MENDEL AND THE GENE ID Introduction 1 Mendel

MENDEL AND THE GENE ID

Introduction 1. Mendel brought an experimental and quantitative approach to genetics 2. By the law of segregation, the two alleles for a character are packaged into separate gametes 3. By the law of independent assortment, each pair of alleles segregates into gametes independently 4. Mendelian inheritance reflects rules of probability 5. Mendel discovered the particulate behavior of genes

Mendel

Introduction • The current theory for the mechanism for the transmission of genetic material was the “blending” hypothesis. • This hypothesis proposes that the genetic material contributed by each parent mixes in a manner analogous to the way blue and yellow paints blend to make green. • Mendel proposed an alternative model, “particulate” inheritance • This proposes that parents pass on discrete heritable units - genes - that retain their separate identities in offspring. • Genes can be sorted and passed on, generation after generation, in undiluted form.

Blending Hypothesis: 1800 s -suggested that traits of parents mix to form intermediate traits in offspring. Parents Offspring Red flower x White flower Pink flower Tall height x Short height Medium height Blue bird x Yellow bird Green birds Fair skin x dark skin Medium skin color If blending always occurred, eventually all extreme characteristics would disappear from the population.

Father of Modern Genetics: Gregor Mendel -abbey garden, where Mendel documented mechanisms of inheritance. 1857

mendel • Mendel grew up on a small farm in what is today the Czech Republic. • In 1843 - entered monastery. • He studied at the University of Vienna from 1851 to 1853. • The monks at this monastery had a long tradition of interest in the breeding of plants, including peas. • Around 1857, Mendel began breeding garden peas to study inheritance.

JOKE BREAK

Pea plants have several advantages for genetics. • Pea plants are available in many varieties with distinct heritable features (characters) with different variants (traits). • Short generation time, little space needed, cross or self fertilize

Overview of Mendel’s experiments. -Mendel looked at 7 traits-

• Cross fertilization and self fertilization

Ø He took a true breeding purple flower plant and crossed it with a true breeding white flower plant. Ø He called these the parent generation (P generation) Ø What do you think the offspring looked like? E L P R U ALL P Pollen hybridize X

F 1 Allowed to self-pollinate F 2 ? ? Y ? ? SA ? ? ? T A H

Mendel concluded: 1. Something is being passed from parent to offspring. • He called these “Factors” 2. Sometimes you can see “it” and sometimes you can’t see “it”. • If you can see it- it is dominant. (T) • If it’s there and you can’t see itit’s recessive. (t) Purple flower is a dominant trait and white flower is a recessive trait.

Each Version of the factor, now known as gene, is called an Allele.

Mendel developed a hypothesis to explain these results that consisted of four related ideas. 1. Alternative versions of genes (different alleles) account for variations in inherited characters. • Different alleles vary somewhat in the sequence of nucleotides at the specific locus of a gene 2. For each character, an organism inherits two alleles, one from each parent. 3. If two alleles differ, then one, the dominant allele, is fully expressed in the organism’s appearance. • The other, the recessive allele, no noticeable effect on the appearance. organism’s has

4. The two alleles for each character segregate (separate) during gamete production (meiosis).

Mendel’s quantitative analysis of F 2 plants revealed the two fundamental principles of heredity: the law of segregation & the law of independent assortment.

Law of segregation- the two alleles for a character are packaged into separate gametes • If the blending model were correct, the F 1 hybrids from a cross between purple-flowered and white-flowered pea plants would have pale purple flowers. • Instead, the F 1 hybrids all have purple flowers, just as purple as the purple-flowered parents.

Law of segregation- the two alleles for a character are packaged into separate gametes • The white trait, absent in the F 1, reappeared in the F 2. • Based on a large sample size, Mendel recorded in the F 2 plants: • 705 purple-flowered • 224 white-flowered • 3: 1 ratio (phenotype) The reappearance of white -flowered plants in the F 2 generation indicated that the heritable factor for the white trait was not diluted or “blended” by coexisting with the purpleflower factor in F 1 hybrids.

Law of segregation-the two alleles for a character are packaged into separate gametes This segregation of alleles corresponds to the distribution of homologous chromosomes to different gametes in meiosis. packaged into separate gametes

USING THE Law of segregation………. . • A Punnett square predicts the results of a genetic cross between individuals of known genotype. PRACTICE: • DRAGON Genetics #1 & 2 monohybrid • Two heterozygotes • 3: 1 ratio of dominant: recessive phenotypes • 1: 3: 1 genotypes

• In cats, long hair is recessive to short hair. A truebreeding (homozygous) short-haired male is mated to a long-haired female. What will their kittens look like?

• In fruit flies, the gene for wing shape has an unusual allele called ‘curly’ (designated ‘Cy’). The normal (wild type) allele is designated ‘cy. ’ A fly homozygous for cy (cy cy) has normal, straight wings. The heterozygote (Cy cy) has wings which curl up on the ends (and, incidentally, can’t really fly). The homozygote for the Cy allele (Cy Cy) never hatches out of the egg. In other words, this allele is lethal in the homozygous condition. If two curly winged flies are mated, and the female lays 100 eggs, predict the following, : • How many eggs will produce living offspring? • How many straight winged flies do you expect among the living offspring? • What percentage of the living offspring do you expect to be curly winged like the parents?

• When crossing two pure breeding plants, Mendel found similar 3 to 1 ratios among F 2 offspring when he conducted crosses for six other characters, each represented by two different varieties.

Predicting the genotype of an organism with the dominant phenotype: what’s the genotype? ? ? • The organism must have at least one dominant allele, but it could be homozygous dominant or heterozygous. A testcross, breeding a homozygous recessive with dominant phenotype, but unknown geneotype, can determine the identity of the unknown allele.

Testcross • A test cross is a way to explore the genotype of an organism. It is used to determine if the genotype of a plant with the dominant phenotype is homozygous or heterozygous. • EX: "What's the genotype of this guinea pig with respect to its fur color? " Homozygous Dominant (Pure Dominant) BB Black Heterozygous (Hybrid) Bb Black Homozygous Recessive bb (Pure Recessive) White

The law of independent assortment- each pair of alleles segregates into gametes independently • Alleles that are not on the same chromosome • Ex. The alleles for height segregate independently from the alleles of a gene for color. As long as the genes are on separate chromosomes, the separation during meiosis is RANDOM.

Random Orientation of Chromosomes During Meiosis The law of independent assortment

The law of independent assortment- each pair of alleles segregates into gametes independently -During meiosis, the chromosomes line up randomly during metaphase II. -So, depending on how they line, this determines the alleles in each gamete.

The law of independent assortment- each pair of alleles segregates into gametes independently • Mendel’s experiments that followed the inheritance of flower color or other characters focused on only a single character via monohybrid crosses. • He conducted other experiments in which he followed the inheritance of two different characters, a dihybrid cross. animation

• If no independent assortment: • The F 2 offspring (in dihybrid) would only produce two phenotypes in a 3: 1 ratio, just like a monohybrid cross. • This was not consistent with Mendel’s results. animation

Mendel crossed true-breeding plants that had yellow, round seeds (YYRR) with truebreeding plants that has green, wrinkled seeds (yyrr). When sperm with four classes of alleles and ova with four classes of alleles combined, there would be 16 equally probable ways in which the alleles can combine in the F 2 generation.

• When crossing a homozygous dominante with a homozygous recessive… These combinations produce four distinct phenotypes in a 9: 3: 3: 1 ratio. • This was consistent with Mendel’s results.

Black-B Brown-b Long hair-S Short-s • Cross two purebred guinea pigs. Brown-long hair x Black-short hair • Give phenotype and genotype ratios. DO IT NOW………

• Complete 1 st pg of • Dragon Genetics……

Mendelian inheritance reflects rules of probability • The probability scale ranged from zero (an event with no chance of occurring) to one (an event that is certain to occur). • The probability of tossing heads with a normal coin is ½. • The probability of rolling a 3 with a six-sided die is 1/6, and the probability of rolling any other number is 1 - 1/6 = 5/6. • SO……. What’s the probablility of getting a dominant allele from a parent that is Rr ? • ½… 50%, right?

• ALSO…When tossing a coin, the outcome of one toss has no impact on the outcome of the next toss. • Each toss is an independent event, just like the distribution of alleles into gametes. • Like a coin toss, each ovum from a heterozygous parent has a ½ chance of carrying the dominant allele and a ½chance of carrying the recessive allele. • The same odds apply to the sperm.

• Rule of multiplication to determine the chance that two or more independent events will occur together in some specific combination. 1. Compute the probability of each independent event. 2. Multiply the individual probabilities. • The probability that two coins tossed at the same time will land heads up is: Ø ½ x ½ = ¼. • What is the probability that two heterozygous pea plants (Pp) will produce a white-flowered offspring (pp) ? • ½x½=¼

Rule of multiplication also applies to dihybrid crosses. • For a heterozygous parent (Yy. Rr) the probability of producing a YR gamete is ½ x ½ = ¼. • We can use this to predict the probability of a particular F 2 genotype without constructing a 16 -part Punnett square. • The probability that an F 2 plant will have a YYRR genotype from heterozygous parents is • 1/16 (¼ chance for a YR ovum and ¼ chance for a YR sperm).

• Examples: • An organism had three independently assorting traits: Aa. Bb. Cc. What fraction of its gamete will contain ABC? • 1/8 • What about an organism with AABc. Cc? What fraction of its gamete will contain ABC? • 1/4

Do these problems now…. 1. If a Yy. Pp plant is crossed to a Yypp plant, what is the probability that the resulting plant will have the genotype Yypp? 2. A true-breeding plant with green seeds and white flowers is crossed with a plant that is heterozygous for both traits. What is the probability that the cross will yield a plant with green seeds and white flowers? (flower color- purple dominant, seed color- yellow dominant)

EXAMPLES 1. If a Yy. Pp plant is crossed to a Yypp plant, what is the probability that the resulting plant will have the genotype Yypp?

2. A true-breeding plant with green seeds and white flowers is crossed with a plant that is heterozygous for both traits. What is the probability that the cross will yield a plant with green seeds and white flowers? (flower color- purple dominant, seed color- yellow dominant)

Rule of addition: When more than one arrangement of the events producing the specified outcome is possible, the individual probabilities are added. EXAMPLE: Determine the probability of an offspring having two recessive phenotypes for at least two of three traits resulting from a trihybrid cross between pea plants that are Pp. Yy. Rr and Ppyyrr. 1 st –Determine the number of possibilities that will fulfill the requirement: • There are five possible genotypes that fulfill this condition: ppyy. Rr, pp. Yyrr, Ppyyrr, PPyyrr, and ppyyrr. • Determine probability of each and then add together.

• For pp. Yyrr: 1/4 × 1/2 = 1/16. • For Ppyyrr: 1/2 × 1/2 = 1/8 or 2/16. • For PPyyrr: 1/4 × 1/2 = 1/16. • For ppyy. Rr 1/4 × 1/2 = 1/16. • Therefore, the chance that a given offspring will have at least two recessive traits is… • 1/16 + 2/16 + 1/16 = 6/16.

Dragon Genetics …. back page If a homozygous dominant long neck, heterozygous fire breathing, purple bodied dragon was crossed with a heterozygous long neck, heterozygous fire breathing, heterozygous green bodied dragon, what would be the chances of an offspring being at least homozygous recessive for 2 of the traits?

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THE CHROMOSOMAL BASIS OF INHERITANCE 1. T. H. Morgan traced a gene to a specific chromosome 2. Linked genes tend to be inherited together because they are located on the same chromosome 3. Independent assortment of chromosomes and crossing over produce genetic recombinants 4. Geneticists use recombination data to map a chromosome’s genetic loci

Introduction • It was not until 1900 that biology finally caught up with Gregor Mendel. • Mendel’s hereditary factors are the genes located on chromosomes.

Mendelian inheritance has its physical basis in the behavior of chromosomes during sexual life cycles • Around 1900, cytologists and geneticists began to see parallels between the behavior of chromosomes and the behavior of Mendel’s factors. Around 1902, Walter Sutton, Theodor Boveri, and others noted these parallels and a chromosome theory of inheritance began to take form.

CHROMOSOMAL THEORY OF INHERITANCE 1. Chromosomes are carriers of traits and each chromosome could carry the genes for MANY traits. 2. Alternate forms or ALLELES of a gene are located on matched pairs of chromosomes. 3. When chromosome pairs separate in meiosis, each chromosome carries its set of alleles to a gamete. 4. Genes of the same chromosome move together; Genes on different chromosomes assort independently.

Thomas Hunt Morgan 1933 Nobel Prize

Morgan traced a gene to a specific chromosome • He was the first to associate specific gene with a specific (sex-linked) chromosome. • Worked with Drosophila melanogaster • Prolific breeders & have a weeks. • 3 pairs of autosomes and a pair of • Produces about 100 offspring per egg lay – good statistics! • Easy/inexpensive to raise • Chromosomes are VERY large and easy to see and locate • Sexes are easily distinguished • --female is larger • --shapes of abdomen identify sexes at a glance generation time of two sex chromosomes a

Morgan spent a year looking for variant individuals among the flies he was breeding. • He discovered a single male fly with white eyes instead of the usual red. • The normal phenotype is the wild type. • Alternative traits mutant phenotypes. • red=w+ • white=w What would you expect from this cross? (*female is Homozygous)…F 1 and F 2….

Morgan’s Experiments & Findings • When Morgan crossed his white-eyed male with a redeyed female, all the F 1 offspring had red eyes, • The red allele appeared dominant to the white allele. • Crosses between the F 1 offspring produced the classic 3: 1 phenotypic ratio in the F 2 offspring. • Surprisingly, the white-eyed trait appeared only in males. • All the females and half the males had red eyes. • Morgan concluded that a fly’s eye color was linked to its sex.

• Morgan deduced that the gene with the white-eyed mutation is on the X chromosome, a sex-linked gene. • Females (XX) may have two red -eyed alleles and have red eyes or may be heterozygous and have red eyes. • Males (XY) have only a single allele and will be red eyed if they have a red-eyed allele or white-eyed if they have a whiteeyed allele.

Sex Chromosomes 1. Sex-linked genes have unique patterns of inheritance 2. The chromosomal basis of sex varies with the organism

• In addition to their role in determining sex, the sex chromosomes, especially the X chromosome, have genes for many characters unrelated to sex. • Males are hemizygous for the X chromosome (XY) Karyotype

males are far more likely to inherit sex-linked recessive disorders • If a sex-linked trait is due to a recessive allele, a female will have this phenotype only if homozygous. • Heterozygous females will be carriers. • Because males have only one X chromosome (hemizygous), any male receiving the recessive allele from his mother will express the trait. • The chance of a female inheriting a double dose of the mutant allele is much less than the chance of a male inheriting a single dose. • Therefore, males are far more likely to inherit sex-linked recessive disorders than are females.

• Hemophilia & Colorblindness are all on the X-chromosome • Several serious human disorders are sex-linked. • Duchenne muscular dystrophy affects one in 3, 500 males born in the United States. • Affected individuals rarely live past their early 20 s. • due to the absence of an X-linked gene for a key muscle protein, called dystrophin. • a progressive weakening of the muscles and a loss of coordination.

X-inactivation Although female mammals inherit two X chromosomes, only one X chromosome is active.

• Therefore, males and females have the same effective dose (one copy ) of genes on the X chromosome. • During female development, one X chromosome per cell condenses into a compact object, a Barr body. • This inactivates most of its genes. • Barr body occurs randomly and independently • If a female is heterozygous for a sex-linked trait, approximately half her cells will express one allele and the other half will express the other allele. • The condensed Barr body chromosome is reactivated in ovarian cells that produce ova. • In humans, this mosaic pattern is evident in women who are heterozygous for a X-linked mutation that prevents the development of sweat glands. • Hypohidrotic ectodermal dysplasia • A heterozygous woman will have patches of normal skin and skin patches lacking sweat glands.

Cool example: the orange and black pattern on tortoiseshell cats is due to patches of cells expressing an orange allele while others have a nonorange allele.

• In human and other mammals, there are two varieties of sex chromosomes, X and Y. • XX= female. XY= male. • This X-Y system of mammals is not the only chromosomal mechanism of determining sex. • Other options include • X-0 system (Grasshoppers, roaches, and other insects) • Z-W system (Birds, insects like butterflies, frogs and some species of fish) • the haploid-diploid system.

• In humans, the anatomical signs of sex first appear when the embryo is about two months old. • In individuals with the SRY gene (sex- determining region of the Y chromosome), the generic embryonic gonads are modified into testes. • In individuals lacking the SRY gene, the generic embryonic gonads develop into ovaries. Only about 450 genes on the Y chromosome

Linked genes tend to be inherited together because they are located on the same chromosome • Chromosome have 100 s or 1000 s of genes. • Genes located on the same chromosome are called genes linked • tend to be inherited together because the chromosome is passed along as a unit. • Results of crosses with linked genes deviate from those expected according to independent assortment.

• Morgan observed this linkage (and its deviations) when he followed the inheritance of characters for body color and wing size. • The wild-type body color is gray (b+) and the mutant black (b). • The wild-type wing size is normal (vg+) and the mutant has vestigial wings (vg). • Morgan crossed F 1 heterozygous females (b+bvg+vg) with homozygous recessive males (bbvgvg). • (according to Mendelian genetics, what genotypes should he get in the F 2 generation? )

• According to independent assortment, this should produce 4 phenotypes in a 1: 1: 1: 1 ratio. • THAT’ NOT WHAT HE OBSERVED • He observed a large number of wild-type (gray-normal) and double-mutant (blackvestigial) just like the parents. • Fewer recombinations than expected!!!!!

Morgan’s conclusions: • He reasoned that body color and wing shape are usually inherited together because their genes are on the same chromosome (Linked). • The other two phenotypes (gray-vestigial and black-normal) were fewer than expected from independent assortment (and totally unexpected from dependent assortment). • These new phenotypic variations must be the result of crossing over. What % were recombinants?

• The results of Morgan’s testcross for body color and wing shape did not conform to either independent assortment or complete linkage. • Under independent assortment the testcross should produce a 1: 1: 1: 1 phenotypic ratio. • If completely linked, we should expect to see a 1: 1: 0: 0 ratio with only parental phenotypes among offspring. • Most of the offspring had parental phenotypes, suggesting linkage between the genes. • However, 17% of the flies were recombinants, suggesting incomplete linkage.

• Morgan proposed that some mechanism occasionally exchanged segments between homologous chromosomes. • This switched alleles between homologous chromosomes. • The actual mechanism, crossing over during prophase I, results in the production of more types of gametes than one would predict by Mendelian rules alone.

• Crossing over accounts for the recombinant phenotypes in Morgan’s testcross.

Geneticists can use recombination data to map a chromosome’s genetic loci • One of Morgan’s students, Alfred Sturtevant, used crossing over of linked genes to develop a method for constructing a genetic map. • This map is an ordered list of the genetic loci along a particular chromosome. Alfred Sturtevant hypothesized that the frequency of recombinant offspring reflected the distances between genes on a chromosome

• The farther apart two genes are, the higher the probability that a crossover will occur between them and therefore a higher recombination frequency. • The greater the distance between two genes, the more points between them where crossing over can occur. • Sturtevant used recombination frequencies from fruit fly crosses to map the relative position of genes along chromosomes, a linkage map.

• Sturtevant used the testcross design to map the relative position of three fruit fly genes, body color (b), wing size (vg), and eye color (cn). • The recombination frequency between cn and b is 9%. • The recombination frequency between cn and vg is 9. 5%. • The recombination frequency between b and vg is 17%. Try and create a map from the above data…. Click to see map • One map unit (sometimes called a centimorgan) is equivalent to a 1% recombination frequency.

A couple of points…. . • Some genes on a chromosome are so far apart that a crossover between them is virtually certain. • In this case, the frequency of recombination reaches is its maximum value of 50% and the genes act as if found on separate chromosomes and are inherited independently. • In fact, several genes studies by Mendel are located on the same chromosome. • For example, seed color and flower color are far enough apart that linkage is not observed. • Plant height and pod shape should show linkage, but Mendel never reported results of this cross.

A couple of points…. . • Genes located far apart on a chromosome are mapped by adding the recombination frequencies between the distant genes and intervening genes. • Sturtevant and his colleagues were able to map the linear positions of genes in Drosophila into four groups, one for each chromosome.

Your turn to do some MAPPING…… Anaylsis: • If the two genes showed independent assortment, we would hypothesize a 1: 1: 1: 1 ratio of phenotypes from the testcross done. • Instead, most resemble the phenotypes of the two original parents (the normal round eyes and no tooth, and the double mutant vertical eyes and tooth). We must conclude that the genes do not assort independently. • How many map units between these 2 linked genes? Hints/Steps; 1 - Calculate % of recombinants.

Solution

Another practice problem Determine the sequence of 4 genes along a chromosome based on the following recombination frequencies: • A-B, 8 map units • A-C, 28 map units • A-D, 25 map units • B-C , 20 map units • B-D, 33 map units.

Solution Steps for solving 1. Choose the greatest map distance (in this case 33) and place the genes involved at opposite ends of a line representing a portion of the chromosome in question 2. Now choose a gene combination with either B or D in it. For example: A-D = 25 map units a) 3. Since A-B is only 8 map units, A must be in the middle Now choose on that has A in it : a) b) You already have A-D and A-B, so all that is left is A-C, but where does it go…which side? Since the distance from A to B is about 8 and the distance from C to B is less than the distance from A to B the C-gene must be to the left of the B-gene.

Practice • A phenotypic wild-type fruit fly (heterozygous for gray body color and red eyes) was mated with a double mutant black fruit fly with purple eyes. • The offspring were as follows: wild-type, 721; black-purple, 751; gray-purple, 49; black-red, 45. (a) What is the recombination frequency between the genes for body color and eye color? Genotypes: • b+ = gray body (b) Are these genes on the same • b = black bodyfar chromosome? If so, how • pr+= red eyes apart are they? • pr = purple eyes

Solution Genotypes: • b+ = gray body • b = black body • pr+= red eyes • pr = purple eyes Total flies = 1566 (a) The percent recombination is therefore 6% 6

Chi Squared Test • A statistical method assessing the goodness of fit between observed values and those expected theoretically. • Example: https: //www. biologycorner. com/2014/01/21/teaching-chisquare-analysis/

An example for Destiny…. . We want to test for whether people with college degrees are more or less likely to identify with one political party over another. The cross tabulation is below…. . | College Degree? partyid | No Yes | Total --------+-------------+-----Democrat | 198 289 | 487 Republican | 128 348 | 476 Independent | 97 285 | 382 --------+-------------+-----Total | 423 922 | 1, 345 • Perform a χ² test… All data are from an ABC News Post-Debate Poll of October 2000

Multiple Alleles genes that exist in more two allelic forms

• Practice Problems

1. Alterations of chromosome number or structure cause some genetic disorders Nondisjunction Aneuploidy Polyploidy

Nondisjunction • Nondisjunction occurs when problems with the meiotic spindle cause errors in daughter cells. • This may occur if tetrad chromosomes do not separate properly during meiosis I. • Alternatively, sister chromatids may fail to separate during meiosis II. Consequencesome gametes receive two of the same type of chromosome and another gamete receives no copy.

Aneuploidy • Aneuploidy =Abnormal chromosome #. • Trisomic cells have 3 copies of a particular chromosome type ØDown syndrome, Klinefelter’s • Monosomic cells have only 1 copy of a particular chromosome type. ØTurner syndrome condition in which a female does not have the usual pair of two X chromosomes • Can occur during failures of the mitotic spindle. Aneuploidy occurs during cell division when the chromosomes don't separate properly (nondisjunction) between the two cells. Chromosome abnormalities occur in 1 of 160 live births, the most common being extra chromosomes 21 (Down syndrome), 18 (Edwards syndrome) and 13 (Patau syndrome).

Aneuploidy-trisomy • Down syndrome, is due to three copies of chromosome 21. • Caused by nondisjunction • It affects one in 700 children born in the United States. • Although chromosome 21 is the smallest human chromosome, it severely alters an individual’s phenotype in specific ways.

e m o r d n y S n w o D

Aneuploidy -trisomy Klinefelter Syndrome • Affects only males. • Caused by nondisjunction. • Males who have Klinefelter syndrome have an extra X chromosome (XXY), giving them a total of 47 instead of the normal 46 chromosomes • Develop as males with subtle characteristics that become apparent during puberty. They are often tall and usually don't develop secondary sex characteristics, such as facial hair or underarm and pubic hair. The extra X chromosome primarily affects the testes, which produce sperm and the male hormone testosterone. • At puberty, men with this syndrome often develop more breast tissue than normal, have a less muscular body, and grow very little facial or body hair. Most are sterile because they cannot produce sperm. Learning disabilities are also a common problem for them. • Hormone replacement: Teenagers are typically given testosterone injections to replace the hormone that would normally be produced by the testes.

Polyploidy • More than two complete sets of chromosomes. • This may occur when a normal gamete fertilizes another gamete in which there has been nondisjunction of all its chromosomes. • The resulting zygote would be triploid ( n).

Polyploidy is relatively common among plants and much less common among animals. • Both fishes and amphibians have polyploid species. • Recently, researchers in Chile have identified a new rodent species that may be the product of polyploidy. Triploid crops: banana, apple, grass carp, ginger Tetraploid crops: durum or macaroni wheat, maize, cotton, potato, cabbage, leek, tobacco, peanut, Pelargonium Hexaploid crops: chrysanthemum, bread wheat, triticale, oat Octaploid crops: strawberry, dahlia, pansies, sugar cane

Charts that show relationships within a family Pedigree 47

• • • 13 How many boys? _________ 12 How many Girls? _________ 3 How many generations? ________ 4 How many with the disorder? ______ 6 How many marriages are shown? ______

1. How many generations? 2. How many carriers? 3. How many affected males? 4. How many affected females? 5. Autosomal or sex-linked?

Autosomal-Dominant What can you tell by the pedigree? Dominant or recessive? Autosomal or sex-linked?

SEX-LINKED What can you tell by the pedigree? Dominant or recessive? Autosomal or sex-linked?

SEX-LINKED What can you tell by the pedigree? Dominant or recessive? Autosomal or sex-linked?

Autosomal Dominant or recessive? -Dominant Autosomal or sex-linked? NO carriers when Dominant.

Autosomal -Recessive Carriers when recessive.

Practice 1) How many unique gametes could be produced through independent assortment by an individual with the genotype Aa. Bb. CCDd. EE? Aa. Bb. Cc. DDEe? 2) Genes A, B, and C are located on the same chromosome. Testcrosses show that the recombination frequency between A and B is 28% and between A and C is 12%. Can you determine the linear order of these genes? 3) Study guide p. 116 “Genetics Problmes #1, 2, 3 6) Study Guide p 117, Questions 7 7) Study Guide p 118 question 16, 17, 18 8) Book page 292 # 8 9) Book page 292 # 9

Video Clips to Watch Cassiopeia Project Genetics Disorders http: //www. pbs. org/wgbh/nova/genome/program_adv. html • http: //highered. mcgrawhill. com/sites/0072437316/student_view 0/chapter 16/anim ations. html