23.1 Mendel’s Laws Gregor Mendel –Investigated inheritance at the organism level (1860’s) –Concluded that plants transmit distinct factors to offspring –Based on his studies, he formulated the law of segregation
23.1 Mendel’s Laws The Law of Segregation –Each individual has two factors for each trait –The factors segregate (separate) during the formation of gametes –Each gamete contains only one factor from each pair of factors –Fertilization gives each new individual two factors for each trait
23.1 Mendel’s Laws Today we know genes determine characteristics of an organism, genes are found on chromosomes Chromosomes that are homologous are members of a pair and carry genes for the same traits in the same order Alleles are alternate forms of a gene for the same trait Alleles are always at the same locus (location) on each chromosome of a homologous pair
23.1 Mendel’s Laws The Inheritance of a Single Trait –Phenotype: physical appearance of the individual with regard to a trait –Genotype: Alleles responsible for a given trait Two alleles for a trait A capital letter symbolizes a dominant allele (W) A lower-case letter symbolizes a recessive allele (w) Dominant refers to the allele that will mask the expression of the alternate (recessive) allele
23.1 Mendel’s Laws
Gamete Formation –During meiosis, homologous chromosomes separate so there is only 1 member of each pair in a gamete –There is one allele for each trait, such as hairline, in each gamete –No two letters in a gamete can be the same letter of the alphabet If genotype is Ww, then gametes from this individual will contain either a W or a w
23.1 Mendel’s Laws One-Trait Cross –A homozygous man with a widow’s peak (X) A woman with a straight hairline
23.1 Mendel’s Laws One-Trait Cross –Two individuals who are both Ww –A Punnett Square is useful to solve this problem
23.1 Mendel’s Laws One-Trait Crosses and Probability –The chance of 2 or more independent events occurring together is the product of their chance of occurring separately –In the cross Ww X Ww, what is the chance of obtaining either a W or a w from a parent? Chance of W = ½ and the chance of w = ½ –Therefore the probability of having these genotypes is as follows Chance of WW= ½ X ½ = ¼ Chance of Ww = ½ X ½ = ¼ Chance of wW= ½ X ½ = ¼ Chance of ww = ½ X ½ = ¼
23.1 Mendel’s Laws The One-Trait Test Cross –Breeders of plants and animals may do a test cross to determine the likely genotype of an individual with the dominant phenotype Cross with a recessive individual-has a known genotype If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous
23.1 Mendel’s Laws Practice Problems –Both a man and a woman are heterozygous for freckles. Freckles are dominant over no freckles. What is the chance that their child will have freckles?
23.1 Mendel’s Laws Practice Problems –Both you and your sibling have attached ear lobes, but your parents have unattached lobes. Unattached earlobes (E) are dominant over attached (e). What are the genotypes of your parents?
23.1 Mendel’s Laws Practice Problems –A father has dimples, the mother of his children does not, and all 5 of their children have dimples. Dimples (D) are dominant over no dimples (d). Give the probable genotypes of all persons concerned.
23.1 Mendel’s Laws The Inheritance of Two Traits –The Law of Independent Assortment: Each pair of factors assorts independently (without regard to how the others separate) All possible combinations of factors can occur in the gametes
The Inheritance of Two Traits
Two-Trait Crosses (Dihybrid Cross)
23.1 Mendel’s Laws Two-Trait Crosses (Dihybrid Cross) WwSs (X) WwSs –Phenotypic Ratio: –9 widow’s peak, short fingers –3 widow’s peak, long fingers –3 straight hairline, short fingers –1 straight hairline, long fingers
23.1 Mendel’s Laws Two-Trait Crosses and Probability –Probability Laws Probability of widow’s peak = ¾ Probability of short fingers= ¾ Probability of straight hairline= ¼ Probability of long fingers= ¼ –Using the Product Rule Probability of widow’s peak and short fingers = ¾ X ¾ = 9/16 Probability of widow’s peak and long fingers = ¾ X ¼ = 3/16 Probability of straight hairline and short fingers = ¼ X ¾ = 3/16 Probability of straight hairline and long fingers = ¼ X ¼ = 1/16
23.1 Mendel’s Laws Practice Problems –Attached earlobes are recessive, What genotype do children have if one parent is homozygous recessive for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline?
23.1 Mendel’s Laws Practice Problems –If an individual from this cross reproduces with another of the same genotype, what are the chances that they will have a child with a straight hairline and attached earlobes?
23.1 Mendel’s Laws Practice Problems –A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant) and a woman who does not. What are the genotypes of all persons concerned?
23.1 Mendel’s Laws Pedigrees –A chart of family’s history with regard to a particular genetic trait Males = Females = –Affected individuals (for a given trait) are shaded
A Human Pedigree
23.2 Beyond Simple Inheritance Patterns Incomplete Dominance –Occurs when the heterozygote is intermediate between the two homozygotes Codominance –Occurs when alleles are equally expressed in a heterozygote –Blood type AB is an example Red blood cells have both Type A and Type B surface antigens
23.2 Beyond Simple Inheritance Patterns Multiple Allele Inheritance –A trait is controlled by multiple alleles, the gene exists in several allelic forms. Each person has only two of the possible alleles.
23.2 Beyond Simple Inheritance Patterns Multiple Allele Inheritance –ABO Blood Types I A = A antigens on red blood cells I B = B antigens on red blood cells i = has neither A nor B antigens on red blood cells Both I A and I B are dominant over i, I A and I B are codominant PhenotypeGenotype AI A I A or I A i BI B I B or I B i ABI A I B Oii
23.2 Beyond Simple Inheritance Patterns Multiple Allele Inheritance –ABO Blood Types –Both I A and I B are dominant over i, I A and I B are codominant –The Rh factor is inherited separately from ABO blood types.
Inheritance of Blood Types
23.2 Beyond Simple Inheritance Patterns Practice Problems –If a person with straight hair marries someone with wavy hair, can they have a child with curly hair?
23.2 Beyond Simple Inheritance Patterns Practice Problems –A child with type O blood is born to a mother with type A blood. What is the genotype of the child? The mother? What are the possible genotypes of the father?
23.2 Beyond Simple Inheritance Patterns Practice Problems –From the following blood types determine which baby belongs to which parents: Baby 1 type O Mrs. Doe type A Mrs. Jones type A Baby 2 type B Mr. Doe type A Mr. Jones type AB
23.2 Beyond Simple Inheritance Patterns Sex-Linked Inheritance –In Humans: 22 pairs of autosomes, 1 pair of sex chromosomes –X and Y »In females, the sex chromosomes are XX »In males, the sex chromosomes are XY »Note that in males the sex chromosomes are not homologous Traits controlled by genes in the sex chromosomes are called sex-linked traits X chromosome has many genes, the Y chromosome does not
23.2 Beyond Simple Inheritance Patterns Sex-Linked Alleles –Red-green colorblindness is X-linked The X chromosome has genes for normal color vision –X B = normal vision –X b – colorblindness –GenotypesPhenotypes X B X B female with normal color vision X B X b carrier female with normal color vision X b X b colorblind female X B Ymale with normal color vision X b Ycolorblind male
Cross involving an X-linked Allele
23.2 Beyond Simple Inheritance Patterns Practice Problems –Both the mother and the father of a colorblind male appear to be normal. From whom did the son inherit the allele for colorblindness? What are the genotypes of the mother, father, and the son?
23.2 Beyond Simple Inheritance Patterns Practice Problems –A woman is colorblind. What are the chances that her son will be colorblind? If she is married to a man with normal vision, what are the chances that her daughters will be colorblind? Will be carriers?
23.2 Beyond Simple Inheritance Patterns Practice Problems –A husband and the wife both have normal vision. The wife gives birth to a colorblind daughter. Is it more likely the father had normal vision or was colorblind? What does this lead you to deduce about the girl’s parentage?
23.2 Beyond Simple Inheritance Patterns Polygenic Inheritance –Occurs when a trait is governed by two or more sets of alleles. –Each dominant allele codes for a product –The effects of the dominant alleles are additive. –The result is continuous variation. –Examples of traits include size or height, shape, weight, and skin color.
23.2 Beyond Simple Inheritance Patterns Practice Problems –A certain polygenic trait is controlled by three pairs of alleles: A vs a, B vs b, and C vs c. What are the extreme genotypes for this trait?
23.3 Environmental Influences Environmental factors can influence the expression of genetic traits. –Examples: Primrose flowers are white at warmer temperatures and red at cooler temperatures Siamese cats and Himalayan rabbits are darker in color where body heat is lost to the environment.
Coat Color in Himalayan Rabbits
23.4 Inheritance of Linked Genes All the alleles on one chromosome form a linkage group. Recall that during meiosis crossing over sometimes occurs If crossing over occurs between two alleles of interest, then four types of gametes are formed instead of two
23.4 Inheritance of Linked Genes The occurrence of crossing-over can help determine the sequence of genes on a chromosome Crossing-over occurs more often between distant genes than genes that are close together In the example below, it is expected that recombinant gametes would include G and z more often than R and s.
23.4 Inheritance of Linked Genes Practice Problems –When AaBb individuals reproduce, the phenotypic ratio is about 3:1. What ratio was expected? What may have caused the observed ratio?
23.4 Inheritance of Linked Genes Practice Problems –The genes for ABO blood type and for fingernails are on the same homologous pair of chromosomes. In an actual family, 45% of offspring have type B blood and no fingernails, and 45% have type O blood and fingernails; 5% have type B blood and fingernails, and 5% have type O blood and no fingernails. What process accounts for the recombinant phenotypes?