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Honors Biology chapter 23 Genetics

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1 Honors Biology chapter 23 Genetics
John Regan Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2 23.1 Mendel’s laws Heredity - Biological inheritance
Passing of traits from parents to offspring Genetics- Study of heredity Gregor Mendel 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 Law of Dominance One form of a factor may be dominant over an alternative form Dominant (capital letter) Prevents the expression of the other form –recessive (small letter)

3 Alleles- alternate forms of a gene for the same trait
Mendel’s laws cont’d. Alleles- alternate forms of a gene for the same trait Capital 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 Genotype- genetic composition of an individual with regard to a specific 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 Phenotype- physical characteristics of a trait - outward appearance of a trait

4 Mendel’s laws 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 The Inheritance of Traits
The parent generation is also known as the P generation. The offspring of this P cross are called the first filial (F1) generation. The second filial (F2) generation is the offspring from the F1 cross.

6 Mendel’s laws cont’d. What we know now
Genes are sections of chromosomes Chromosomes come in pairs called homologous pairs Homologous pairs have genes controlling the same traits Genes are 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: blues vs. brown eyes) Alternative forms of a gene for a trait are called alleles

7 Gene locus Fig. 23.2

8 Mendel’s laws cont’d. The inheritance of a single trait cont’d.
Phenotype- physical appearance of the individual with regard to a trait Homozygous dominant individual and heterozygous individual will have the same genotype in this instance Homozygous recessive individual will have a different phenotype Example- inheritance of a widow’s peak vs. a straight hairline in people 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

9 Widow’s peak Fig. 23.3

10 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

11 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

12 Genotype related to phenotype
Table 23.1

13 Mendel’s laws cont’d. One trait crosses-monohybrid cross
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 Ww

14 Mendel’s laws cont’d. One-trait cross cont’d.
In this example there was only 1 combination of eggs and sperm possible What if the parents were each Ww? Then sperm could have either W or w, and eggs also could have W or w One way to figure out the possible combinations of eggs and sperm that could occur is to use a Punnett square This is illustrated on the following slide

15 A punnett square is an easy way to figure out all possible combinations of eggs and sperm

16 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 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

17 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

18 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 is the sum of the individual chances 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%

19 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 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-has a known genotype If there are any offspring produced with the recessive phenotype, then the dominant parent must be heterozygous

20 One-trait testcross Fig. 23.5

21 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? 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.

22 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

23 Segregation and independent assortment
Fig. 23.6

24 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

25 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 slide Figure the genotypes of the offspring Phenotypic ratio: 9 widow’s peak, short fingers: 3 widow’s peak, long fingers; 3 straight hairline, short fingers; 1 straight hairline, long fingers

26 Dihybrid cross Fig. 23.7

27 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

28 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

29 Two-trait testcross Fig. 23.8

30 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?

31 23.2 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

32 Polygenic inheritance
Fig. 23.9

33 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 AABB Very dark AABb or AaBB Dark AaBb, AAbb, aaBB Medium brown Aabb or aaBb Light aabb Very light

34 Beyond simple inheritance patterns cont’d.
Environmental influences Environment can influence gene expression and therefore phenotype Ex: sunlight exposure on skin-suntan; coat color in Himalayan rabbits only make black pigment where body temp. is colder- extremities 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

35 Coat color in Himalayan rabbits

36 Beyond simple inheritance patterns cont’d.
Incomplete dominance and codominance Incomplete dominance Heterozygous individual has a phenotype intermediate to the two homozygous individuals Ex: 4 o’clock flowers red incompletely dominant with white; heterozygote is pink 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 Codominance Occurs when both alleles are equally expressed Ex: horses:chestnut codominant w/cream; heterozygote is palomino Ex: cattle: red codominant w/ white; heterozygote is roan Ex: type AB blood has both A antigens and B antigens on red blood cells

37 Incomplete dominance Fig

38 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 IA = A antigens on RBCs IB = B antigens on RBCs i = has neither A nor B antigens on RBCs Both IA and IB are dominant over I, IA and IB are codominant Phenotype Genotype A IAIA or IAi B IBIB or IBi AB IAIB O ii

39 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

40 Inheritance of blood type

41 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 O Mrs. Doe type A Mrs. Jones type A Baby 2 type B Mr. Doe type A Mr. Jones type AB

42 23.3 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

43 Sex-linked inheritance cont’d.
X-linked traits Red-green colorblindness is X-linked The X chromosome has genes for normal color vision XB = normal vision Xb – colorblindness Genotypes Phenotypes XBXB female with normal color vision XBXb carrier female with normal color vision XbXb colorblind female XBY male with normal color vision XbY colorblind male

44 Cross involving an X-linked allele

45 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?

46 23.4 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

47 Linkage group Fig

48 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?

49 Chapter 23 Objectives Know vocabulary terms like allele, homozygous, heterozygous, genotype, phenotype. Know Mendel’s laws. Be able to solve a 1-trait genetics problem for simple dominance using a Punnett square and use it to determine chances of a trait appearing in the offspring. Be able to determine genotype of members of a family from phenotypic information. Be able to solve incomplete dominance and codominace problems. Be able to solve a 2-trait genetics problem. Know that the phenotypic ratio for a dihybrid cross is 9:3:3:1. Be able to solve ABO blood type problems. Know how multiple alleles and polygenic traits affect inheritance. Know that environmental factors can affect phenotypes. Be able to solve problems for sex-linked traits.

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