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Topic 10.2.  10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.  10.2.2.

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Presentation on theme: "Topic 10.2.  10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.  10.2.2."— Presentation transcript:

1 Topic 10.2

2  10.2.1 Calculate and predict the genotypic and phenotypic ratio of offspring of dihybrid crosses involving unlinked autosomal genes.  10.2.2 Distinguish between autosomes and sex chromosomes.  10.2.3 Explain how crossing over between non-sister chromatids of a homologous pair in prophase I can result in an exchange of alleles.  10.2.4 Define linkage group.  10.2.5 Explain an example of a cross between two linked genes.  10.2.6 Identify which of the offspring are recombinants in a dihybrid cross involving linked genes.

3  Observations: Seed shape – some round, others wrinkled (allele for round is dominant) Seed colour – some green, others yellow (allele for yellow is dominant)  Mendel crossed true breeding plants with each other One parent: homozygous dominant for both traits (round and yellow seeds) Other parent: homozygous recessive for both traits (wrinkled and green)

4  R = allele for round peas  r = allele for wrinkled peas  Y = allele for yellow peas  y = allele for green peas Parent phenotypes Round yellow Green wrinkled Parent genotypes RRYYrryy Parent gametes RYry F1 genotypes RrYy F1 phenotypes Round yellow

5  Allowing heterozygous offspring to self- pollinate yields following proportions: Round and yellow 56.6% Wrinkled and yellow 18.2% Round and green 19.4% Wrinkled and green 8%  Ratio 9:3:3:1

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7  Sex chromosomes: X and Y (one pair)  Autosomes: any chromosome not X or Y (22 pairs)  Sex-linked gene or trait is located on a sex chromosome.  Autosomal gene or trait is located on one of the autosomes.  On which type of chromosome is the gene for protein production in the testes found?  So, the gene is known as ___.  On which type of chromosome is the gene for protein production in the pancreas found?  So, the gene is known as ___.

8  Two non-sister chromatids can swap segments of their DNA  A maternal chromosome can end up with a segment of a paternal chromosome and vice versa

9  Any two genes which are found on the same chromosome are said to be linked to each other.  Linked genes are usually passed on to the next generation together.  Linkage group - groups of genes on the same chromosome inherited together  Linked genes are the exception to Mendel’s law of independent assortment

10  Fruit fly gene for body color is in the same linkage group as the gene for wing length  G – grey body  g – black body  L – long wings  l – short wings  GGLL – gray body, long wings  ggll – black body, short wings  G L  Two horizontal bars symbolize homologous chromosomes and that the locus of G is on the same chromosome as L

11  A cross between homozygous dominant true-breeding fruit fly (GGLL) and a homozygous recessive true-breeding fruit fly (ggll) would result in flies which were all heterozygous for both of the traits (GgLl)  If heterozygotes for these traits are mixed with homozygous dominant flies, then one could not determine the genotype by sight.  What to do?

12  Cross “mystery” fly with homozygous recessive (ggll)  Assuming the mystery fly is heterozygous (GgLl) for both traits: GLGlgLgl GgLlGgllggLlggll  A new shuffling of the alleles has created a new combination which does not match either of the parents’ genotypes  The term recombinant is used to describe both the new chromosome and the resulting organism.  Recombinants form through the process of crossing over

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14 Topic 10.3

15  10.3.1 Define polygenic inheritance.  10.3.2 Explain that polygenic inheritance can contribute to continuous variation using two examples, one of which must be human skin colour.

16  Involves two or more genes influencing the expression of one trait  Increased number of possible genotypes  Believed that most human traits are too complex to be determined by one gene

17  When an array of possible phenotypes can be produced, it is called continuous variation Examples: skin color, height, body shape, and intelligence These traits are also influenced by environmental conditions  When only a number of phenotypes can be produced, it is called discontinuous variation Examples: earlobe attachment, widow’s peak, blood type

18 CONTINUOUS VARIATION DISCONTINUOUS VARIATION

19  Iris is made up of zones, rings, streaks or speckles of different colored pigments with varying intensities  What color are your eyes, really?  Since there is so much variety, eye color must be influenced by multiple alleles and has continuous variation.

20  Many societies feel the need to label people with categories  Oversimplification serves more administrative purposes than any biological purpose  So what is the purpose of skin?!  PROTECTION FROM THE SUN’S HARMFUL ULTRAVIOLET RADIATION!!!

21  Melanin protects from serious sunburn which can lead to melanoma  Melanin is found in all individuals except albinos  Concentration of melanin varies  How can one increase the melanin level in their skin?  EXPOSE ONESELF TO SUNLIGHT  Tanning is a natural defense against the negative effects of excess sunlight

22  UVB radiation helps the skin to make vitamin D  Vitamin D is necessary for proper growth and bone formation  It is good to have a MODERATE amount of sunlight  In regions of low Sun exposure people need light-colored skin  Why?

23  How do people of varying degrees of skin color relocated to parts of the world that receive differing amounts of sunlight get vitamin D? How do others fight off the sun?  Should there be equal esteem for all humans?  Why is human diversity so often used to divide and discriminate, rather than be appreciated, respected, and celebrated?


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