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23-1 Inheritance Overview. 23-2 23.1 Mendel’s laws Gregor Mendel –An Austrian monk, who combined his farmer’s skills with his training in mathematics.

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Presentation on theme: "23-1 Inheritance Overview. 23-2 23.1 Mendel’s laws Gregor Mendel –An Austrian monk, who combined his farmer’s skills with his training in mathematics."— Presentation transcript:

1 23-1 Inheritance Overview

2 Mendel’s laws Gregor Mendel –An Austrian monk, who combined his farmer’s skills with his training in mathematics and developed certain laws of heredity in 1860 after doing crosses between garden pea plants. –Published a paper 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

3 23-3 His research and his findings led to develop what we call 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 Mendel’s laws cont’d.

4 23-4 Mendel’s laws cont’d. Mendel’s law of segregation 1. Each individual has two factors (genes) for each trait. We now know that these factors are what we call genes and that genes are sections of chromosomes. We have a pair of each chromosome, known as homologous pairs which have genes controlling the same traits located at the same point, or locus, on each member of the pair Information contained within the homologous genes is not necessarily the same (ex: widow’s peak vs. straight hairline) –Alternative forms of a gene for a specific trait are called alleles

5 23-5 Mendel’s laws cont’d. 2. The factors segregate (separate) during the formation of gametes. This is due to Independent Assortment of Homologous Chromosomes during Meiosis.

6 23-6 Mendel’s laws cont’d. 3. Each gamete contains only one factor from each pair of factors. Gametes are haploid, which means they only have 1 chromosome of each pair, therefore each gamete has only one allele of the pair of alleles for a trait

7 23-7 Mendel’s laws cont’d. 4. Fertilization gives each new individual 2 factors for each trait. 2 Haploid Gametes combine to form a Diploid Zygote. Diploid means that it has a pair of each chromosome and therefore a pair of alleles for a given trait.

8 23-8 Gene locus

9 23-9 Mendel’s laws cont’d. Inheritance of a single trait In simple dominance, one form of the allele is dominant over the other form which is recessive. –Capitol letter- represents the dominant allele –Small letter- represents the recessive allele Dominant-a certain trait will result if the individual has at least 1 dominant allele Recessive- for a recessive trait to result the individual must have 2 copies of the recessive allele

10 23-10 –Genotype- genetic composition of an individual with regard to a specific trait Homozygous - 2 copies of the same allele for a specific trait Heterozygous – 1 of each of the 2 copies of the allele for that trait 2 copies of the dominant allele- homozygous dominant 1 copy of the dominant allele and 1 of the recessive- heterozygous 2 copies of the recessive allele- homozygous recessive The inheritance of a single trait cont’d. –Phenotype- physical appearance of the individual with regard to a trait.

11 23-11 Mendel’s laws cont’d. The inheritance of a single trait cont’d. Phenotype vs. Genotype –Homozygous dominant individual and heterozygous individual. These 2 individuals will have the same phenotype but different gentoypes. –Homozygous recessive individual will have a different phenotype than the other 2 individuals. –Example- inheritance of a widow’s peak vs. a straight hairline in humans Alternative forms of alleles for hairline shape Widow’s peak is dominant to straight –W=allele for widow’s peak –w= allele for straight hairline

12 23-12 Widow’s peak

13 23-13 Mendel’s laws cont’d. Inheritance of a single trait cont’d. –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 –If the genotype is WwLl (looking at 2 traits), gametes can contain any of the following combinations »WL, Wl, wL, or wl

14 23-14 Mendel’s laws 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

15 23-15 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Tt Ttgg AaBb –For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

16 23-16 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Ttgg AaBb –For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

17 23-17 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg AaBb –For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

18 23-18 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg Gametes – Tg or tg AaBb –For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

19 23-19 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg Gametes – Tg or tg AaBb Gametes – AB or aB or Ab or ab –For each of the following, state whether the genotype or a gamete is represented D Ll Pw LlGg

20 23-20 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg Gametes – Tg or tg AaBb Gametes – AB or aB or Ab or ab –For each of the following, state whether the genotype or a gamete is represented D Gamete (has one allele for a trait) Ll Pw LlGg

21 23-21 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg Gametes – Tg or tg AaBb Gametes – AB or aB or Ab or ab –For each of the following, state whether the genotype or a gamete is represented D Gamete (has one allele for a trait) Ll Genotype (has 2 alleles for a trait) Pw LlGg

22 23-22 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg Gametes – Tg or tg AaBb Gametes – AB or aB or Ab or ab –For each of the following, state whether the genotype or a gamete is represented D Gamete (has one allele for a trait) Ll Genotype (has 2 alleles for a trait) Pw Gamete (has one allele for each of 2 traits) LlGg

23 23-23 Mendel’s laws cont’d. Practice problems –For each of the following genotypes, give all possible gametes WW Gametes – All W WWSs Gametes – WS or Ws Tt Gametes – T or t Ttgg Gametes – Tg or tg AaBb Gametes – AB or aB or Ab or ab –For each of the following, state whether the genotype or a gamete is represented D Gamete (has one allele for a trait) Ll Genotype (has 2 alleles for a trait) Pw Gamete (has one allele for each of 2 traits) LlGg Genotype (has 2 alleles for each of 2 traits)

24 23-24 Genotype related to phenotype

25 23-25 Mendel’s laws cont’d. One trait crosses-monohybrid cross In one-trait crosses, only one trait such as type of hairline is being considered. –Let’s consider a specific cross 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? –Use W= widow’s peak, w=straight hairline – figure out the genotype of each parent: »Woman has a widow’s peak and we are told she is homozygous so she is WW »Man has a straight hairline so he must be ww –Determine their gametes: her eggs will all have W and his sperm will all have w –Put together all possible combinations: all offspring will receive a W from her and a w from him so all will have Ww

26 23-26 Mendel’s laws cont’d. One-trait cross cont’d. –In this example there was only 1 combination of eggs and sperm possible. The children are all monohybrids, they are heterozygous for one pair of alleles. –What if the parents were each Monohybrids (Ww)? Then sperm could have either W or w, and eggs also could have W or w –One way to figure out all of the possible combinations of eggs and sperm that can occur is to use a Punnett square –This is illustrated on the following slide

27 23-27

28 23-28 Mendel’s laws cont’d. One-trait cross cont’d. –After the genotypes and phenotypes of offspring are determined, we can determine the ratios In our previous example, in the Punnett square we had the following offspring: WW, Ww, Ww, and ww –The genotypic ratio is 1 WW:2 Ww:1 ww –The phenotypic ratio is 3 individuals with a widow’s peak to 1 individual with a straight hairline. –When a monohybrid reproduces with a monohybrid (a monohybrid cross) the ratio of expected phenotypes expressed in the offspring is 3 : 1. –Another way to phrase the phenotypic ratio is in terms of probability –This couple has a 75% chance of producing a child with a widow’s peak and a 25% chance of producing a child with a straight hairline The probability will be the same for each pregnancy between this couple

29 23-29 Mendel’s laws cont’d. One-trait crosses and probability –Product rule of 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

30 23-30 Mendel’s laws cont’d. One-trait crosses and probability cont’d. –Sum rule of probability-the chance of an event that can occur in more than one way occurring a certain way is the sum of the individual chances of it occurring that way. To calculate the chance of an offspring having a widow’s peak, add the chances of WW, Ww, or wW from the preceding slide ¼ + ¼ + ¼ = ¾ or 75%

31 23-31 Mendel’s laws cont’d. The one-trait test cross –You cannot distinguish between a homozygous dominant individual and a heterozygous individual just by looking because they are phenotypically the same –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 which has a known genotype - homozygous recessive. If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous

32 23-32 One-trait testcross When crossed with Recessive all resulting offspring show the dominant trait. Therefore the dominant parent must be Homozygous Dominant

33 23-33 One-trait testcross When crossed with Recessive some resulting offspring show the dominant trait and some show the recessive trait. Therefore the dominant parent must be Heterozygous Dominant

34 23-34 Mendel’s laws 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? Mother - _____________ Father - _____________

35 23-35 Mother - _____ Ff ________ Father - ___ Ff __________ 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?

36 23-36 Ff F f Mother - _____ Ff ________ Father - ___ Ff __________ 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?

37 23-37 Ff FFF f Mother - _____ Ff ________ Father - ___ Ff __________ 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?

38 23-38 Ff FFFFf f Mother - _____ Ff ________ Father - ___ Ff __________ 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?

39 23-39 Ff FFFFf ffF Mother - _____ Ff ________ Father - ___ Ff __________ 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?

40 23-40 Ff FFFFf ffFff Mother - _____ Ff ________ Father - ___ Ff __________ 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?

41 23-41 Ff FFF Freckles Ff Freckles ffF Freckles ff NO Freckles Mother - _____ Ff ________ Father - ___ Ff __________ 3:1 Ratio Freckles:No Freckles 75% Chance child will have Freckles 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?

42 23-42 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? Mother - _____________ Father - _____________

43 23-43 ee Attached Mother - _____________ Father - _____________ 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?

44 23-44 e eee Attached Mother - _____________ Unattached Earlobes Father - _____________ Unattached Earlobes 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?

45 23-45 Ee E eee Attached Mother - _____________ Unattached Earlobes Father - _____________ Unattached Earlobes 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?

46 23-46 Ee EEE unattached Ee unattached eeE unattached ee Attached Mother - ____Ee_________ Unattached Earlobes Father - ___Ee__________ Unattached Earlobes Parents had to both be (Ee) Heterozygous Dominant to have a recessive offspring 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?

47 23-47 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. Mother - _____________ Father - _____________

48 23-48 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. Dimples Mother - _____________ Father - _____________

49 23-49 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. Dimples Mother - _____________ No Dimples Father - _____________ Dimples

50 23-50 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. Dimples Mother - __dd_________ No Dimples Father - _____________ Dimples

51 23-51 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. dd Dimples Mother - __dd_________ No Dimples Father - _____________ Dimples

52 23-52 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. dd Dd Dimples Dd Dimples Dd Dimples Dd Dimples Mother - __dd_________ No Dimples Father - _____________ Dimples

53 23-53 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. dd D Dd Dimples Dd Dimples D Dd Dimples Dd Dimples Mother - __dd_________ No Dimples Father - _____________ Dimples

54 23-54 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. dd D Dd Dimples Dd Dimples D Dd Dimples Dd Dimples Mother - __dd_________ No Dimples Father - _____DD________ Dimples Father has to be Homozygous Dominant or approx. ½ of the children won’t have dimples

55 23-55 Mendel’s laws cont’d. The inheritance of two traits –Mendel reasoned from the results of his pea plant crosses that each pair of factors assorts independently into gametes If one pair of homologous chromosomes has a W on one and w on the other, and a second pair has an R on one and an r on the other, when gametes form all possible combinations of these letters are possible: WR, Wr, wR, and wr Called the Law of Independent Assortment

56 23-56 Segregation and independent assortment Fig. 23.6

57 23-57 Mendel’s laws cont’d. The law of independent assortment states : –Each pairs of factors assorts independently (without regard to how the others separate) –All possible combinations of factors can occur in the gametes

58 23-58 Mendel’s laws cont’d. Two-trait crosses-dihybrid cross –Let’s consider two traits, hairline and finger length We’ll use the W and w as before, and now S=short fingers and s=long fingers –A person who is WwSs widow’s peak and short fingers and a person who is also WwSs have children Figure the gametes for each parent –WS, Ws, wS, and ws for both Make a big Punnett square as illustrated on the following slides Figure the genotypes of the offspring

59 23-59 In two-trait crosses, genotypes of the parents require four letters because there is an allelic pair for each trait. Gametes will contain one letter of each kind in every possible combination.

60 23-60 Dihybrid cross

61 23-61 Dihybrid cross

62 23-62 Phenotypic ratio: 9 widow’s peak, short fingers: 3 widow’s peak, long fingers; 3 straight hairline, short fingers; 1 straight hairline, long fingers When a dihybrid (heterozygous for 2 traits) reproduces with a dihybrid the ratio of expected phenotypes expressed in the offspring are 9 : 3 : 3 : 1.

63 23-63 Mendel’s laws cont’d. 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

64 23-64 Mendel’s laws cont’d. The two-trait testcross –Cross an individual with the dominant phenotype for each trait with an individual with the recessive phenotype of both traits If the dominant individual is homozygous for both traits, all offspring will show the dominant phenotypes If the dominant individual is heterozygous, then a Punnett square shows the following phenotypic ratio of offspring –1 dominant for both: 1 dominant for first trait recessive for second: 1 recessive for first trait dominant for second: and 1 recessive for both

65 23-65 Two-trait testcross

66 23-66 Mendel’s laws Practice problems –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) and a woman who does not. What are the genotypes of all persons concerned?

67 Beyond simple inheritance patterns Polygenic inheritance –Controlled by 2 or more sets of alleles –Each dominant allele codes for a product and effects are additive –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

68 23-68 Skin Color The inheritance of skin color, determined by an unknown number of gene pairs, is a classic example of polygenic inheritance. A range of phenotypes exist and several possible phenotypes fall between the two extremes of very dark and very light. The distribution of these phenotypes follows a bell- shaped curve.

69 23-69 Polygenic inheritance Fig. 23.9

70 23-70 Beyond simple inheritance patterns cont’d. Polygenic inheritance cont’d. –Skin color Controlled by many gene pairs and many alleles Assuming the simplest model, 2 alleles at loci –A and B –When an AaBb person has children with another AaBb person, children can range from very light to very dark AABBVery dark AABb or AaBBDark AaBb, AAbb, aaBB Medium brown Aabb or aaBbLight aabbVery light

71 23-71 Beyond simple inheritance patterns cont’d. Environmental influences –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

72 23-72 Coat color in Himalayan rabbits Fig

73 23-73 Beyond simple inheritance patterns cont’d. Incomplete dominance and codominance –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 Such a cross would produce a phenotypic ratio of 1 : 2 : 1.

74 23-74 Incomplete dominance

75 23-75 ABO Blood Types –Codominance Occurs when both alleles are equally expressed Ex: type AB blood has both A antigens and B antigens on red blood cells A person can have an allele for an A antigen (blood type A) or a B antigen (blood type B), both A and B antigens (blood type AB), or no antigen (blood type O) on the red blood cells. Human blood types can be type A (I A I A or I A i), type B (I B I B or I B i), type AB (I A I B ), or type 0 (ii).

76 23-76 Beyond simple inheritance patterns cont’d. Multiple allele inheritance –The gene exists in several allelic forms, but each person still has only 2 of the possible alleles –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 ABI A I B Oii

77 23-77 Inheritance of blood type

78 23-78 Beyond simple inheritance patterns cont’d. ABO blood types 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

79 23-79 Beyond simple inheritance patterns cont’d. 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

80 Sex-linked inheritance Sex chromosomes –22 pairs of autosomes, 1 pair of sex chromosomes X and y - X chromosome has many genes, the Y chromosome does not –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

81 23-81 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 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 A female would have to have two recessive genes to express the trait; a male would only need one.

82 23-82 X-Linked Alleles The key for an X-linked problem shows the allele attached to the X as in: X B = normal vision X b = color blindness. Females with the genotype X B X b are carriers because they appear to be normal but each son has a 50% chance of being color blind depending on which allele the son receives. X b X b and X b Y are both colorblind.

83 23-83 Cross involving an X-linked allele

84 23-84 Color Blindness and X- Linked Recessive Disorders Three types of cones are in the retina detecting red, green, or blue. Genes for blue cones are autosomal; those for red and green cones are on the X chromosome. Males are much more likely to have red-green color blindness than females. About 8% of Caucasian men have red-green color blindness. In pedigree charts that show the inheritance pattern for X-linked recessive disorders, more males than females have the trait because recessive alleles on the X chromosome are expressed in males. A grandfather passes an X-linked recessive disorder to a grandson through a carrier daughter. X-linked recessive disorders include red-green color blindness, muscular dystrophy, and hemophilia.

85 23-85 Sex-linked inheritance- 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?

86 Inheritance of linked genes A single chromosome has many genes –The sequence 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 Or, if very few recombined phenotypes appear in offspring, linkage is also suspected –Recall that during meiosis crossing over sometimes occurs If crossing over occurs between 2 alleles of interest, then 4 types of gametes are formed instead of 2 –Fig on the following slide illustrates linkage

87 23-87 Linkage group

88 23-88 Inheritance of linked genes cont’d. Ocurrence of crossing over can indicate the sequence of genes on a chromosome –Crossing over occurs more frequently between distant genes than between those close together on a chromosome 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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