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PATTERNS OF INHERITANCE Chapter 23 23-1. Gregor Mendel  1822 – 1884  Austrian monk  Experimented with garden peas  Provided a basis for understanding.

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Presentation on theme: "PATTERNS OF INHERITANCE Chapter 23 23-1. Gregor Mendel  1822 – 1884  Austrian monk  Experimented with garden peas  Provided a basis for understanding."— Presentation transcript:

1 PATTERNS OF INHERITANCE Chapter

2 Gregor Mendel  1822 – 1884  Austrian monk  Experimented with garden peas  Provided a basis for understanding heredity 23-2

3 Mendel cont’d 23-3  Published a paper in 1866 stating that parents pass discrete heritable factors on to their offspring  Factors retain individuality generation after generation  Identified that each trait is inherited by a pair of factors, one from each parent One form of a factor may be dominant over an alternative form Reasoned that each egg and sperm must contain only 1 copy of a factor for each trait

4 Mendel cont’d  Mendel’s law of segregation  Each individual has two factors (genes) 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 2 factors for each trait

5 Modern Genetics – Genes 23-5  Sections of chromosomes which give instructions for one characteristic or protein.  Located at the same point or locus, on each member of a homologous pair  All together make up the organism’s genome.  Controls the physical characteristics of a species

6 Modern Genetics - Alleles 23-6  Alternative forms of the same gene on each chromosome  One allele comes from each parent  Dominant allele Masks other traits present Only 1 dominant allele needs to be present for a certain trait to be expressed  Represented by a capital letter  Recessive Allele  2 copies of the recessive allele need to be present for trait to be expressed  Represented by a lower case letter

7 Modern Genetics – Alleles cont’d 23-7  Gene locus  Fig. 23.2

8 Modern Genetics – Alleles cont’d  Genotype  Genetic composition of a specific trait Homozygous dominant – 2 dominant alleles Heterozygous – 1 dominant allele and 1 recessive allele Homozygous recessive 2 recessive allele  Phenotype  Physical expression of a specific trait Homozygous dominant or heterozygous  Dominant Trait Homozygous recessive  recessive trait 23-8

9 Single Gene Inheritance 23-9  Simplest situation  One gene carries all the information responsible for one trait  Widow’s Peak  Alternative forms of alleles for hairline shape  Widow’s peak is dominant to straight hair line W=allele for widow’s peak w= allele for straight hairline

10 Widow’s peak  Fig. 23.3

11 Genotype related to phenotype  Table 23.1

12 Single Gene Inheritance cont’d.  Monohybrid cross  Looks at inheritance of one trait only  A punnett squares used to find all possible combinations of alleles

13 Single Gene Inheritance cont’d.  Example 1:  If a homozygous woman with a widow’s peak reproduces with a man with a straight hairline, what kind of hairline will their children have? 23-13

14 Single Gene Inheritance cont’d.  Example:  If two heterozygous parents reproduce what kind of hairline will their children have? Ww WWWWw w ww

15 Single Gene Inheritance cont’d  Genetic Ratios  Express ratio’s of possible outcomes Genotypic ratio Homozygous Dominant: Heterozygous: Homozygous Recessive Phenotypic ratio Dominant trait : recessive trait Often expressed as probability The probability resets and is the same for each pregnancy! Having one child with a trait has no effect on future children.

16 Single Gene Inheritance cont’d. Look back to example 1 –What are the genotypic and phenotypic ratio’s? –WW x ww Genotypic Ratio –0 : 4 : 0 Phenotypic Ratio –1:0 –100% Widow’s peak –Ww x Ww Genotypic Ratio –1: 2: 1 Phenotypic Ratio –3 : 1 –Probability is ¾ Widow’s Peak »75% –Probability ¼ Straight »25% 23-16

17 Single Gene Inheritance cont’d  Determining Genotype  No way to distinguish between a homozygous dominant individual and a heterozygous individual just by looking They are phenotypically the same  Test cross may help us determine Used in breeders of plants and animals Cross unknown with a recessive individual We know one parent genotype, this will help us determine the other genotype If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous

18 Single Gene Inheritance cont’d  Example : In rats, large ears is dominant to small ears. A rat breeder has a female rat with large ears, which she breeds with a male rat with small ears. In the first litter, all rats are born with large ears.  What is the genotype of the female rat?  In a second litter from the same parents, 4 baby rats have large ears, one has small ears.  What is the genotype of the female rat? 23-18

19 Single Gene Inheritance cont’d  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?  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?  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.

20 Homework (WHAT??? It’s BIO!)  Bikini Bottom Beach Genetics 23-20

21 Independent Assortment  Mendel reasoned from the results of his pea plant crosses that each pair of factors assorts independently into gametes  Each trait is passed down individually. The allele you receive for any one gene is not related to any other alleles you receive. We can now explain this through independent alignment and crossing over in meiosis. (Fig 23.6)

22 Independent Assortment cont’d  Fig. 23.6

23 Independent Assortment cont’d  Called the Law of Independent Assortment Each pairs of factors assorts independently (without regard to how the others separate) All possible combinations of factors can occur in the gametes 23-23

24 Independent Assortment cont’d  Practice problems  For each of the following genotypes, give all possible gametes WW WWSs Tt Ttgg AaBb  For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

25 Dihybrid Cross  Punnet squares, considering two-trait crosses at one time.  Example:  The traits for hairline and finger length are both single gene traits. As before, widows peak is dominant over straight hairline. Having short fingers is considered dominant over long fingers.  Two parents who are both heterozygous for both traits have children. Determine the genotypic and phenotypic ratios for their children

26 Dihybrid cross cont’d  Possible Gametes?  Genotypic Ratio?  Phenotypic Ratio? 23-26

27 Dihybrid Crosses cont’d  Determining Ratio’s  Product rule of probability The chance of 2 or more independent events occurring together is the product of their chance of occurring separately  In our example: Probability of widow’s peak = ¾ Probability of short fingers= ¾ What is the probability of widow’s peak AND short fingers? ¾ x ¾ = 9/

28 Dihybrid Crosses cont’d  Recall from our single trait crosses  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 = X =  Probability of widow’s peak and long fingers = X = 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 These values are standard for all heterozygous crosses! You don’t need to memorize them, but should be able to figure them out in your head!

29 Dihybrid Crosses cont’d  Using the product rule  Probability of widow’s peak and short fingers  Probability of widow’s peak and long fingers  Probability of straight hairline and short fingers  Probability of straight hairline and long fingers

30 Dihybrid Crosses cont’d  Two-trait test cross  Cross an individual with the dominant phenotype for each trait with an individual with the recessive phenotype of both traits W?S? x wwss WSW??S?? wwssWwSsWs?s?wSs?w?s

31 Dihybrid Cross cont’d  Attached earlobes are recessive, What genotype do children have if one parent is homozygous for earlobes and homozygous dominant for hairline, and the other is homozygous dominant for unattached earlobes and homozygous recessive for hairline?  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?  A child who does not have dimples or freckles is born to a man who has dimples and freckles (both dominant traits) and a woman who does not. What are the genotypes of all persons concerned?

32 Polygenic Inheritance  Not fully understood by geneticists.  Generally:  One trait controlled by 2 or more genes at different loci  The higher the number of dominant alleles you possess, the stronger the expression of the trait.  Result is a continuous range of phenotypes  Distribution resembles a bell curve  The more gene pairs involved, the more continuous the pattern of variation  Ex: human height, skin pigmentation, eye colour

33 Polygenic inheritance cont’d  Fig

34 Polygenic Inheritance cont’d  Skin color  Controlled by many gene pairs and many alleles  Let’s assuming a simple model of two alleles at 2 loci A and B If two heterozygous parents have children, children can range from very light to very dark GenotypePhenotype AABBVery Dark Skin AABb or AaBBDark Skin AaBb, AAbb, or aaBB Medium brown skin Aabb, or aaBbLight Skin aabbVery light skin 23-34

35 Polygenic Inheritance cont’d  Eye colour is controlled by 3 genes we have identified  We suspect there are more  Not a clear dominant and recessive  Brown allele 23-35

36 Environmental Influences on Inheritance  Environment can influence gene expression and therefore phenotype  Ex: sunlight exposure on skin; coat color in Himalayan rabbits  Human twin studies  Polygenic traits are most influenced  “nature vs. nurture”  Identical twins separated at birth are studied If they share a trait in common even though raised in different environments, it is likely genetic

37 Coat color in Himalayan rabbits  Fig

38 Incomplete Dominance  Incomplete dominance  Heterozygous individual has a phenotype intermediate to the two homozygous individuals  Ex: Curly-haired Caucasian woman and a straight- haired Caucasian man produce wavy-haired children When 2 wavy-haired people have children, the phenotypic ratio is 1 curly: 2 wavy: 1 straight 23-38

39 Incomplete dominance  Fig

40 Codominance  Multiple allele inheritance  The gene exists in several allelic forms, but each person still has only 2 of the possible alleles  Occurs when both alleles are equally expressed  Ex: type AB blood has both A antigens and B antigens on red blood cells 23-40

41 Codominance cont’d.  ABO blood types  I A = A antigens on RBCs  I B = B antigens on RBCs  i = has neither A nor B antigens on RBCs  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 ABIAIBIAIB Oii

42 Codominance cont’d.  Paternity testing- ABO blood groups often used  Can disprove paternity but not prove it  Rh factor- another antigen on RBCs  Rh positive people have the antigen  Rh negative people lack it There are multiple alleles for Rh negative, but all are recessive to Rh positive 23-42

43 Inheritance of blood type  Fig

44 Practice Problems  A polygenic trait is controlled by three pairs of alleles. What are the two extreme genotypes for this trait?  What is the genotype of the lightest child that could result from a mating between two medium-brown individuals?  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?  From the following blood types determine which baby belongs to which parents:  Baby 1 type OMrs. Doe type A Mrs. Jones type A  Baby 2 type BMr. Doe type A Mr. Jones type AB 23-44

45 Sex-linked inheritance  Sex chromosomes  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 23-45

46 Sex-linked inheritance cont’d.  X-linked traits  Red-green colorblindness is X- linked The X chromosome has genes for normal color vision X B = normal vision X b – colorblindness GenotypesPhenotypes XBXBXBXB female with normal vision XBXbXBXb carrier female, normal vision XbXbXbXb colorblind female XBYXBYmale with normal vision XbYXbYcolorblind male 23-46

47 Cross involving an X-linked allele  Fig

48 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?  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?  Both the husband and the wife 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?  What is the genotype of a colorblind male with long fingers is s=long fingers? If all his children have normal vision and short fingers, what is the likely genotype of the mother? 23-48

49 Inheritance of linked genes  The sequence of individual genes on a chromosome is fixed because each allele has a specific locus  All genes on a single chromosome form a linkage group  When linkage is complete, a dihybrid produces only 2 types of gametes  Any time traits are inherited together, a linkage group is suspected  If very few recombined phenotypes appear in offspring, linkage is also suspected 23-49

50 Inheritance of linked genes  Crossing over between 2 alleles of interest, can result in 4 types of gametes  Occurrence of crossing over can indicate the sequence of genes on a chromosome  More frequent between distant genes Fig

51 Practice Problems  When AaBb individuals reproduce, the phenotypic ratio is about 3:1. What ratio was expected? What may have caused the observed ratio?  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?


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