Test to see problem with tables
[qwiz]
[q multiple_choice=”true” unit=”Information_Flow” topic=”Genetics”] Blood type inheritance involves three alleles. While there are several notation systems for representing these alleles, a common one has alleles IA and IB as codominant to one another, with each of these dominant over i. If a child has type O blood (genotype ii), which of the following blood types is impossible in either parents?
[c]IFR5cGUgTw==[Qq]
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Cg==| [Qq] | ||
| ii |
If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.
| ___ | i | |
| ___ | ||
| i | ii |
Each parent is __i. It’s completely possible that the child could have a parent that’s type O (genotype ii). But what genotype is impossible for either of the parents to have?
[c]IFR5cGUgQg==[Qq]
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Cg==| [Qq] | ||
| ii |
If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.
| ___ | i | |
| ___ | ||
| i | ii |
Each parent is __i. It’s completely possible that the child could have a parent that’s type B (genotype IBi). But what genotype is impossible for either of the parents to have?
[c]IFR5cGUgQQ==[Qq]
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Cg==| [Qq] | ||
| ii |
If the child is ii, then each of their parents must have given them an i allele, as shown in the slightly more complete Punnett square shown below.
| ___ | i | |
| ___ | ||
| i | ii |
Each parent is __i. It’s completely possible that the child could have a parent that’s type A (genotype IAi). But what genotype is impossible for either of the parents to have?
[c]IFR5cG UgQUI=[Qq]
[f]IFdheSB0byBnby4gR3JlZ29yIE1lbmRlbCB3b3VsZCBiZSBwcm91ZC4gQSBjaGlsZCB3aXRoIHR5cGUgTyBibG9vZCBuZWVkcyBlYWNoIG9mIHRoZWlyIHBhcmVudHMgdG8gZ2l2ZSB0aGVtIGE=IGk=IGFsbGVsZS4gR2l2ZW4gdGhhdCwgdHlwZSBBQiBpcyBpbXBvc3NpYmxlIChiZWNhdXNlIGEgdHlwZSBBQiBwYXJlbnQgY291bGQgZ2l2ZSB0aGVpciBjaGlsZCBlaXRoZXLCoA==SQ==QcKgwqA=[Qq]or IB, but not a i.
[q multiple_choice=”true” unit=”Genetics”] Blood type inheritance involves three alleles. While there are several notation systems for representing these alleles, a common one has alleles IA and IB as codominant to one another, with each of these dominant over i. If a man with type B blood has children with a woman with type O blood, which of the following blood types would be impossible in their children.
[c]IFR5cGUgTw==[Qq]
[f]IE5vLiBUaGUgbWFuIHdpdGggdHlwZSBCIGJsb29kIGNvdWxkIGhhdmUgb25lIG9mIHR3byBnZW5vdHlwZXM6wqA=SQ==Qg==aSA=KGlmIGhlIHdlcmUgaGV0ZXJvenlnb3VzKSw=[Qq] or IBIB, (if he were homozygous type B). The woman is type O, so her genotype has to be ii. If you can set up two Punnett squares on your own, then do so, and you’ll see the two genotypes that are impossible. If not, then continue reading below.
Here’s the Punnett square that shows the father as a heterozygote:
| IB | i | |
| i | ||
| i | ii |
Here’s the Punnett square that shows the father as a homozygote:
| IB | IB | |
| i | ||
| i | ii |
Complete both Punnett squares by bringing the alleles down and over. As you can see by the first Punnett square, a type O child is possible. But which blood type is impossible?
[c]IFR5cGUgQg==[Qq]
[f]IE5vLiBUaGUgbWFuIHdpdGggdHlwZSBCIGJsb29kIGNvdWxkIGhhdmUgb25lIG9mIHR3byBnZW5vdHlwZXM6wqA=SQ==Qg==aSA=KGlmIGhlIHdlcmUgaGV0ZXJvenlnb3VzKSw=[Qq] or IBIB, (if he were homozygous type B). The woman is type O, so here genotype has to be ii. If you can set up two Punnett squares on your own, then do so, and you’ll see the two genotypes that are impossible. If not, then continue reading below.
Here’s the Punnett square that shows the father as a heterozygote:
| IB | i | |
| i | ||
| i | ii |
Here’s the Punnett square that shows the father as a homozygote:
| IB | IB | |
| i | ||
| i | ii |
Complete both Punnett squares by bringing the alleles down and over. As you can see, in either Punnett square, a type B child is possible. But which blood type is impossible?
[c]IFR5cG UgQUI=[Qq]
[f]IENvcnJlY3QhwqBUaGUgbWFuIHdpdGggdHlwZSBCIGJsb29kIGNvdWxkIGhhdmUgb25lIG9mIHR3byBnZW5vdHlwZXM6wqA=SQ==Qg==aSA=KGlmIGhlIHdlcmUgaGV0ZXJvenlnb3VzKSw=[Qq] or IBIB, (if he were homozygous type B). The woman is type O, so her genotype has to be ii. The offspring will all be type B or type O. Type AB is impossible (as is type A).
[q multiple_choice=”true”] In Drosophila, allele W codes for normal wings, while allele w codes for vestigial wings. Allele G codes for normal body color, while allele g codes for ebony coloring. A heterozygous normal winged, ebony male is crossed with a vestigial winged, heterozygous normal body color female. What ratio of of phenotypes is expected in the offspring?
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[f]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
Cg==| [Qq] | wG | wg |
| Wg | WgGg | |
| wg |
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Cg==| [Qq] | wG | wg |
| Wg | WgGg | |
| wg |
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[f]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
Cg==| [Qq] | wG | wg |
| Wg | WgGg | |
| wg |
[q multiple_choice=”true”] A primatologist studying lowland gorillas has developed a scheme for classifying gorilla noses as long, medium, or short in length, and either wide or narrow. She observes the results of several naturally-occurring matings, which are recorded below.
- Long, wide X long, wide | All long, wide
- Short, narrow X long, narrow | all medium, narrow
- Medium, wide X medium, narrow | 1/4 long wide, 1/2 medium wide, 1/4 short wide
- Short, wide X medium, narrow | 1/2 medium, wide; 1/2 short, wide
- Short, wide X short, narrow | all short, wide
Which of the following best describes the underlying genetics for inheritance of nose length.
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[c]IFR3byBsb2NpIGdvdmVybiBub3NlIGxlbmd0aC4gSWYgdGhlcmUgYXJlIGRvbWluYW50IGFsbGVsZXMgYXQgYm90aCBsb2NpLCB0aGUgbm9zZSB3aWxsIGJlIExvbmcuIElmIG9uZSBsb2N1cyBoYXMgYSBkb21pbmFudCBhbGxlbGUsIHRoZSBub3NlIHdpbGwgYmUgc2hvcnQuIElmIHRoZXJlIGFyZSBubyBkb21pbmFudCBhbGxlbGVzIGF0IGVpdGhlciBsb2N1cywgdGhlIG5vc2Ugd2lsbCBiZSBtZWRpdW0u[Qq]
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[c]IE9uZSBsb2N1cyBnb3Zlcm5zIG5vc2UgbGVuZ3RoLiBUaGUgbG9jdXMgaGFzIHR3byBhbGxlbGVzLiBBIGdvcmlsbGEgY2FuIGJlIG hvbW96eWdvdXMgbG9uZyBvciBob21venlnb3VzIHNob3J0LiBIZXRlcm96eWdvdGVzIGhhdmUgbWVkaXVtIG5vc2UgbGVuZ3RoLg==[Qq]
[f]IEV4Y2VsbGVudC4gVGhpcyBtb2RlbCB3b3JrcyBwZXJmZWN0bHkgZm9yIGRlc2NyaWJpbmcgaW5oZXJpdGFuY2Ugb2Ygbm9zZSBsZW5ndGgu[Qq]
[q multiple_choice=”true”] A research team is studying the genetics of beak length in house sparrows, where beak length has two phenotypes, long and short. Pedigree studies have shown that this phenotype is under the control of a single autosomal gene with two alleles. The DNA for each allele is isolated and sequenced, and it’s determined that a single nucleotide difference accounts for the short mutation.
When the DNA of both alleles is subjected to restriction fragment analysis, the pattern below is produced.
Based on the diagram, the most reasonable explanation for the different RFLPs associated with the long and short beaked alleles is that a mutation in the long allele DNA
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[q multiple_choice=”true”] In Drosophila, allele W codes for normal wings, while allele w codes for vestigial wings. Allele G codes for normal body color, while allele g codes for ebony. A heterozygous normal winged, ebony male is crossed with a vestigial winged, heterozygous normal body color female.
What is the genotype of the male fly?
[c]IFd3R2c=[Qq]
[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LiBXd0dnIHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBoZXRlcm96eWdvdXMgbm9ybWFsIHdpbmdlZCwgaGV0ZXJvenlnb3VzIG5vcm1hbCBib2R5IGNvbG9yLiBXaGF0IGdlbm90eXBlIHdvdWxkIGEgZmx5IGhhdmUgdG8gaGF2ZSB0byBiZSBlYm9ueT8=[Qq]
[c]IFdXR0c=[Qq]
[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LldXR0cgd291bGQgYmUgZGVzY3JpYmVkIGFzIGhvbW96eWdvdXMgbm9ybWFsIHdpbmdlZCwgaG9tb3p5Z291cyBub3JtYWwgYm9keSBjb2xvci4gV2hhdCBnZW5vdHlwZSB3b3VsZCBhIGZseSBoYXZlIHRvIGJlIGhldGVyb3p5Z291cyBmb3Igbm9ybWFsIHdpbmdzLCBhbmQgZWJvbnkgYm9kaWVkLg==[Qq]
[c]IHd3Z2c=[Qq]
[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LlRoZSBnZW5vdHlwZSB3d2dnIHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBob21venlnb3VzIHZlc3RpZ2lhbCB3aW5nZWQsIGhvbW96eWdvdXMgZWJvbnkgYm9keSBjb2xvciAoc28geW91IGdvdCB0aGUgc2Vjb25kIHBhcnQgcmlnaHQpIFdoYXQgZ2Vub3R5cGUgd291bGQgYSBmbHkgaGF2ZSB0byBiZSBoZXRlcm96eWdvdXMgZm9yIG5vcm1hbCB3aW5ncz8=[Qq]
[c]IHd3R0c=[Qq]
[f]IE5vLiBUaGUgbWFsZSBpcyBkZXNjcmliZWQgYXMgaGV0ZXJvenlnb3VzIG5vcm1hbCB3aW5nZWQsIGVib255LlRoZSBnZW5vdHlwZSB3d0dHIHdvdWxkIGJlIGRlc2NyaWJlZCBhcyBob21venlnb3VzIHZlc3RpZ2lhbCB3aW5nZWQsIGhvbW96eWdvdXMgbm9ybWFsIGJvZHkgY29sb3IuIFdoYXQgZ2Vub3R5cGUgd291bGQgYSBmbHkgaGF2ZSB0byBoYXZlIHRvIGJlIGhldGVyb3p5Z291cyBmb3Igbm9ybWFsIHdpbmdzLCBhbmQgZWJvbnkgYm9kaWVkIChhbmQgcmVtZW1iZXIgdGhhdCBlYm9ueSBib2RpZWQgaXMgYSByZWNlc3NpdmUgdHJhaXQpLg==[Qq]
[c]IFd3 Z2c=[Qq]
[f]IEZhYnVsb3VzLiBBIGdlbm90eXBlIG9mIFd3Z2cgbWVhbnMgdGhhdCB0aGUgZmx5IGlzIGhldGVyb3p5Z291cyBmb3Igbm9ybWFsIHdpbmdzLCBhbmQgZWJvbnkgY29sb3JlZC4=[Qq]
[q multiple_choice=”true”] Imagine that an error has during DNA replication in a cell. The error substitutes a G in place of a T in one of the cell’s genes. Which of the following is the most likely effect that this will have on the cell?
[c]IEVhY2ggb2YgdGhlIGNlbGwmIzgyMTc7cyBwcm90ZWlucyB3aWxsIGNvbnRhaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQu[Qq]
[f]IE5vLiBUaGUgbWFpbiBwcm9ibGVtIHdpdGggdGhpcyBjaG9pY2UgaXMgdGhhdCB0aGUgbXV0YXRpb24gd2lsbCBhZmZlY3Qgb25seSBhIHNpbmdsZSBwcm90ZWluIChub3QgZWFjaCBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zKS4gV2hlbiB5b3UgbmV4dCBzZWUgdGhpcyBxdWVzdGlvbiwgdGhpbmsgb2YgaG93IGluZm9ybWF0aW9uIGluIEROQSBnZXRzIHRyYW5zbGF0ZWQgaW50byBwcm90ZWluLCBhbmQgY29uc2lkZXIgdGhlIGxpa2VseSBlZmZlY3Qgb2YgYSBzaW5nbGUgYmFzZSBwYWlyIHN1YnN0aXR1dGlvbi4=[Qq]
[c]IFRoZSBhbWlubyBhY2lkIHNlcXVlbmNlIG9mIG9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zIHdpbGwgYmUgY29tcGxldGVseSBhbHRlcmVkLg==[Qq]
[f]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[Qq]
[c]IE9uZSBhbWlubyBhY2lkIHdpbGwgYmUgbWlzc2luZyBmcm9tIG9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zLA==[Qq]
[f]IE5vLiBBbGwgdGhhdCB0aGlzIG11dGF0aW9uIGRvZXMgaXMgc3Vic3RpdHV0ZSBvbmUgbnVjbGVvdGlkZSBmb3IgYW5vdGhlci4gVGhhdCBtaWdodCByZXN1bHQgaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQsIGJ1dCBpdCB3b3VsZG4mIzgyMTc7dCByZXN1bHQgaW4gYSBtaXNzaW5nIGFtaW5vIGFjaWQu[Qq]
[c]IE9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW5zIG1p Z2h0IGNvbnRhaW4gYW4gaW5jb3JyZWN0IGFtaW5vIGFjaWQu[Qq]
[f]IENvcnJlY3QuIEEgbGlrZWx5IGVmZmVjdCBvZiBhIHNpbmdsZSBudWNsZW90aWRlIGNoYW5nZSBpcyB0aGF0IG9uZSBvZiB0aGUgY2VsbCYjODIxNztzIHByb3RlaW4gbWlnaHQgY29udGFpbiBhbiBpbmNvcnJlY3QgYW1pbm8gYWNpZC4=[Qq]
[q multiple_choice=”true”] A team of biologists is studying gene expression in rat liver tissue. They find that two different proteins with different structures were translated from two different mRNAs. When they traced the source of the mRNA, however, they found the both mRNAs were transcribed from the same gene in the cell’s nucleus. Which of the following explanation best accounts for what the team has uncovered?
[c]IFRoZSBnZW5lIG1pZ2h0IGhhdmUgYmVlbiBhbHRlcmVkIGJ5IGEgbXV0YXRpb24u[Qq]
[f]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[Qq]
[c]IFRoZSBkaWZmZXJlbnQgZnVuY3Rpb25zIG9mIGVhY2ggcHJvdGVpbiByZXF1aXJlIHRoZW0gdG8gaGF2ZSBkaWZmZXJlbnQgdW5kZXJseWluZyBSTkFzIHdoaWNoIGNvZGUgZm9yIHRoZW0u[Qq]
[f]IE5vLiBSZW1lbWJlciB0aGUgY2VudHJhbCBkb2dtYSBvZiBtb2xlY3VsYXIgYmlvbG9neTogRE5BIG1ha2VzIFJOQSBtYWtlcyBwcm90ZWluLiBUaGlzIGNob2ljZSBpcyBwcm9wb3NpbmcgYSBwcm9jZXNzIHRoYXQgdmlvbGF0ZXMgdGhpcyBrZXkgaWRlYS7CoFRoZSBuZXh0IHRpbWUgeW91IHNlZSB0aGlzIHF1ZXN0aW9uLCB0aGluayBhYm91dCBnZW5lIGV4cHJlc3Npb24gaW4gZXVrYXJ5b3RlcywgYW5kIHNlZSBpZiB5b3UgY2FuIHJlbWVtYmVyIGEgbWVjaGFuaXNtIGJ5IHdoaWNoIHR3byBSTkFzIGNvdWxkIHJlc3VsdCBmcm9tIG9uZSBnZW5lLg==[Qq]
[c]IERpZmZlcmVudCBzeXN0ZW1zIG9mIGhpc3RvbmUgYWNldHlsYXRpb24gY291bGQgcmVzdWx0IGluIHR3byBkaWZmZXJlbnQgbVJOQXMu[Qq]
[f]IE5vLiBEaWZmZXJlbnQgcGF0dGVybnMgb2YgaGlzdG9uZSBhY2V0eWxhdGlvbiBjYW4gcmVzdWx0IGluIGRpZmZlcmVudCBwYXR0ZXJucyBvZiBnZW5lIGV4cHJlc3Npb24sIGJ1dCBpdCB3b3VsZG4mIzgyMTc7dCBwcm9kdWNlIHR3byBtUk5BcyBmcm9tIG9uZSBnZW5lLsKgVGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24sIHRoaW5rIGFib3V0IGdlbmUgZXhwcmVzc2lvbiBpbiBldWthcnlvdGVzLCBhbmQgc2VlIGlmIHlvdSBjYW4gcmVtZW1iZXIgYSBtZWNoYW5pc20gYnkgd2hpY2ggdHdvIFJOQXMgY291bGQgcmVzdWx0IGZyb20gb25lIGdlbmUu[Qq]
[c]IER1cmluZyBwcmUtbVJOQSBwcm9jZXNzaW5nLCBleG9ucyBjb3VsZCBiZSBzcGxpY2Vk IHRvZ2V0aGVyIGluIHZhcmlvdXMgd2F5cyB0byBtYWtlIGRpZmZlcmVudCBtUk5Bcy4=[Qq]
[f]IFdheSB0byBnbyEgSWYgb25lIGdlbmUgaXMgcmVzdWx0aW5nIGluIHR3byBtUk5BcywgdGhlbiB3aGF0JiM4MjE3O3MgaGFwcGVuaW5nIGlzIGFsdGVybmF0aXZlIHNwbGljaW5nIG9mIFJOQS4gVGhlc2UgZGlmZmVyZW50IFJOQXMgd291bGQgdGhlbiBnbyBvbiB0byBiZSB0cmFuc2xhdGVkIGFzIGRpc3RpbmN0IHByb3RlaW5zLg==[Qq]
[c]IE9uZSBzZXQgb2Ygcmlib3NvbWVzIHJlYWQgdGhlIG1STkEgaW4gdGhlIDUmIzgyNDI7IHRvIDMmIzgyNDI7IGRpcmVjdGlvbi4gQW5vdGhlciBzZXQgcmVhZHMgdGhlbSBpbiB0aGUgMyYjODI0MjsgdG8gNSYjODI0MjsgZGlyZWN0aW9uLg==[Qq]
[f]IE5vLiBBbGwgcmlib3NvbWVzICh3aGV0aGVyIGluIGFuwqA=RS4gY29saQ==IGJhY3Rlcml1bSBvciBhbiBvcmNoaWQsIHJlYWQgbVJOQSBpbiB0aGUgNSYjODI0MjsgdG8gMyYjODI0MjsgZGlyZWN0aW9uLsKgVGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24sIHRoaW5rIGFib3V0IGdlbmUgZXhwcmVzc2lvbiBpbiBldWthcnlvdGVzLCBhbmQgc2VlIGlmIHlvdSBjYW4gcmVtZW1iZXIgYSBtZWNoYW5pc20gYnkgd2hpY2ggdHdvIFJOQXMgY291bGQgcmVzdWx0IGZyb20gb25lIGdlbmUu[Qq]
[q multiple_choice=”true”] In shorthorn cattle, coat color can be solid red, solid white, or red-roan (a mixture of white and pigmented hairs). When a true breeding solid red male is crossed with a white female, the offspring are roan. This demonstrates that
[c]IGNvYXQgY29sb3IgaXMgY29udHJvbGxlZCBieSBhdCBsZWFzdCB0aHJlZSBhbGxlbGVzLg==[Qq]
[f]IE5vLiBZb3UgbWlnaHQgYmUgdGhpbmtpbmcgb2Ygc3lzdGVtcyBsaWtlIGluaGVyaXRhbmNlIG9mIGJsb29kIHR5cGUsIHdoZXJlIHRoZSBBIGFsbGVsZSBhbmQgdGhlIEIgYWxsZWxlIHByb2R1Y2UgYmxvb2QgdHlwZSBBQi4gQnV0IG5vdGUgdGhhdCB5b3UgY2FuIGFjaGlldmUgdGhlIGludGVybWVkaWF0ZSBwaGVub3R5cGUgd2l0aCBqdXN0IHR3byBhbGxlbGVzIChhcyBvcHBvc2VkIHRvIHRocmVlKS4gS2VlcCB0aGF0IGluIG1pbmQgd2hlbiB5b3UgbmV4dCBzZWUgdGhpcyBxdWVzdGlvbi4=[Qq]
[c]IHJlZCBhbmQgd2hpdGUgc2hvdyBp bmNvbXBsZXRlIGRvbWluYW5jZS4=[Qq]
[f]IFllcy4gSWYgcg==IGlzIHRoZSBhbGxlbGUgZm9yIFJlZCBhbmQgdw==IGlzIHRoZSBhbGxlbGUgZm9yIHdoaXRlLCB0aGVuIHRoZSBjcm9zcyBhYm92ZSBjb3VsZCBiZSByZXByZXNlbnRlZCBhcyA=[Qq]rr x ww, with all the offspring having the genotype rw, and the roan phenotype.
[c]IHRoZSByZWQgYWxsZWxlIGlzIGRvbWluYW50IG92ZXIgdGhlIHdoaXRlIGFsbGVsZS4=[Qq]
[f]IE5vLiBJZiA=Ug==IHJlcHJlc2VudGVkIHRoZSBkb21pbmFudCBhbGxlbGUgZm9yIHJlZCwgYW5kIA==cg==IHJlcHJlc2VudGVkIHRoZSByZWNlc3NpdmUgYWxsZWxlIGZvciB3aGl0ZSwgdGhlbiBhIGNyb3NzIGJldHdlZW4gdHJ1ZSBicmVlZGluZyByZWQgYW5kIHdoaXRlIHdvdWxkIGJlIFJSIHggcnIuIEFsbCB0aGUgb2Zmc3ByaW5nIHdvdWxkIGJlIA==[Qq]Rr, and we’d expect them to be red, not roan. What kind of inheritance system would result in pure bred red crossed with purebred white resulting in an intermediate phenotype?
[c]IENyb3NzaW5nIG92ZXIgYW5kIHJlY29tYmluYXRpb24gb2NjdXJyaW5nIGR1cmluZyBtZWlvc2lzLg==[Qq]
[f]IE5vLiBDcm9zc2luZyBvdmVyIGFuZCByZWNvbWJpbmF0aW9uIGFsbW9zdCBjZXJ0YWlubHkgZGlkIG9jY3VyIGR1cmluZyBtZWlvc2lzLCBidXQgaXQgd291bGRuJiM4MjE3O3QgYWNoaWV2ZSB0aGUgcmVzdWx0IGRlc2NyaWJlZCBhYm92ZS7CoFdoYXQga2luZCBvZiBpbmhlcml0YW5jZSBzeXN0ZW0gd291bGQgcmVzdWx0IGluIHB1cmUgYnJlZCByZWQgY3Jvc3NlZCB3aXRoIHB1cmVicmVkIHdoaXRlIHJlc3VsdGluZyBpbiBhbiBpbnRlcm1lZGlhdGUgcGhlbm90eXBlPw==[Qq]
[q multiple_choice=”true”] Down syndrome and and Klinefelter’s syndrome have in which of the following in common?
[c]IEJvdGggaW52b2x2ZSB0aGUgWCBvciBZIGNocm9tb3NvbWVz[Qq]
[f]IE5vLiBXaGlsZSBLbGluZWZlbHRlciYjODIxNztzIHN5bmRyb21lIGRvZXMgaW52b2x2ZSB0aGUgWCBjaHJvbW9zb21lLCBEb3duIHN5bmRyb21lIGhhcyBubyBjb25uZWN0aW9uIHRvIHRoZSBYIG9yIFkgY2hyb21vc29tZS7CoCBEb3duIHN5bmRyb21lIG9jY3VycyB3aGVuIGEgcGVyc29uIHJlY2VpdmVzIHRocmVlIGNvcGllcyBvZiBjaHJvbW9zb21lIDIxLCByYXRoZXIgdGhhbiB0d28uIEtsaW5lZmVsdGVyJiM4MjE3O3Mgc3luZHJvbWUgb2NjdXJzIHdoZW4gYSBjaHJvbW9zb21hbGx5IG1hbGUgcGVyc29uIGhhcyBhbiBleHRyYSBYIGNocm9tb3NvbWUuIE5leHQgdGltZSwgY2hvb3NlIGFub3RoZXIgYW5zd2VyIChhbmQgbm90ZSB0aGF0IEkmIzgyMTc7dmUganVzdCBnaXZlbiB5b3UgYSBodWdlIGhpbnQgYXMgdG8gd2hhdCB0aGUgYW5zd2VyIGlzKS4=[Qq]
[c]IEJvdGggYXJlIHRoZSByZXN1bHQgb2YgYSBwb2ludCBtdXRhdGlvbnMu[Qq]
[f]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[Qq]
[c]IEJvdGggYXJlIGFzc29jaWF0ZWQgd2l0aCBvZi BjaGFuZ2VzIGluIGNocm9tb3NvbWUgbnVtYmVy[Qq]
[f]IFllcy4gQm90aCBEb3duIHN5bmRyb21lIGFuZCBLbGluZWZlbHRlciYjODIxNztzIHN5bmRyb21lIGludm9sdmUgY2hhbmdlcyBpbiBjaHJvbW9zb21lIG51bWJlci7CoERvd24gU3luZHJvbWUgb2NjdXJzIHdoZW4gYSBwZXJzb24gcmVjZWl2ZXMgdGhyZWUgY29waWVzIG9mIGNocm9tb3NvbWUgMjEsIHJhdGhlciB0aGFuIHR3by4gS2xpbmVmZWx0ZXImIzgyMTc7cyBzeW5kcm9tZSBvY2N1cnMgd2hlbiBhIGNocm9tb3NvbWFsbHkgbWFsZSBwZXJzb24gaGFzIGFuIGV4dHJhIFggY2hyb21vc29tZS4=[Qq]
[c]IEJvdGggY29uZGl0aW9ucyBhcmUgY2F1c2VkIGJ5IGF1dG9zb21hbCBkb21pbmFudCBhbGxlbGVz[Qq]
[f]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[Qq]
[q multiple_choice=”true”] Conditions like hemophilia, Red-green colorblindness, and Duchenne’s muscular dystrophy are inherited a sex-linked recessive disorders. The term “sex-linked” means:
[c]IFRoZSBhbGxlbGUgaXMgY2FycmllZCBvbiBhbiBhdXRvc29tZQ==[Qq]
[f]IE5vLiBUaGUgcmVhc29uIHRoZXNlIGNvbmRpdGlvbnMgYXJlICYjODIyMDtzZXggbGlua2VkJiM4MjIxOyBpcyBiZWNhdXNlIHRoZXkmIzgyMTc7cmUgbm90IGNhcnJpZWQgb24gYXV0b3NvbWVzLiBJZiB0aGV5IHdlcmUgY2FycmllZCBvbiBhdXRvc29tZXMsIHRoZWlyIGluaGVyaXRhbmNlIHdvdWxkIG5vdCBiZSBsaW5rZWQgdG8gc2V4LiBXaGF0IGNocm9tb3NvbWVzIHdvdWxkIGEgY29uZGl0aW9uIGhhdmUgdG8gYmUgbGlua2VkIHRvIGluIG9yZGVyIHRvIGJlIGxpbmtlZCB0byBzZXg/[Qq]
[c]IFRoZSBhbGxlbGUgaXMgY2FycmllZCBvbiB0aGUgWSBjaHJvbW9zb21l[Qq]
[f]IE5vLCBidXQgeW91JiM4MjE3O3JlIHRoaW5raW5nIGFib3V0IHRoaXMgdGhlIHJpZ2h0IHdheS4gSGVtb3BoaWxpYSwgcmVkLWdyZWVuIGNvbG9yYmxpbmRuZXNzLCBhbmQgRHVjaGVubmUmIzgyMTc7cyBtdXNjdWxhciBkeXN0cm9waHkgc2hvdyB1cCBtb3JlIG9mdGVuIGluIG1hbGVzIHRoYW4gZmVtYWxlcywgYnV0IG5vdCBiZWNhdXNlIHRoZXkgYXJlIG9uIHRoZSBZIGNocm9tb3NvbWUuwqBXaGF0IGNocm9tb3NvbWUgd291bGQgYSBjb25kaXRpb24gaGF2ZSB0byBiZSBsaW5rZWQgdG8gaW4gb3JkZXIgdG8gYmUgbGlua2VkIHRvIHNleD8=[Qq]
[c]IFRoZSBhbGxlbGUgaXMgY2Fycmll ZCBvbiB0aGUgWH5jaHJvbW9zb21l[Qq]
[f]IEZhYnVsb3VzISBUaGUgYWxsZWxlcyBmb3IgaGVtb3BoaWxpYSwgcmVkLWdyZWVuIGNvbG9yYmxpbmRuZXNzLCBhbmQgRHVjaGVubmUmIzgyMTc7cyBtdXNjdWxhciBkeXN0cm9waHkgYXJlIGFsbCBmb3VuZCBvbiB0aGUgWCBjaHJvbW9zb21lLCBtYWtpbmcgdGhlbSBzZXggbGlua2VkLg==[Qq]
[c]IFRoZSBhbGxlbGUgaXMgY2FycmllZCBpbiB0aGUgbWl0b2Nob25kcmlhbCBETkE=[Qq]
[f]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[Qq]
[q multiple_choice=”true”] A man and a woman, both with normal vision, have a son who is red-green colorblind. Which of the following is the most likely explanation.
[c]IEJvdGggbWF0ZXJuYWwgZ3JhbmRwYXJlbnRzIGFyZSBjb2xvcmJsaW5k[Qq]
[f]IE5vLiBUaGUga2V5IHRoaW5nIHRvIHJlbWVtYmVyIGhlcmUgaXMgdGhhdCByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUsIHNleC1saW5rZWQgY29uZGl0aW9uLCB3aXRoIHRoZSBhbGxlbGUgZm9yIHRoZSBjb25kaXRpb24gZm91bmQgb24gdGhlIFggY2hyb21vc29tZS4gVGhlIGdlbm90eXBlIG9mIGEgY29sb3JibGluZCBtYWxlIGlzIHJlcHJlc2VudGVkIGFzIFg=Yw==WSwgd2hlcmVhcyBhIG1hbGUgd2l0aCBub3JtYWwgdmlzaW9uIHdvdWxkIGhhdmUgdGhlIGdlbm90eXBlIFg=Qw==WS4gTm93LCB0aGluayBhYm91dCB3aGF0IHlvdSBrbm93IGFib3V0IGluaGVyaXRhbmNlLCBhbmQgYXNrIHlvdXJzZWxmIHdobyB0aGF0IGNvbG9yYmxpbmQgbWFsZSB3b3VsZCBoYXZlIGluaGVyaXRlZCBoaXMgWCBjaHJvbW9zb21lIGZyb20uIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]
[c]IFRoZSBtb3RoZXIgY2Fycmllcy BhIHJlY2Vzc2l2ZSBhbGxlbGU=[Qq]
[f]IEV4Y2VsbGVudCEgQmVjYXVzZSByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUgc2V4IGxpbmtlZCBjb25kaXRpb24sIGl0JiM4MjE3O3MgYWx3YXlzIHRyYW5zbWl0dGVkIGJ5IHRoZSBtb3RoZXIsIHdobyBwYXNzZXMgYSBjb3B5IG9mIHRoZSByZWNlc3NpdmUgYWxsZWxlIHRvIGhlciBzb25zIHRocm91Z2ggYW4gWCBjaHJvbW9zb21lIGluIGhlciBlZ2cu[Qq]
[c]IFRoZSBmYXRoZXIgcGFzc2VkIGhpcyBhbGxlbGUgZm9yIGNvbG9yYmxpbmRuZXNzIHRvIGhpcyBzb24u[Qq]
[f]IE5vLiBUaGUga2V5IHRoaW5nIHRvIHJlbWVtYmVyIGhlcmUgaXMgdGhhdCByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUsIHNleC1saW5rZWQgY29uZGl0aW9uLCB3aXRoIHRoZSBhbGxlbGUgZm9yIHRoZSBjb25kaXRpb24gZm91bmQgb24gdGhlIFggY2hyb21vc29tZS4gVGhlIGdlbm90eXBlIG9mIGEgY29sb3JibGluZCBtYWxlIGlzIHJlcHJlc2VudGVkIGFzIFg=Yw==WSwgd2hlcmVhcyBhIG1hbGUgd2l0aCBub3JtYWwgdmlzaW9uIHdvdWxkIGhhdmUgdGhlIGdlbm90eXBlIFg=Qw==WS4gTm93LCB0aGluayBhYm91dCB3aGF0IHlvdSBrbm93IGFib3V0IGluaGVyaXRhbmNlLCBhbmQgYXNrIHlvdXJzZWxmIHdobyB0aGF0IGNvbG9yYmxpbmQgbWFsZSB3b3VsZCBoYXZlIGluaGVyaXRlZCBoaXMgWCBjaHJvbW9zb21lIGZyb20uIFJlbWVtYmVyIHRoYXQgdGhlIG5leHQgdGltZSB5b3Ugc2VlIHRoaXMgcXVlc3Rpb24u[Qq]
[c]IEJvdGggdGhlIG1vdGhlciBhbmQgdGhlIGZhdGhlciB3ZXJlIGhldGVyb3p5Z291cyBmb3IgdGhlIGNvbG9yLWJsaW5kIGdlbmUu[Qq]
[f]IE5vLiBUaGUga2V5IHRoaW5nIHRvIHJlbWVtYmVyIGhlcmUgaXMgdGhhdCByZWQtZ3JlZW4gY29sb3JibGluZG5lc3MgaXMgYSByZWNlc3NpdmUsIHNleC1saW5rZWQgY29uZGl0aW9uLCB3aXRoIHRoZSBhbGxlbGUgZm9yIHRoZSBjb25kaXRpb24gZm91bmQgb24gdGhlIFggY2hyb21vc29tZS4gVGhlIGdlbm90eXBlIG9mIGEgY29sb3JibGluZCBtYWxlIGlzIHJlcHJlc2VudGVkIGFzIFg=Yw==WSwgd2hlcmVhcyBhIG1hbGUgd2l0aCBub3JtYWwgdmlzaW9uIHdvdWxkIGhhdmUgdGhlIGdlbm90eXBlIFg=Qw==WS4gV2hpbGUgYSBmZW1hbGUgY2FuIGJlIGhldGVyb3p5Z291cyBub3JtYWwsIHdpdGggdGhlIGdlbm90eXBlwqDCoFg=[Qq]CXc, a male can’t be heterozygous. He either inherits the normal allele, or the recessive one (with no countervailing allele on his Y chromosome)
[q multiple_choice=”true”] The karyotype shown below is an example of a
[c]IHRyaX NvbXku[Qq]
[f]IE5pY2Ugam9iISBBcyB5b3Ugb2JzZXJ2ZWQsIHRoZXJlIGFyZSB0aHJlZSBjb3BpZXMgb2YgY2hyb21vc29tZSAxNSwgY3JlYXRpbmcgYSB0cmlzb215Lg==[Qq]
[c]IG1vbm9zb215Lg==[Qq]
[f]IE5vLiBBIG1vbm9zb215IHdvdWxkIGhhdmUganVzdCBvbmUgY2hyb21vc29tZSwgaW5zdGVhZCBvZiBhIHBhaXIuIFlvdSBtaWdodCBoYXZlIGJlZW4gY29uZnVzZWQgYnkgdGhlIHNpbmdsZSBYIGFuZCBZIGNocm9tb3NvbWVzLCBidXQgdGhhdCYjODIxNztzIHRoZSBwYWlyIHRoYXQgY3JlYXRlcyBhIGNocm9tb3NvbWFsbHkgbWFsZSBtYW1tYWwuIFRha2UgYSBjbG9zZXIgbG9vayBhdCB0aGUgY2hyb21vc29tZXMsIGFuZCBzZWUgaWYgeW91IGNhbiBmaW5kIGFub3RoZXIgdHlwZSBvZiBjaHJvbW9zb21hbCBkaWZmZXJlbmNlIGluIHRoaXMga2FyeW90eXBlLg==[Qq]
[c]IHBvbHlwbG9pZHku[Qq]
[f]IE5vLiBQb2x5cGxvaWR5IGludm9sdmVzIGR1cGxpY2F0aW9uIG9mIGVudGlyZSBjaHJvbW9zb21lIHNldHMuIFRoaXMgaXMgYSBzaW5nbGUgc2V0LsKgVGFrZSBhIGNsb3NlciBsb29rIGF0IHRoZSBjaHJvbW9zb21lcywgYW5kIHNlZSBpZiB5b3UgY2FuIGZpbmQgc29tZSB0eXBlIG9mIGNocm9tb3NvbWFsIGRpZmZlcmVuY2UgdGhhdCBjb3JyZXNwb25kcyB0byBvbmUgb2YgdGhlIGFuc3dlcnM=[Qq]
[c]IGNocm9tb3NvbWFsIGludmVyc2lvbi4=[Qq]
[f]IE5vLiBBIGNocm9tb3NvbWFsIGludmVyc2lvbiB3b3VsZG4mIzgyMTc7dCBiZSBwb3NzaWJsZSB0byBkaXNjZXJuIGluIHRoaXMgdHlwZSBvZiBibGFjayBhbmQgd2hpdGUga2FyeW90eXBlLiBBbiBpbnZlcnNpb24gcmVzdWx0cyB3aGVuIGNocm9tb3NvbWFsIHNlZ21lbnRzIGFyZSBmbGlwcGVkLCBhcyBpcyBzaG93biBpbiB0aGUgZGlhZ3JhbSBiZWxvdyAoZnJvbSBub2JlbHByaXplLm9yZyk=
Cg==Cg==[Qq]Take a closer look at the chromosomes, and see if you can find some type of chromosomal difference that corresponds to one of the other answers.
[q multiple_choice=”true”] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype.
What is the frequency of the recessive allele for this trait?
[c]IDAuMzY=[Qq]
[f]IE5vLiBUaGlzIGlzIGEgcG9wdWxhdGlvbiBnZW5ldGljcyBwcm9ibGVtLCBhbmQgdGhlIGVhc2llc3Qgd2F5IHRvIHNvbHZlIHRoaXMgcHJvYmxlbSBpcyB0byB1c2UgYSBjcm9zcyBtdWx0aXBsaWNhdGlvbiB0YWJsZSAoYSBraW5kIG9mIFB1bm5ldHQgc3F1YXJlKS4gWW91IHN0YXJ0IGZyb20gdGhlIGlkZWEgdGhhdCA=cCArIHEgPSAxLCB3aGVyZQ==IHA=IHJlcHJlc2VudHMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hbnQgYWxsZWxlLCBhbmQg[Qq]q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.
[c]IDAuNzQ=[Qq]
[f]IE5vLiBUaGlzIGlzIGEgcG9wdWxhdGlvbiBnZW5ldGljcyBwcm9ibGVtLCBhbmQgdGhlIGVhc2llc3Qgd2F5IHRvIHNvbHZlIHRoaXMgcHJvYmxlbSBpcyB0byB1c2UgYSBjcm9zcyBtdWx0aXBsaWNhdGlvbiB0YWJsZSAoYSBraW5kIG9mIFB1bm5ldHQgc3F1YXJlKS4gWW91IHN0YXJ0IGZyb20gdGhlIGlkZWEgdGhhdCA=cCArIHEgPSAxLCB3aGVyZQ==IHA=IHJlcHJlc2VudHMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hbnQgYWxsZWxlLCBhbmQg[Qq]q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.
[c]IDAu Ng==[Qq]
[f]IEZhYnVsb3VzISBIYXJkeSBhbmQgV2VpbmJlcmcgKHdobyBkZXZlbG9wZWQgdGhlIG1hdGggZm9yIHRoaXMpIHdvdWxkIGJlIHByb3VkLiBJZiB0aGUgZnJlcXVlbmN5IG9mIHRoZSByZWNlc3NpdmUgcGhlbm90eXBlIGlzIDAuMzYsIHRoZW4gKGFzc3VtaW5nIHRoZSBwb3B1bGF0aW9uIGlzIGluIEhhcmR5LVdlaW5iZXJnIGVxdWlsaWJyaXVtKSwgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgcmVjZXNzaXZlIGFsbGVsZSB3aWxsIGJlIDAuNi4=[Qq]
[c]IDAuNA==[Qq]
[f]IE5vLiBUaGlzIGlzIGEgcG9wdWxhdGlvbiBnZW5ldGljcyBwcm9ibGVtLCBhbmQgdGhlIGVhc2llc3Qgd2F5IHRvIHNvbHZlIHRoaXMgcHJvYmxlbSBpcyB0byB1c2UgYSBjcm9zcyBtdWx0aXBsaWNhdGlvbiB0YWJsZSAoYSBraW5kIG9mIFB1bm5ldHQgc3F1YXJlKS4gWW91IHN0YXJ0IGZyb20gdGhlIGlkZWEgdGhhdCA=cCArIHEgPSAxLCB3aGVyZQ==IHA=IHJlcHJlc2VudHMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hbnQgYWxsZWxlLCBhbmQg[Qq]q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.
[c]IDAuNDg=[Qq]
[f]IE5vLiBUaGlzIGlzIGEgcG9wdWxhdGlvbiBnZW5ldGljcyBwcm9ibGVtLCBhbmQgdGhlIGVhc2llc3Qgd2F5IHRvIHNvbHZlIHRoaXMgcHJvYmxlbSBpcyB0byB1c2UgYSBjcm9zcyBtdWx0aXBsaWNhdGlvbiB0YWJsZSAoYSBraW5kIG9mIFB1bm5ldHQgc3F1YXJlKS4gWW91IHN0YXJ0IGZyb20gdGhlIGlkZWEgdGhhdCA=cCArIHEgPSAxLCB3aGVyZQ==IHA=IHJlcHJlc2VudHMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hbnQgYWxsZWxlLCBhbmQg[Qq]q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). I’ll leave that step to you.
[q multiple_choice=”true”] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype. What is the frequency of the dominant allele for this trait?
[c]IDAuMzY=[Qq]
[f]IE5vLiBUaGlzIGlzIGEgcG9wdWxhdGlvbiBnZW5ldGljcyBwcm9ibGVtLCBhbmQgdGhlIGVhc2llc3Qgd2F5IHRvIHNvbHZlIHRoaXMgcHJvYmxlbSBpcyB0byB1c2UgYSBjcm9zcyBtdWx0aXBsaWNhdGlvbiB0YWJsZSAoYSBraW5kIG9mIFB1bm5ldHQgc3F1YXJlKS4gWW91IHN0YXJ0IGZyb20gdGhlIGlkZWEgdGhhdCA=cCArIHEgPSAxLCB3aGVyZQ==IHA=IHJlcHJlc2VudHMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hbnQgYWxsZWxlLCBhbmQg[Qq]q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?
[c]IDAuNzQ=[Qq]
[f]IE5vLiBUaGlzIGlzIGEgcG9wdWxhdGlvbiBnZW5ldGljcyBwcm9ibGVtLCBhbmQgdGhlIGVhc2llc3Qgd2F5IHRvIHNvbHZlIHRoaXMgcHJvYmxlbSBpcyB0byB1c2UgYSBjcm9zcyBtdWx0aXBsaWNhdGlvbiB0YWJsZSAoYSBraW5kIG9mIFB1bm5ldHQgc3F1YXJlKS4gWW91IHN0YXJ0IGZyb20gdGhlIGlkZWEgdGhhdCA=cCArIHEgPSAxLCB3aGVyZQ==IHA=IHJlcHJlc2VudHMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hbnQgYWxsZWxlLCBhbmQg[Qq]q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?
[c]IDAuNg==[Qq]
[f]IE5vLiAwLjYgaXMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgcmVjZXNzaXZlIGFsbGVsZS4gSSYjODIxNzttIG5vdCBzdXJlIHdoZXJlIHlvdSB3ZW50IHdyb25nLCBzbyBJJiM4MjE3O2xsIHdhbGsgeW91IHRocm91Z2ggdGhlIHdob2xlIHRoaW5nLg==
Cg==VGhpcyBpcyBhIHBvcHVsYXRpb24gZ2VuZXRpY3MgcHJvYmxlbSwgYW5kIHRoZSBlYXNpZXN0IHdheSB0byBzb2x2ZSB0aGlzIHByb2JsZW0gaXMgdG8gdXNlIGEgY3Jvc3MgbXVsdGlwbGljYXRpb24gdGFibGUgKGEga2luZCBvZiBQdW5uZXR0IHNxdWFyZSkuIFlvdSBzdGFydCBmcm9tIHRoZSBpZGVhIHRoYXQgcCArIHEgPSAxLCB3aGVyZQ==[Qq] p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?
[c]IDAu NA==[Qq]
[f]IEZhYnVsb3VzIcKgIElmIHRoZSBmcmVxdWVuY3kgb2YgdGhlIHJlY2Vzc2l2ZSBwaGVub3R5cGUgaXMgMC4zNiwgdGhlbiAoYXNzdW1pbmcgdGhlIHBvcHVsYXRpb24gaXMgaW4gSGFyZHktV2VpbmJlcmcgZXF1aWxpYnJpdW0pLCB0aGVuIHRoZSBmcmVxdWVuY3kgb2YgdGhlIHJlY2Vzc2l2ZSBhbGxlbGUgd2lsbCBiZSAwLjYuIFNpbmNlIHAgKyBxID0gMSwgdGhlbiB5b3Ugc2ltcGx5IHN1YnRyYWN0IDEgJiM4MjExOyAwLjYgdG8gZ2V0IHRoZSBhbnN3ZXIsIDAuNC4=[Qq]
[c]IDAuNDg=[Qq]
[f]IE5vLiAwLjQ4IGlzIHRoZSBmcmVxdWVuY3kgb2YgaGV0ZXJvenlnb3RlcyBpbiB0aGlzIHBvcHVsYXRpb24uIEkmIzgyMTc7bSBub3Qgc3VyZSB3aGVyZSB5b3Ugd2VudCB3cm9uZywgc28gSSYjODIxNztsbCB3YWxrIHlvdSB0aHJvdWdoIHRoZSB3aG9sZSB0aGluZy4=
Cg==VGhpcyBpcyBhIHBvcHVsYXRpb24gZ2VuZXRpY3MgcHJvYmxlbSwgYW5kIHRoZSBlYXNpZXN0IHdheSB0byBzb2x2ZSB0aGlzIHByb2JsZW0gaXMgdG8gdXNlIGEgY3Jvc3MgbXVsdGlwbGljYXRpb24gdGFibGUgKGEga2luZCBvZiBQdW5uZXR0IHNxdWFyZSkuIFlvdSBzdGFydCBmcm9tIHRoZSBpZGVhIHRoYXQgcCArIHEgPSAxLCB3aGVyZQ==[Qq] p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele. If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q. To find the frequency of p, just remember that p = q = 1. That means that 1 – q = p. What’s 1 – 0.6?
[q multiple_choice=”true”] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype. What is the frequency of the heterozygotes within this population?
[c]IDAuMzY=[Qq]
[f]IE5vLiAwLjM2IGlzIHRoZSBmcmVxdWVuY3kgb2YgaG9tb3p5Z291cyBkb21pbmFudHMgaW4gdGhpcyBwb3B1bGF0aW9uLsKgSSYjODIxNzttIG5vdCBzdXJlIHdoZXJlIHlvdSB3ZW50IHdyb25nLCBzbyBJJiM4MjE3O2xsIHdhbGsgeW91IHRocm91Z2ggdGhlIHdob2xlIHRoaW5nLg==
Cg==VGhpcyBpcyBhIHBvcHVsYXRpb24gZ2VuZXRpY3MgcHJvYmxlbSwgYW5kIHRoZSBlYXNpZXN0IHdheSB0byBzb2x2ZSB0aGlzIHByb2JsZW0gaXMgdG8gdXNlIGEgY3Jvc3MgbXVsdGlwbGljYXRpb24gdGFibGUgKGEga2luZCBvZiBQdW5uZXR0IHNxdWFyZSkuIFlvdSBzdGFydCBmcm9tIHRoZSBpZGVhIHRoYXQgcCArIHEgPSAxLCB3aGVyZQ==[Qq] p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.
If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.
There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.
[c]IDAuNzQ=[Qq]
[f]IE5vLiBJJiM4MjE3O20gbm90IHN1cmUgd2hlcmUgeW91IHdlbnQgd3JvbmcsIHNvIEkmIzgyMTc7bGwgd2FsayB5b3UgdGhyb3VnaCB0aGUgd2hvbGUgdGhpbmcu
Cg==VGhpcyBpcyBhIHBvcHVsYXRpb24gZ2VuZXRpY3MgcHJvYmxlbSwgYW5kIHRoZSBlYXNpZXN0IHdheSB0byBzb2x2ZSB0aGlzIHByb2JsZW0gaXMgdG8gdXNlIGEgY3Jvc3MgbXVsdGlwbGljYXRpb24gdGFibGUgKGEga2luZCBvZiBQdW5uZXR0IHNxdWFyZSkuIFlvdSBzdGFydCBmcm9tIHRoZSBpZGVhIHRoYXQgcCArIHEgPSAxLCB3aGVyZQ==[Qq] p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.
If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.
There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.
[c]IDAuNg==[Qq]
[f]IE5vLiAwLjYgaXMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgcmVjZXNzaXZlIGFsbGVsZS4gVGhhdCYjODIxNztzIGEgZ29vZCBzdGFydCwgYnV0IHNpbmNlIEkmIzgyMTc7bSBub3Qgc3VyZSB3aGVyZSB5b3Ugd2VudCB3cm9uZywgSSYjODIxNztsbCB3YWxrIHlvdSB0aHJvdWdoIHRoZSB3aG9sZSB0aGluZy4=
Cg==VGhpcyBpcyBhIHBvcHVsYXRpb24gZ2VuZXRpY3MgcHJvYmxlbSwgYW5kIHRoZSBlYXNpZXN0IHdheSB0byBzb2x2ZSB0aGlzIHByb2JsZW0gaXMgdG8gdXNlIGEgY3Jvc3MgbXVsdGlwbGljYXRpb24gdGFibGUgKGEga2luZCBvZiBQdW5uZXR0IHNxdWFyZSkuIFlvdSBzdGFydCBmcm9tIHRoZSBpZGVhIHRoYXQgcCArIHEgPSAxLCB3aGVyZQ==[Qq] p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.
If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.
There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.
[c]IDAuNA==[Qq]
[f]IE5vLiAwLjQgaXMgdGhlIGZyZXF1ZW5jeSBvZiB0aGUgZG9taW5hdCBhbGxlbGUuIFRoYXQmIzgyMTc7cyBhIGdvb2Qgc3RhcnQsIGJ1dCBzaW5jZSBJJiM4MjE3O20gbm90IHN1cmUgd2hlcmUgeW91IHdlbnQgd3JvbmcsIEkmIzgyMTc7bGwgd2FsayB5b3UgdGhyb3VnaCB0aGUgd2hvbGUgdGhpbmcu
Cg==VGhpcyBpcyBhIHBvcHVsYXRpb24gZ2VuZXRpY3MgcHJvYmxlbSwgYW5kIHRoZSBlYXNpZXN0IHdheSB0byBzb2x2ZSB0aGlzIHByb2JsZW0gaXMgdG8gdXNlIGEgY3Jvc3MgbXVsdGlwbGljYXRpb24gdGFibGUgKGEga2luZCBvZiBQdW5uZXR0IHNxdWFyZSkuIFlvdSBzdGFydCBmcm9tIHRoZSBpZGVhIHRoYXQgcCArIHEgPSAxLCB3aGVyZQ==[Qq] p represents the frequency of the dominant allele, and q represents the frequency of the recessive allele.
If you plugged these values into a cross multiplication table, you’d get:
| p | q | |
| p | p2 | pq |
| q | pq | q2 |
You’re told in the problem that 0.36 (36 percent) of the population has this recessive phenotype. So plug that value into the lower right corner of the square.Note that q2 represents the frequency of individuals with the recessive phenotype.
| p | q | |
| p | p2 | pq |
| q | pq | .36 |
To find the frequency of the recessive allele (q), just take the square root of q2 (in other words, take the square root of 0.36). That gives you the value of q, which in this case is equal to 0.6. If q = 0.6, then p = 1 – 0.6, or 0.4.
There’s just one more step. The frequency of heterozygotes in a population is 2pq. So, multiply 2 times p times q and you’ll have your answer.
[c]IDAu NDg=[Qq]
[f]IEV4Y2VsbGVudCEgVGhlIHBlcmNlbnRhZ2Ugb2YgaGV0ZXJvenlnb3RlcyBpbiB0aGUgcG9wdWxhdGlvbiBpcyAwLjQ4Lg==[Qq]
[q multiple_choice=”true”] A population geneticist is studying tail feather length in a population of wrens. Within this population, 36% of the sampled individuals have a homozygous recessive phenotype. What percentage of individuals will have the dominant phenotype?
[c]IDAuMzY=[Qq]
[f]IE5vLiAwLjM2IGlzIHRoZSBmcmVxdWVuY3kgb2YgdGhlIGluZGl2aWR1YWxzIHdpdGggdGhlIA==cmVjZXNzaXZlIHBoZW5vdHlwZS4gWW91JiM4MjE3O3JlIGxvb2tpbmcgZm9yIHRoZSBwZXJjZW50YWdlIG9mIGluZGl2aWR1YWxzIHdpdGggdGhlIA==ZG9taW5hbnQ=IHBoZW5vdHlwZS4=[Qq]
[c]IDAuNzQ=[Qq]
[f]IE5vLiBZb3UmIzgyMTc7cmUgdG9sZCBpbiB0aGUgcHJvYmxlbSB0aGF0IDAuMzYgKDM2IHBlcmNlbnQpIG9mIHRoZSBwb3B1bGF0aW9uIGhhcyB0aGlzIHJlY2Vzc2l2ZSBwaGVub3R5cGUuIElmIDM2JSBvZiB0aGUgcG9wdWxhdGlvbiBoYXMgdGhlIHJlY2Vzc2l2ZSBwaGVub3R5cGUsIHdoYXQgcGVyY2VudGFnZSBoYXMgdGhlIGRvbWluYW50IHBoZW5vdHlwZT8=[Qq]
[c]IDAu NjQ=[Qq]
[f]IEV4Y2VsbGVudC4gSWYgMzYlIG9mIHRoZSBwb3B1bGF0aW9uIGhhcyB0aGUgcmVjZXNzaXZlIHBoZW5vdHlwZSwgdGhlbiBldmVyeWJvZHkgZWxzZSBpbiB0aGUgcG9wdWxhdGlvbiBoYXMgdG8gaGF2ZSB0aGUgZG9taW5hbnQgcGhlbm90eXBlLiAxICYjODIxMTsgMC4zNiA9IDAuNjQu[Qq]
[c]IDAuNDA=[Qq]
[f]IE5vLiBZb3UmIzgyMTc7cmUgdG9sZCBpbiB0aGUgcHJvYmxlbSB0aGF0IDAuMzYgKDM2IHBlcmNlbnQpIG9mIHRoZSBwb3B1bGF0aW9uIGhhcyB0aGlzIHJlY2Vzc2l2ZSBwaGVub3R5cGUuIElmIDM2JSBvZiB0aGUgcG9wdWxhdGlvbiBoYXMgdGhlIHJlY2Vzc2l2ZSBwaGVub3R5cGUsIHdoYXQgcGVyY2VudGFnZSBoYXMgdGhlIGRvbWluYW50IHBoZW5vdHlwZT8=[Qq]
[c]IDAuNDg=[Qq]
[f]IE5vLiAwLjQ4IGlzIHRoZSBmcmVxdWVuY3kgb2YgaGV0ZXJvenlnb3Rlcy4gWW91IGxvb2tpbmcgZm9yIHRoZSBudW1iZXIgb2YgaW5kaXZpZHVhbHMgd2l0aCB0aGUgZG9taW5hbnQgcGhlbm90eXBlLiBZb3UmIzgyMTc7cmUgdG9sZCBpbiB0aGUgcHJvYmxlbSB0aGF0IDAuMzYgKDM2IHBlcmNlbnQpIG9mIHRoZSBwb3B1bGF0aW9uIGhhcyB0aGlzIHJlY2Vzc2l2ZSBwaGVub3R5cGUuIElmIDM2JSBvZiB0aGUgcG9wdWxhdGlvbiBoYXMgdGhlIHJlY2Vzc2l2ZSBwaGVub3R5cGUsIHdoYXQgcGVyY2VudGFnZSBoYXMgdGhlIGRvbWluYW50IHBoZW5vdHlwZT8=[Qq]
[/qwiz]