or not such a variegated coat pattern could occur in marsupials.

Note: If you don't know the marsupial mechanism you didn't try very hard to solve problem 5 on the practice problems for the first exam. marsupial dosage compensation problem.

Calico cats are heterozygous for the O and o alleles of the X-linked orange gene. Because of random X inactivation some cells express ( O ) Orange fur color and some cells ( o ) express black fur color. Because the inactivation process is random no two such cats look alike.

In marsupials, the paternally-derived X is always inactivated. This means that females will always express from the maternal chromosome so any individual animal would have a uniform coat color for an X-linked gene.

(b) In fruit flies, both eyes of females heterozygous for functional, partially functional, or non-functional white alleles are of uniform color. This is true regardless of the parental origin of the alleles. Does this exclude the common mammalian mechanism of X dosage compensation? How about the marsupial mechanism? If the mechanism is excluded, explain why if not, explain and offer other evidence in favor of, or in opposition to, the proposed mechanisms.

The mammalian mechanism is excluded. Since white is cell autonomous, female flies would express a mixture of white and pigmented eyes if one of the X's were randomly inactivated.

The marsupial mechanism is possible given only the information in this problem. If females always inactivated the male X (and thus the white gene) they eyes would be a uniform eye color. However, from numerous examples in the text and lecture, you know that crosses between w/w females and +/Y males produce red eyed progeny which would exclude inactivation of the paternal X.

Extra credit--up to 4 pts . Two identical twin males marry identical twin women. What are the relationships between the couple's children if you're:

(c) a headline writer for the Post?

6) 18 pts . You wish to map the location of a new Drosophila X chromosome alteration called inv (for inviable ) that was isolated after treatment with g -rays. You decide to map the lesion genetically. You set up the following preliminary cross:

The yellow gene ( y ) maps at the left end of the X chromosome at 0.0 cM and the Bar eye ( B ) gene near the right end at 57.0 cM. The Bar allele is a dominant mutation that changes the eye from its normal smooth oval shape to a bar shaped structure.

The cross produced the following male progeny:

(a) What can you say about inviable ? Is there anything odd about its position? Show, or briefly explain, how you arrived at your answer.

As the name implies, inviable seems to be lethal. This is evidenced by the recovery of large numbers of the inv + parental chromosomes (Bar phenotype) and very few of the Yellow phenotype expected for the other parental inv chromosome. Thus the recombinant progeny types must have had the following origins:

There were 15 + 50 y-inv recombinants out of 500 = 13% recombination

There were 105 + 15 inv-Bar recombinants out of 500 = 24 % recombination.

This would put inv 13 mu to the right of y at 13 cM and 24 mu to the left of Bar at

57-24 = 33 cM. Something is wrong if the lethal maps to two different locations.

The sum of the map distances between y and Bar in this experiment is 13 + 24 = 37 mu which is less than the expected 57 units. This implies that something was present that inhibited recombination. In theory, this could be a deletion or an inversion. Since the region is so large, 20 mu, a deletion would be lethal in the heterozygous state so inv is most likely an inv ersion with one breakpoint at

(b) Predict the number and phenotypes of the female progeny.

If there were no inversion one would have expected independent assortment for genes this far apart. Since all females would have been viable you should have seen 250 yellow, 250 wt, 250 Bar and 250 yellow Bar females.

Since recombination is suppressed in the inverted region, you expect 37% recombination between y and Bar . Thus, you would predict 630 parental types and 370 recombinant types or 315 yellow, 315 Bar, 185 wild-type and 185 yellow Bar recombinants.

7) 24 pts . The insect compound eye is made up of a repeating pattern of hexagonal facets called ommatidia. A number of mutations can alter the structure of the ommatidia causing a disorganized eye phenotype that changes the normal smooth texture of the eye to a rough one.

Two new recessive mutations with identical "rough eye" phenotypes have been isolated. rufX maps on the X and ruf2 on the 2nd chromosome. The rufX and ruf2 mutations are thought to affect signaling molecules. One gene is predicted to produce an extracellular signaling ligand that instructs cells to adopt their proper eye fate. The other gene encodes a cell surface receptor thought to be required to sense the extracellular signal.

To test this hypothesis, the two ruf mutations were analyzed in somatic mosaics. For rufX , the X-linked white gene was used to mark the rufX mutant tissue.

(a) Diagram how you would arrange the rufX and white alleles on the homologous chromosomes so that you could generate the rufX clones and identify them by their eye color phenotype. Explain your answer (a diagram may be most convenient.)

Map of the X chromosome, white maps at 1.5 cM, rufX at 5.0 cM.

To generate somatic rufX clones that are marked with white - in a wild-type background the chromosomes should have been arranged as follows: w rufX / + + .

(b) The following figure shows the eye phenotype of a typical rufX mosaic. Is rufX likely to encode the extracellular signaling molecule or the receptor for that signal? Explain your answer.

rufX most likely encodes the receptor molecule. All rufX mutant cells have the mutant phenotype suggesting the wild-type gene product acts cell autonomously. If rufX encoded the extracellular signaling molecule, mutant cells at the edge of the clone should have had the wild-type phenotype as their wild-type neighbors could have provided a functional signal.

This might have looked like:

c) The ruf2 gene maps to the left arm of chromosome 2 where there are no eye color genes suitable for use in mosaic analysis. Can you think of a way that you could use white to mark ruf2 clones? You may explain with words or a diagram. Be sure to include all relevant genotypes. You may invent any reasonable genetic tool to assist you.

The simplest way to mark the second would be to use a duplication of the w+ gene on the second chromosome. Such a duplication could have arisen from an X:2 translocation. The chromosomes are diagrammed below. The key difference with part (a) is that the X chromosomes must carry w - alleles so that the white ruf2 clones can be identified.