By James N. Thompson Jr, Jenna J. Hellack, Gerald Braver
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Extra info for Primer of Genetic Analysis: A Problems Approach
Example text
Perhaps the most frequent problem that students have in Mendelian genetics is understanding the implications of segregation. As can be seen from meiosis, each parent transmits only one allele of each gene to an offspring. If the parent is homozygous, each gamete will be identical for that gene, but if the parent is heterozygous, two different gamete types will be produced, each of equal frequency. For a dihybrid (for example, AaBb), a single allele of each gene must be included in each gamete. If you use a Punnett square to summarize the outcomes of a cross, be sure to double-check that the gametes you list are the correct products of segregation.
From the July 31, 2007 35 20:12 P1: KAE 9780521842686c04 36 CUFX162/Thompson 0 521 84268 9 July 31, 2007 BASIC MENDELIAN GENETICS Punnett square, we can see that only 1/4 will have vision (Nnss); therefore, the answer is that 3/4 will be blind. 12. Since all genes are segregating independently, we can consider them one at a time. From the Ss × SS part of the cross, all offspring will inherit an S allele from the second parent and, thus, have straight hair. From Gg × Gg there will be a 3:1 ratio of golden brown fur to dark brown fur.
There is, therefore, a probability that 1/4 of their children will be homozygous (phenotypically albino). In other words, there is a 1/4 probability that any one of their children will be albino. 8. (b) The parents are Cc ch and c ch c ch . Since a heterozygous chinchilla mouse carries the dominant C allele, in addition to c ch , the progeny consist of half with normal pigmentation (Cc ch ) and half with chinchilla color. The cross of the normal F1 mice to chinchilla mice is therefore Cc ch to c ch c ch and the expected genotypic ratio is half Cc ch and half c ch c ch or 1:1.