Genetics Exam 2

Discussion Question

1. I have a pure breed yellow Labrador Retriever that happens to be black. My dog should be yellow. Can you explain how I know I should have a yellow dog? 2. Based upon acceptable Labrador genetics it is not possible to get a black dog and therefore mis-mating must have occurred. According to the breeder, mis-mating did not occur. According to the literature there are too many of these "mis-matings" to be accidental. Can you explain possible reasons (other than mis-mating) for my dog to be black?

Autosomal Recessive trait (table typed out)

1. Usually appears in both sexes with equal frequency 2. tends to skip generations 3. Affected offspring are usually born to unaffected parents 4. When both parent are heterzygous, approximately one-fourth of the offspring will be affected. 5. appears more frequendly among the children of consanguine marriages.

X-linked recessive trait (table typed out)

1. Usually more males than females are affected 2. affected sons are usually born to unaffected mother; thus, the trait skips generations 3. Approximately half of a carrier (heterozygous) mother's sons are affected 4. Never passed from father to son 5. All daughters of affected fathers are carriers

X-linked dominant trait (table typed out)

1. both males and females are usually affected; often more females than males are affected 2. does not skip generations. Affected sons must have an affected mother; affected daughters must have either an affected mother or father 3. affected fathers will pass the trait on to all their daughters 4. affected mothers (if heterozygous) will pass the trait on to half of their sons and half daughters

Y-Linked trait

1. only males are affected 2. passed from father to all sons 3. does not skip generations

common reasons for seeking genetic counseling

1. person knows of a genetic disease in the family 2. a couple has given birth to a child with a genetic disease, birth defect, or chromosomal abnormality. 3. a couple has a child who is intellectually disabled or has a close relative who is intellectually disabled. 4. an older woman becomes pregnant or wants to become pregnant. There is a disagreement about the age at which a prospective mother who has no other risk factor should seek genetic counseling; many experts suggest 35 5. husband and wife are closely related 6. A couple experiences difficulties achieving a successful pregnancy 7. a pregnant woman is concerned about exposure to an environmental substance (drug, chemical, or virus) that causes birth defects 8. a couple needs assistance in interpreting the results of a prenatal or other test 9. Both prospective parents are known carriers for a recessive genetic disease or both belong to an ethnic group with a high frequency of genetic disease

Characteristics of cytoplasmically inherited traits

1. present in males and females. 2. usually inherited from on parent, typically the maternal parent 3. Reciprocal crosses give different resultts 4. Exhibit extensive phenotypic variation, even with a single family.

Autosomal dominant trait (table typed out)

1. usually appears in both sexes with equal frequency 2. both sexes transmit the trait to their offspring 3. does not skip generations 4. affected offspring must have an affected parent unless they possess a new mutation. 5. when one parent (heterozygous) and the other parent is unaffected, approximately half of the offspring will be affected 6. unaffected parents do not transmit the trait

Heterozygote X testcross

A heterozygote for two linked genes crossed to a test cross produces more "Parental Type" offspring than "Recombinant Type offspring" M= normal, m = mottled, D= normal and d= dwarf. MD/md X md/md MD/md > ¼ Parental Type Md/md > ¼ Parental Type Md/md < ¼ Recombinant Type mD/md < ¼ Recombinant Type This is illustrated in the next two slide

ABO Blood Group

A person with blood type A has at least one A allele (iA). A person with blood type B has at least one B allele (iB). A person with blood type AB has one A allele (iA) and one B allele (iB). A and B ore co-dominant and each is expressed when present. The recessive allele (i) codes the O allele. A person with O blood type has an ii genotype because the i is recessive to the co-dominant iA and iB alleles. A person will illicit an immune response against a blood transfusion with the wrong type blood. Universal recipients (iAiB individuals) can get blood from any of the other genotypes. An ii individual is a universal donor and can give blood to any of the other genotypes but can only receive blood from ii genotypes. Please note there are other blood factors that can cause immune response issues. What are the possible genotypes for B blood type, AB blood type and O blood Type? The next slide illustrates all possible genotypes and the consequences for mixing the wrong blood groups.

What is heterogametic sex?

An individual that has two different types of sex chromosomes (XY)

What is homogametic sex?

An individual that has two of the same sex chromosomes (XX)

Are X and Y homologous??????

Are the X and Y homologous? Given 1) We have defined homologous chromosomes as identical in structure and similar in content. 2) We claim that homologous chromosomes attach to each other through the synaptonemeal complex and therefore are able to line up on the metaphase plate during Meiosis I. 3) The X chromosome is a metacentric chromosome. The Y chromosome is acrocentric. 4) The X and Y chromosomes have very few genes in common. Answer: Yes and no. The X and Y chromosomes do have some homologous regions called the psuedoautosomal regions that allow them to form synaptonemeal complexes. They are different at other regions. They do pair and line up on the Metaphase I plate and they separate during Anaphase I.

Co-dominance

Both allele are completely expressed

What does SRY do?

Causes testes to form which produces testosterone. Testosterone normally causes male development. XY individuals are male.

Complete verses Incomplete Dominance

Complete Dominance is when one allele is expressed when present and the other allele is only observed in the homozygote recessive state Incomplete Dominance is when the expression of one allele does not completely mask the expression of the second allele. The phenotype is intermediate between the phenotype of either homozygote. If a pure breeding red flower snap dragon mates with a pure breeding white flower snap dragon the offspring have pink flowers. The pink is intermediate between red and white. In the case of pink snap dragons, looking at the pigment of their individual cells, one observes red cells and white cells which appear pink to our naked eyes.

Complete verses Co-Dominance

Complete Dominance is when one allele is expressed when present and the other allele is only observed in the homozygote recessive state. Co-Dominance is when both alleles are completely expressed. This often appears as a Blended Inheritance even though it is not. The Eggplant in the next slide shows that when two heterozygotes are mated the offspring produce heterozygotes with the same characteristics (violet) or pure breds with different characteristics (purple or white). The heterozygote is expressing both alleles equally in all cells in which the gene is active.

What organism's sex is determined by an environmental factor- the position in the stack?

Crepidula Fornicata

Sex-Linked traits: XY organisms - Experiment example part II

Differences between the numbers of male and female offspring indicate sex linkage. If you complete the two crosses on the previous slide you notice that a white eyed homozygous recessive female crossed to a red eyed hemizygous male gives all white eyed sons and all red eyed daughters. This indicates the trait is sex-linked. The father gives his X chromosomes with the red allele to his daughters. Since red is dominant his daughters express the red eye color. The mother gives her X chromosomes with the white allele to her children. This is then expressed in her hemizygous sons since they do not have a comparable allele from their father. They inherit the Y chromosome from their father. Note the reciprocal cross yielded all red eyed offspring. From this cross you can not tell sex-linkage until the F2 generation. The next slide illustrates the F1 and F2 generations of these reciprocal crosses

What is unique about the duckbill platypus?

During meosis, the sex chromosomes form chainlike structures.

Epistasis

Epistasis is when many genes influence one characteristic. In Labrador retrievers there are black, chocolate and yellow labs. Black labs make the black pigment (B--) and lay down the pigment in their hair follicles (E--). Chocolate labs do not make the black color (bb) but do lay pigment into the hair follicles (E--). Yellow labs with black noses make the pigment (B--) but can not place it in the follicle (ee). Yellow labs with brown noses do not make the black pigment nor lay down any pigment (bbee). The ability to lay down the pigment is epistatic making the pigment. This is recessive epistasis since yellow labs must be homozygous recessive for the ee locus.

Expressivity and Penetrance

Expressivity represents the degree to which individuals with a particular genotype demonstrate that genotype. If a trait is 100% expressive then the characteristic of that trait appears as all with anyone having the genotype for the characteristic or none with anyone lacking the alleles for the characteristic. Penetrance represents the number of individuals who have a particular genotype demonstrating that genotype. If a trait is 100% penetrant then everyone with that genotype demonstrates the genotype. If it is 50% penetrant then half the people with the genotype demonstrates that genotype. Notice: I have used the word demonstrates instead of expresses. Expresses is the correct terminology but gets confusing when you want to differentiate the two conditions Many genes have variable degrees of expressivity and penetrance. The next slide shows an individual with polydactyly (extra digits). This disorder shows variable expressivity and incomplete penetrance which means that a person with the dominant allele for polydactyly may not demonstrate polydactyly or may have a complete extra finger or a partial extra finger. Expressivity and penetrance are influenced by epigenetic factors. Epigenetic factors are factors other than the immediate genes that regulates the expression of these genes. Epigenetic factors are often environmental factors. Environmental factors can be external or internal factors. Temperature on the sexual phenotype of turtles is an external environmental factor. Modifiers, enhancers or silencers are internal environmental factors. Often these are proteins that influence the expression of other genes. They can enhance expression producing more gene product. They can silence expression preventing the gene from transcribing. They modify the gene product influencing its function or half life.

Genetic Counseling

Genetic counselors take to individuals about their probability of having a special child. This can be done before the couple have their first child or after the first afflicted child is born. The slide is a table as to why people seek genetic counselors. Genetic counselors can give couples probabilities and other materials relating to the disorder but will never input their personal opinions. The decision to have a child is up to the parents.

Complete, Incomplete and Co-Dominance concepts

Important concepts to note is: 1. that these definitions are for the expression of the genes. The genes are inherited as monohybrid crosses via Mendelian Genetics. 2. that the definitions for complete, incomplete or co-dominance for any given gene seem to change at different levels of the organism. It may appear as co-dominant on the molecular or cellular level, incomplete dominance on the tissue level and complete dominance on the organismic level.

Sex determination: Birds

In Birds, snakes, some amphibians, and fish the males are ZZ and females are ZW. This is the opposite for other organisms so we use to ZW symbols as opposed to XY symbols.

Sex determination: Drosophila

In Drosophila and many other insects the ratio of X chromosomes to sets of autosomes determines if the individual is male (one X: two sets of autosomes) or female (two X: two sets of autosomes). The default pathway is male. The X:A ratio

Review Independent Assortment

In independent assortment genes on different chromosomes line up independently from one another. This allows for you to have 4 potential gametes in equal numbers for two genes (remember 2n). See next slide for review

Linkage

In linkage the genes are located on the same chromosome. These genes will segregate but are not independent from each other. We still expect 4 potential gametes but no longer in equal numbers. If two genes are very close to each other on the same chromosome there is a high probability that they will remain on the same sister during meiosis. As the distance between two genes increases so does the probability that they will crossover with the other sister chromatids. Crossover with a sister on the same chromosome does not lead to an exchange of different alleles and therefore is not noticeable. Crossover with one of the sisters of the homologue can lead to an exchange of different alleles. If the alleles are exchanged we have recombination of the genes. Crossover events are repaired correctly 50% of the time and incorrectly 50% of the time. Therefore for each crossover event there is a 50% chance of recombination occurring.

Sex linked (X-chromosome) Pedigree

In sex-linked dominant pedigrees the dominant allele is passed from father to daughter. Therefore all the daughters and non of the sons of an afflicted father will have the disorder. If the mother is afflicted half of her children (regardless of gender) would be afflicted). A sex-linked dominant pedigree is illustrated in the next slide. In sex-linked recessive pedigree an afflicted father (with a genotypically normal spouse) passes his X chromosome to his daughters making them carriers. The sons will not be afflicted. A carrier mother (with a genotypically normal spouse) will pass the bad X to half of her sons and half her daughters. These sons would be afflicted and these daughters would be carriers. An afflicted mother (with a genotypically normal spouse) will have all afflicted sons and all carrier daughters. A sex-linked recessive pedigree is illustrated in two slides.

Sex-Linked Traits ZW organisms

In the previous slide (sex determination in birds) we discussed that male birds were the homogametic sex and females were the heterogametic sex. The next slide illustrates a cross and its reciprocal for Sex-linked Traits in birds. Note: The method for working out the problem is the same as it would be for mammals except males are ZZ and females are ZW 4.15 The cameo phenotype in Indian blue peafowl is inherited as a Z-linked recessive trait. (a) Blue female crossed with cameo male. (b) Reciprocal cross of cameo female crossed with homozygous blue male.

What is sex determination?

It is the determination of the species sex. There are many mechanisms that determine the individual members of a species.

Creating a Genetic Map

Let us say we have 4 genes and we know the distances between some of these genes. A.Rose and Purple are 2 map units apart B. Purple and Grainy are 7 map units apart C. Rose and Waxy are 6 map units apart D. Rose and Grainy are 9 map units apart E. Waxy and Grainy are 3 map units apart Draw a genetic map. For my map I will use the first letter of each gene as a symbol First take the largest distance between two genes that has a third gene that is linked to them. Second place that gene in relation to the first two. Third find a fourth gene and place it in relation to any two already in place. Make sure you repeat until all statements are used. Note: Waxy was 3 from grainy. It could be three to the left or three to the right. Purple was 6 from waxy. It could be 6 to the right or six to the left. The point where the two meet is 6 to the right of purple and three to the left of grainy. Therefore Waxy was placed between purple and grainy.

Potential Gametes

Males produce sperm that can carry either an X chromosome or a Y chromosome. These are produced in equal numbers. As a result, the male gives his X chromosome to his daughters and his Y chromosome to his sons. The male determines the sex of the child (barring mutaions) Females produce ova that contain one X chromosome. As a result the female gives an X chromosome to each of her children.

Dosage Compensation Mechanisms

Mammals: X-inactivation - The all but one rule says that all but one X-chromosome will be inactivated in any given cell. Inactivated X chromosomes are heterochromatic and are referred to as Barr bodies. In normal males their single X chromosome is functional in each cell. In normal females one X-chromosome is functional and one is inactivated randomly in each cell. The next slide shows a normal female cat which is heterozygous for orange and black coat color. Coat color is located on the X chromosome in cats. Dosage compensation is a mechanism by which species compensate for the differences between the sexes at least with regard to the number of X-linked transcripts. Males and females may have a different number of X chromosomes but they produce the same number of X linked transcripts. It would be lethal for the species if the genes on the X chromosome were not regulated since the number of transcripts synthesized correlate to the number of proteins produced which in turn have specific functions in the cell. If you have too many copies in the female and too few copies in the male it upsets the balance. These differences must be compensated. Dosage compensation is a mechanism by which species compensate for the differences between the sexes at least with regard to the number of X-linked transcripts. Males and females may have a different number of X chromosomes but they produce the same number of X linked transcripts. It would be lethal for the species if the genes on the X chromosome were not regulated since the number of transcripts synthesized correlate to the number of proteins produced which in turn have specific functions in the cell. If you have too many copies in the female and too few copies in the male it upsets the balance. These differences must be compensated. Mammals: X-inactivation - The all but one rule says that all but one X-chromosome will be inactivated in any given cell. Inactivated X chromosomes are heterochromatic and are referred to as Barr bodies. In normal males their single X chromosome is functional in each cell. In normal females one X-chromosome is functional and one is inactivated randomly in each cell. The next slide shows a normal female cat which is heterozygous for orange and black coat color. Coat color is located on the X chromosome in cats. In individuals who have too many X chromosomes such as a male with Klinefelters syndrome all but one X-is inactivated. In a female with only one X and no Y chromosome (Turners syndrome) the single X chromosome is functional in all cells. It should be noted that Barr bodies are "leaky" which means that there are some genes that are transcribed completely and some others to low levels. It is due to this "leakiness" that the phenotypic characteristics of individuals with too many X chromosomes are observed. The next slide is a table that lists various syndromes with the number of Barr bodies - Note the all but one rule applies Drosophila: Hyperactivation of the single X-chromosomes in males. Male flies transcribe their X-chromosome at exactly 2x the rate that females transcribe each of their two X-chromosomes. End result is the same number of X-linked transcripts. C. Elegans; Hypoactivation of the X chromosomes. The female X chromosomes are hypoactivated so that the transcription rate of each X-chromosome is one half the rate of the males single X-chromosome.

Gene Interactions

Many biochemical reactions occur in tandem to produce an end product. These reactions often use different enzymes at each step. If one of the enzymes is mutated then the reaction stops. There may be one phenotype produced or different phenotypes produced if the stoppage occurs at an intermediate reaction. In the case of peppers the intermediates have different phenotypes as illustrated in the next slide

Epistatsis

Many genes influence one characteristic

Multiple Alleles

Multiple alleles at one locus is a very common situation and adds to the diversity of the phenotype. The formula [n(n+1)]/2 allows you to calculate the number of possible genotypes. If there are 2 alleles then [2(2+1)]/2 Is equal to [2(3)]/2 or 6/2 or 3 genotypes. These are the homozygous dominant, homozygous recessive and heterozygous conditions. What if there are 3 alleles? The next slide illustrates the relationship of 3 possible alleles governing the feather pattern of Mallard ducks. In this situation the restricted pattern (MR) is dominant to both Mallard (M) and Dusky (md) patterns. Mallard is dominant to Dusky The following represents the possible genotypes and their corresponding phenotypes M^R M^R :restricted M^R M: restricted M^R m^d: restricted MM: mallard M m^d: mallard m^d m^d: dusky

Pleiotropy

One gene influence many characteristics

Three Point Linkage

Three point linkage is when you examine three different linked genes. To determine the map distance between them. You can do this using only one cross. A heterozygous parent for all three genes crossed to a test cross parent. The next three slides illustrate the different the types of crossovers generically, resulting phenotypes, and the types of crossovers specifically, respectively.

What is Klinefelter Syndrome?

Persons with Klinefelter syndrome have a Y chromosome and two or more X chromosomes in their cells.

What is Turner Syndrome?

Persons with Turner syndrome have a single X chromosome in their cells.

Pleiotropy

Pleiotropy - one gene influences many characteristics. An example of this is the yellow locus in Drosophila. Individuals who are homozygous recessive or hemizygous for yellow (XYXY or XYY) have a yellow cuticle. The larvae have yellow mouthparts which makes them easy to spot. These individuals also have problems with their malpighian tubules and vision.

Prenatal/Postnatal Genetic Testing

Prenatal: The next three slides illustrate prenatal genetic testing. The first is a table of some diseases that can be tested, then a picture of an ultrasound of a in utero fetus, then the procedure for amniocentesis, and the procedure for chorionic villus sampling (CVS). Postnatal: Individuals from newborns to adults can be tested for many diseases. These can include biochemical tests or genetic screening

What makes a platypus unique?

Sex in the platypus is determined by sex chromosomes. Females have 10 X chromosomes and males have 5 X and 5 Y chromosomes.

What is SRY?

Sex-determining region on the Y chromosome

Sex Influences on Heredity

Sex-linked characteristic- genes located on the sex chromosome Sex-influenced characteristic- genes on autosomal chromosomes that are more readily expressed in one sex. Sex-limited characteristic- Autosomal genes whose expression is limited to one sex Genetic maternal efffect- Nuclear genotype of the maternal parent Cytoplasmic inheritance- cytoplasmic genes, which are usually inherited entirely from one parent Genomic imprinting- genes whose expression is affected by the sex of the transmitting parent

What's the difference between an X and Y chromosome

Size and shape. X is much larger and shaped like an X.

Complementation Test

The complementation test allows you to determine if a phenotype is due to a mutation at one location or if multiple locations are involved. If a mutation has occurred in one location then a homozygous for the mutation crossed to homozygous always produce homozygous offspring. If two locations are involved then a homozygous for a mutation crossed to a homozygous for the second mutation shows the dominant phenotype. This is illustrated in the next slide with an example on the following slide Clot and safranin are two separate genes encoding eye color in Drosophila. Both produce brown eyes in the homozygous recessive state. If you have a pure breeding clot strain and cross it to a pure breeding safranin strain the flies have red eyes. This demonstrates complementation

Configuration of Linked Genes

The configuration of linked genes could be MD/md or Md/mD Both these individuals are heterozygotes for the two genes. You do not have to have dominants on one chromosome and recessives on the other. Conversely, you can have dominants on one and recessives on the other. ******If you do not know the genotype of the heterozygote parent then take the parental type chromosome from the two most numerous offspring. The parental type chromosome was given intact to those offspring. One parental type chromosome to each type of two most numerous offspring. Example: If the parent of interest was Md/mD then a mottled Tall offspring was given a mD from the parent of interest and an md from the testcross parent. The other most numerous offspring would be a normal (non mottled) dwarf or Md/md. Therefore the parent of interest was Md/mD******

Expressivity

The degree to which individuals with a particular genotype demonstrate that genotype.

Working Three Point Linkage Problems

The easiest way to do three point linkage is by setting up a chart. The chart will compare the characteristics of the recombinant type offspring to the parental type offspring for any two traits. You will have three such comparisons (columns). Indicate in the chart P for parental and R (#) for recombinant where # represents that number of offspring. Note you should have P's in all three columns for your Parental type offspring and P in one column for each recombinant type. You should have P in 50% of your rows for any given column. Remember you are comparing the two genes listed at the top of the column for each recombinant type to the two parental types. Finally for each column calculate your map units by adding all your recombinants, dividing by total number of offspring X 100 Remember 1. The eight offspring shown in the previous slide are the same eight you would expect to get for independent assortment except you do not have a 1:1:1:1:1:1:1:1 ratio. 2. The two most numerous types are always your parental type. Do not go by order of listing 3. The genotype for any type of offspring would be the entire chromosome given by the parent of interest (in correct gene order) over the entire chromosome given by the test cross parent. If you do not do this then you are not correct. If you list just the information from the parent of interest then you are indicating haploid. If you list the like genes together you are indicating independent assortment.

Incomplete Dominance

The expression of one allele does not completely mask the expression of the second allele

Egg VS Sperm

The human egg is much larger than the human sperm. This is due to two phenomenon. 1. The oocyte is enlarged due to unequal cytokinesis during meiosis 2. The sperm is reorganized and compacted during the maturation process. Most internal organelles have been rearranged or removed to form the tightly packed head of the sperm which contains the male pronucleus (haploid) and an acrosome (helps in fertilization). There is significantly less cytoplasm in a mature sperm than in other cells.

Color Blindness

The most common form of color blindness in humans is a sex linked trait. This slide illustrates the mating between a color blind male and a normal female as well as the reciprocal mating

Drosophila melanogaster a model organism

The name Drosophila means dew lover and melanogaster means black bellied. These small organisms are a model system for studying genetics.

Imprinting

The next slide illustrates imprinting. The environment (egg or sperm) will influence the expression of a given gene. In humans an example of this would be Angelman Syndrome vs Prader Willi Syndrome. The mutation is the same the phenotypes are very different. The only environmental difference is paternal or maternal origin. Angelman Syndrome: a rare congenital disorder characterized by mental disability and a tendency toward jerky movement, caused by the absence of certain genes normally present on the copy of chromosome 15 inherited from the mother Prader Willi Syndrome: Prader-Willi syndrome is a genetic disorder usually caused by deletion of a part of chromosome 15 passed down by the father.

Maternal Effect

The next slide illustrates maternal effect. The egg actively transcribed and translated information from a diploid cell. The egg undergoes meiosis keeping this information. This information is used in early development to get the zygote rapidly developing into an embryo. This information is used very early in development before the zygotes genome starts to transcribe and translate its own information. The end result is that the offspring develops based upon the genotype of its mother. If it is a female its own offspring will develop based upon the females genotype. In Drosophila the daughterless mutation (da) is a maternal effect mutation. A da da mother can not produce daughters. A DA da mother crossed to a da da father will have both Da da and da da sons and daughters in equal number. The da da daughters can not produce the daughterless protein and therefore can not count the X-chromosomes. Male is the default pathway in flies and therefore these XX individuals attempt to hyperactivate both their X chromosmes and die early.

Temperature Sensitive Mutations

The next slide illustrates the difference between Himalayan rabbits reared at 20 degrees or below (permissive temperatures) and those reared above 30 degrees (non-permissive temperatures). Outdoor Siamese cats are darker in the winter than the summer and are darker in their extremities and ears than on the core body. 70% of individuals with cystic fibrosis have a temperature sensitive form of the mutation. Unfortunately the permissive temperature is below normal body. In each case stated above the lower temperature gives the mutant enzyme time to fold correctly and therefore function. The non-permissive temperature causes the enzyme to fold too rapidly and therefore is non-functional. Note the permissive temperature does not have to be colder than the non-permissive temperature.

Autosomal Dominant Pedigree

The next slide illustrates the pedigree for an autosomal dominant disorder. Notice: A person never gets this disorder unless a parent had the disorder (barring spontaneous mutations) The slide after that shows a child with Waardenburg syndrome - an autosomal dominant disorder.

Autosomal Recessive Pedigree

The next slide illustrates the pedigree for an autosomal recessive disorder. Notice: The parents maybe heterozygotes or the disorder and therefore may not demonstrate the disorder i.e. cystic fibrosis. If the disorder is not lethal then two afflicted individuals would have an afflicted offspring.

Distance Between Genes

The next slide illustrates the recombination rate between two genes on the same chromosome. This rate is correlated with the distance between the two genes. Total number of recombinants/Total number of offspring = frequency of recombination. Frequency of recombination X 100 = % recombination 1% recombination = 1 map unit. If two genes are 10 map units apart then they have a percent recombination of 10%. The total number of recombinants are 10 out of every 100 or 10/100 = 0.1 frequency. Note frequency is less than 1. There are two recombinants for every four possible offspring so the frequency is 0.05 for each type of recombinant How many should be recombinant if you have 200 offspring?

Completely Linked vs Independent Assortment

The next slide illustrates two genes for leaf type and plant height M= normal, m=mottled, D= normal, d = dwarf. If the genes are completely linked a heterozygote (MD/md) X homozygous recessive (md/md) produces ½ MD/md and ½ md/md. If the genes are on different chromosomes are independent a heterozygote (Mm Dd) X homozygous recessive (mm dd) produces ¼ Mm Dd, ¼ Mm dd, ¼ mm Dd, and ¼ mm dd. Note: All the genes for one chromosome are written together over the genes for its homologue. Genes on more than one chromosome are written separate. For linked genes the MD are written together over another md. For independent assortment it would be written Mm Dd

Review of Pedigree Analysis

The next slide is a table that compares the pedigree analysis for autosomal dominant, autosomal recessive, sex-linked (X-linked) dominant, sex-linked (X-linked) recessive or Y-linked

Pedigree Analysis

The next slide is the key used for human pedigrees. For this class either you will be provided a key or you can provide your own key for any genetic problem. Pedigree analysis allows the geneticist the opportunity to visually examine multiple generations of the same family to look for a pattern of inheritance. Patterns of inheritance are autosomal dominant, autosomal recessive, sex-linked (X-linked) dominant, sex-linked (X-linked) recessive or Y-linked

Cytoplasmic (Maternal) Inheritance

The next slide represents cytoplasmic or maternal Inheritance. Maternal inheritance is due to the mitochondria or chloroplast being present in the egg. These organelles do not enter via the sperm. The organism therefore will have the same mitochondrial genes as its mother, maternal grandmother, maternal uncle but different from its own father.

Example: Calculating Map Distance

The next slide shows a cross between a Brown body, straight winged cockroach (heterozygote) and a yellow body, curved wing cockroach (test cross). What are the dominant traits? The offspring are 63 Brown body, Straight wing 28 brown body curved wing 33 yellow body straight wing 77 yellow body curved wing Are these genes linked? You expect a 1:1:1:1 ratio of the offspring. If the ratios are not close then the quick answer would be yes. Even if the ratios are close work out the problem. Note if the genes are assorting independently you expect 50% parental type offspring and 50% recombinant type. I. Assume linkage to set up the problem. What is the genotype of the parent of interest? You are not given this in linkage form. Find your two most numerous offspring and figure out their genotype. These are our parental type offspring. I am using the key given in the figure for my symbols. 63 Brown body, Straight wing y+ cv+/y cv 77 yellow body curved wing y cv/y cv So my parental type chromosomes are y+ cv+/y cv, I took the first chromosome from each of the parental types. The other chromosome was provided by the test cross parent and just happens to be the same as one of the parental types. II. Add up the total number of recombinants 28 brown body curved wing y+ cv/ y cv 33 yellow body straight wing y cv+/ y cv 61 III. Add up total number of offspring 63 Brown body, Straight wing 28 brown body curved wing 33 yellow body straight wing 77 yellow body curved wing 201 Total number recombinant/total number offspring = Frequency of recombination 61/201 =0.30 frequency % recombination = 0.30 X100 = 30% or 30 map units 30 is less than 50 and therefore the genes are linked. Note: Crossover events are repaired correctly 50% of the time. Also Independent assortment would give us 50%. Therefore at 50 map units we can not differentiate linkage from independent assortment. We assume independent assortment unless shown otherwise. If two genes are greater than 50 map units apart we expect to get a mathematical number of 50. We can then show linkage only if another gene is between these two genes. If both genes are linked to this third gene then they must be linked on the same chromosome The more offspring you have the more reliable are your results. In a genetics lab the Chi Square test is used to determine if linkage or independent assortment is accepted. We will not go through the Chi Square test and I will not ask you questions on this statistical procedure

Crossing Over at the Chromosome Level

The next slide shows crossing over at the chromosome level. Note: in order for crossing over to occur the genes must be incompletely linked.

Cross Involving Linked Genes

The next slide shows two generations of crosses involving linked genes. Note the expected 9:3:3:1 ratio is messed up. A better way to examine linked genes would be to use a parent of interest crossed to a double homozygous recessive (test cross). With a test cross you would expect a 1:1:1:1 ratio. You would be able to tell by the phenotype of the offspring what the parent of interest gave to the offspring.

Sex determination in mammals

The presence of the Y chromosome tends to denote maleness. Female is the default pathway for mammals. Individuals with one or more X chromosomes but no Y chromosomes are females.

Exceptions: XX Males

There are cases in which XX individuals are male. These individuals have the SRY gene transposed to the X chromosome. These individual will be phenotypic males but genotypic females.

Exceptions: XY females

There are cases in which XY individuals are female. These individuals either do not make androgenic hormones or are insensitive to these hormones. 1) If an individual has a null mutation for the SRY gene then they do not form testes and do not produce testosterone. These individuals will be phenotypic females but genotypic males. 2) A gene on the X chromosome codes for androgen receptors. If this gene is mutated then the androgen receptors are insensitive to the testosterone that is produced. These individuals are phenotypically female but genotypically male with functioning internal testes.

Multiple Crossovers

There can be multiple crossovers between genes. An odd number of crossovers will lead to recombinant type chromatids. An even number of crossovers will lead to parental type chromatids. This is illustrated on the next slide Note: Multiple crossovers are rare since the probability of getting one crossover is a certain amount. The probability of getting a second one is the probability of the first times the probability of the second. Let us say the distance between two genes is 10 map units or 10% recombination or 0.1 frequency of recombination. A double crossover between those two genes would be 0.1 X 0.1 = 0.01 frequency or 1 % A triple crossover would be 0.1 X 0.1 X 0.1 = 0.001 (1 X10 -3) or 0.1%

Fertilization

This illustration shows the cycle of fertilization forming a diploid organism and the diploid organism forming haploid gametes 4.3 In most eukaryotic organisms, sexual reproduction consists of an alternation of haploid (1n) and diploid (2n) cells.

Sex-Linked traits: XY organisms - Experiment example

This slide illustrates reciprocal crosses in which homozygous white eyed females are crossed to hemizygous red eyed males and vice versa. Note the symbol used to indicate red eyes is X+ not XW. The + symbol represents wildtype which is the type most commonly found in nature. The wildtype does not have to be the dominant but most often it is the dominant. In the case of red eyed flies the wildtype is the dominant. It is not wrong to use XW. Remember you define the symbol in your key.

Who is Thomas Hunt Morgan?

Thomas Hunt Morgan's work with Drosophila helped unravel many basic principles of genetics, including X-linked inheritance.

Non-Mendelian Classical Genetics Homozygous Dominant Lethality

When you mate two colored mice you obtain all colored offspring. The colored mouse strain is a pure breeding strain. When you cross two yellow mice you will obtain 2/3 yellow and 1/3 colored offspring. The yellow allele must be the dominant allele which masks the expression of the colored phenotype. The yellow mice must be heterozygote. Why do two yellow mice always produce some colored offspring? Why do we get a 2:1 ratio of offspring not a 3:1 ratio? Answers to questions on previous slide: It seems yellow mice can never breed true. They always produce 1/3 colored offspring. If you set up a Punnett square for two heterozygotes you expect to get 1/4 homozygous dominant, 2/4 heterozygotes and 1/4 homozygous recessive. The colored offspring represent the 1/4 homozygous recessive. The yellow offspring represent the 2/4 heterozygotes. The 1/4 homozygous dominant is missing. This explains the 2 yellow to 1 colored offspring Where is the homozygous dominant . It never developed. Having two dominant alleles for yellow is lethal.

Y-linked pedigrees

Y-linked traits are passed from father to son and is expressed in all the males. This is illustrated in the next slide.

Sex-Linked organisms: XY organisms

Y-linked traits are passed from father to son. X-linked traits (do not have a comparable Y-counterpart) are passed from father to daughter and mother to child. The expression of the traits in the daughter is dependent on the same factors as for any autosomal trait. The expression in the son is that the trait will be expressed when present. Determination if a trait is sex-linked. If you have an expression of the trait differently between male and female offspring than the trait is sex-linked. Whenever you see difference in the number of males verses females for a given trait you know the trait is sex-linked. When using symbols to represent sex-linkage use the X and a superscript for X-linked alleles. Also use the Y to represent the Y chromosome. For instance the white locus in Drosophila is an X-linked recessive trait. Xw Xw represents a homozygous recessive white eyed female. Note: The white allele is recessive so I use the script form of the symbol (w). I tend to use lower case script for recessives and print upper case for the dominant form of the gene. This allows me to easily differentiate the recessive from the dominant rather than relying on size. You can use any form of the symbols you want as long as you have a key to indicate the form.

Lethal

any allele that can cause the premature death of the organism. It can be due to a dominant or recessive allele. Death can occur at any stage of development from fertilization to relatively old age

Penetrance

the number of individuals who have a particular genotype demonstrating that genotype