Presentation is loading. Please wait.

Presentation is loading. Please wait.


Similar presentations

Presentation on theme: "NONMENDELIAN GENETICS."— Presentation transcript:


2 INCOMPLETE DOMINANCE DEFINITION: When neither allele is dominant over the other. When the two alleles are found together, they result in an intermediate trait.


4 LET’S TRY A PROBLEM: In Mountain Boomers, the genes for length of tail exhibit “incomplete” dominance. Use a Punnett Square to predict the result of a cross between a homozygous long-tailed and a homozygous short-tailed Mountain Boomer. What do the offspring look like? DETERMINE THE LETTERS YOU WILL USE TO REPRESENT THE TRAITS: L = long-tail S = short tail LS = medium-tail DETERMINE THE GENOTYPES OF THE PARENTS THAT ARE BEING CROSSED: Long-tailed Parent = LL Short-tailed parent = SS

LL = GAMETE L SS = GAMETE S DRAW A PUNNET SQUARE AND INSERT THE GAMETES ON EACH SIDE. S L LS ANSWER: OFFSPRING WILL ALL BE MEDIUM-TAILED How many offspring will be medium-tailed if you cross two medium-tailed lizards?

6 MULTIPLE ALLELES DEFINITION: When not just two but THREE or more alleles of the same gene code for a single trait. EXAMPLE: ABO BLOOD TYPE In humans, the A allele for blood type and the B allele for blood type are both dominant over the O allele for blood type. Neither A nor B blood types is dominant over the other, and when they mix you get AB blood type. What are the four types of human blood? Type A Type B Type AB Type O What are the possible genotypes for human blood? AA AO BB BO AB OO

7 What are the possible genotypes of the parents?
PROBLEM: If a man with type A blood marries a woman with type B blood, can they have a child with type O blood? What are the possible genotypes of the parents? Man = AA or AO Woman = BB or BO A O B O Draw Punnett Squares to represent the different Possibilities: AB BO AO OO A A AB AB A O A A B AB BO B O B AB AB AO AO

8 ANSWER = NO, he can only have a type B or O child.
ANSWER: ONLY IF BOTH PARENTS ARE HETEROZYGOUS FOR THEIR BLOOD TYPE CAN THEY HAVE A CHILD WITH TYPE O BLOOD. PROBLEM: CAN A MALE HETEROZYGOUS FOR BLOOD TYPE B, BE THE FATHER OF A CHILD WHO HAS TYPE A BLOOD IF THE MOTHER IS TYPE O? 1) Determine the possible genotypes of the parents and child.. DAD B O Dad = BO Mom = OO Child = AA or AO 2) Place possible gametes in Punnett Square and multiply. M O O M BO OO Dad = B or O; Mom = O ANSWER = NO, he can only have a type B or O child.

9 Some genes alter the affects of other genes
EPISTASIS occurs when the phenotype of one gene is affected by another gene. For example: The dominant coat color in mice is gray(B); the recessive black (b). However, another gene on a different chromosome allows for color. A = normal color pigment, a = no color. In order for a gray mouse to be produced there must be both a dominant gray gene (B) and a dominant color gene (A). If a mouse is recessive for color pigment (aa), it will be albino.

10 Parent 1 =AB or Ab or aB or ab
Try this problem: If a mouse with the genotype AaBb (B = gray color, b = black color; A = color pigment, a= no color pigment) mates with a mouse with the genotype aaBb. What are the phenotypes of their offspring? 1) Determine the cross. AaBb x aaBb Parent 1 =AB or Ab or aB or ab Parent 2 = aB or ab 2) Determine the possible gametes for each parent. 3) Place each parents gametes in the Punnett square and multiply. AB Ab aB ab aB ab AaBB AaBb aaBB aaBb AaBb Aabb aaBb aabb

11 4) Look at all the genotypes and determine the phenotypes.
All the A_B_ = ? All the aa B_ = ? All the A_ bb = ? All the aabb = ? Results : _____gray; _____ black: _____ albino 1/8 1/2 3/8

DEFINITION: When several genes work together to control the expression of a trait, causing the trait to appear in a wide variety of forms. Examples: Hair color, eye color, skin color, height

13 SEX DETERMINATION In humans the 23rd pair of chromosomes = Sex chromosomes These chromosomes determine if the child will be a male or female. In MALES the 23rd pair are not HOMOLOGOUS = X and Y allele. In FEMALES the 23rd pair are HOMOLOGOUS = X and X allele. The FATHER determines the sex of the baby. LET’S SEE HOW THIS WORKS: X X 1. Each parent can pass on only ONE of their alleles. Females only an X; Males an X or Y. Place them in the square and see the results. X Y XX XX XY XY 2. 50% chance of male or female

14 EXAMPLE: Color Blindness
SEX LINKED GENES Sex Linked Genes are genes located on the X or Y chromosome. There are very few traits on the Y chromosome, so most males get only ONE allele for these traits because they get only 1 X. Females get two alleles, one on each X. THUS, males have a greater chance of getting a disease that is found on the sex chromosomes. EXAMPLE: Color Blindness Color Blindness (red-green) is a recessive trait carried on the X chromosome. Only females can be carriers. Males either have the disease or they don’t.

15 Male Genotype = XcY0 Female = XCXc
SAMPLE PROBLEM: If a red-green colorblind man marries a woman with normal color vision whose father was colorblind, what will be the expected phenotypic results of their children? 1) To begin the problem – DETERMINE the GENOTYPES of the P1 generation (parents). Because the trait is on the X chromosome we write it a little differently. To write the genotype we must include the X and Y allele and a superscript for the colorblindness allele above each X allele. The Y allele does not carry the colorblindness allele. You include a 0 superscript above the Y. Male Genotype = XcY0 Female = XCXc 2) Next determine the possible gametes each parent can pass on to their offspring. Male gametes = Xc or Y0 Female = XC or Xc

16 ½ the males and ½ the females will be colorblind
3) Place the possible gametes for each parent in the Punnett Square and multiply to get the possible offspring from this cross. Male Xc Y0 Female XC Xc XCXc XCY0 Carrier Normal male XcXc XcY0 colorblind colorblind male Female 4) Answer the question: ½ the males and ½ the females will be colorblind ½ the females will be carriers of the disease

One pair of genes for coat color in cats is sex-linked. The gene B produces yellow coat, b produces black coat, and the heterozygous (carrier) Bb produces tortoise-shell coat [CALICO]. What kind of offspring will result from the mating of a black male and a calico female? XBY0 XBXb Gametes = Male _________ Female _________ XB Yo Offspring = ½ females calico, ½ black; ½ males black, ½ yellow Can you get a calico male? XBXB XBY0 XBXb XbY0 XB Xb

18 Linked Genes When two or more genes are on the same chromosome, we say they are linked. Linked genes are not inherited by independent assortment. They usually split together during meiosis (not randomly) because they are on the same chromosome. You can tell if two or more genes are linked by looking at the results of a particular cross.

19 Draw the Chromosomes Here: See white board.
Sample Problem If the genes G or g and W or w are on the same gene, what type of genotypic results would you get from the following cross: GgWw x GgWw In unlinked genes, the gametes an individual could pass on with these genotypes would be GW or Gw or gW or gw With linked genes the possible gametes is different because they pull together during meiosis. The gametes would be: GW or gw. Draw the Chromosomes Here: See white board.

20 Compare the two Punnett Squares – Linked and Unlinked UNLINKED
GAMETES = GW OR Gw OR gW OR gw Punnett Square LINKED GAMETES = GW or gw GW Gw gW gw GW Gw gW gw GGWW GGWw GgWW Ggww GGWw GGww GgWw Ggww GgWW GgWw ggWW ggWw GgWw Ggww ggWw ggww GW gw GGWW GgWw GgWw ggww GW gw Classic phenotypic numbers = 9 dominant for both traits; 3 dominant for first, recessive for second; 3 recessive for first, dominant for second; and 1 recessive for both traits Classic phenotypic numbers = 3 dominant for both traits; 1 recessive for both traits

21 If the genes on the same chromosome cross over with their homologous pair the number results may even appear more unusual (not typical 3:1). AS A GENERAL RULE: The CLOSER together two genes are the less likely they will crossover. The FURTHER APART two genes are the more they are likely to cross over.

22 Baldness is dominant in males and recessive in females.
SEX INFLUENCED TRAITS DEFINITION: Certain traits that are dominant in one sex and recessive in the other. Certain chemicals cause these differences. EXAMPLE: BALDNESS Baldness is dominant in males and recessive in females. SAMPLE PROBLEM: If a man who is not bald mates with a female that is not bald but whose father was bald, what % of their children will be bald?

23 Bb bb Determine the genotypes: Male Genotype: bb Female Genotype: Bb
Cross the parents: bb x Bb using a Punnett Square Determine the possible offspring: b Bb bb B b How many boys will be bald? How many girls will be bald? Can women be bald? Who determines if male offspring will be bald?

24 MUTATIONS DEFINITION: A mistake during DNA replication causing a change in either an individual gene or the whole chromosome. TYPES OF MUTATIONS: 1) CHROMOSOME – PART OF MANY GENES OR AN ENTIRE CHROMOSOME IS AFFECTED. A) Deletion: piece of chromosome breaks off Example: Cri du chat – Mental retardation, small head, unusual face features, cat like cry B) Duplication: Segment of chromosome repeats itself. Example: Fragile X syndrome – Mental retardation C) Inversion: Piece breaks and rejoins in reverse order on the same chromosome

25 D) Translocation: When a piece breaks and rejoins to different chromosome

26 E) Nondisjunction: A chromosome pair fails to separate during meiosis; leads to polyploidy (extra numbers of chromosomes) or aneuploidy (too few chromosomes)

27 Types of Diseases Caused by Nondisjunction
DOWN’S SYNDROME = Extra 21st Chromosome; 2n = 47

28 Found in many criminals, fertile.
TURNER’S SYNDROME = XO Female whose sex organs don’t develop, sterile, no mental retardation, short stature SUPER MALE = XYY Found in many criminals, fertile. KLINEFELTER’S – XXY Large breasts, high voice, sterile, retarded male TRISOMY X = XXX NORMAL

29 2) GENE MUTATIONS = Change in the DNA at one gene site
POINT MUTATION – Change in just one base in a single gene. May be addition, deletion or substitution of a base. 1)Frameshift = Type of point mutation - Only addition or deletion (more disastrous) Example = Sickle Cell Anemia Mutations can occur in either somatic or sex cells. Those that occur in somatic cells affect only the individual. Those that occur in sex cells affect the offspring too. CAUSES OF MUTATIONS: 1) Unknown 2) Age 3) Mutagens – viruses, radiation, air pollution, foods and their additives, chemicals (thalidimide, DDT)



Similar presentations

Ads by Google